Is the homogenization of an irreducible polynomial irreducible?

This statement is something I have thought to be true for a long time. However, lately I realized that I did not have any reference for it, so I was a bit unsure. I could find an MSE question, whose answer matched what I thought, and a couple of friends came up with the same arguments, which made me believe the statement. I also think that this fact is available in a book by Cox, Little, and O'Shea (Exercise 7 on p.392), although many places in this book assumes that the ambient field $k$ is algebraically closed. Thus, I wanted to record this here for later purpose.

Why do I care? I care about this statement mostly because I want to think of the notion of hypersurfaces of $\mathbb{P}^{n}$ over a field $k$ somewhat intrinsically. That is, if I have an integral closed subscheme $H \hookrightarrow \mathbb{P}^{n}$ with codimension $1$ and degree $d,$ then I want to say there is a homogenous polynomial $f \in k[x_{0}, \dots, x_{n}]$ of degree $d$ such that $$H \simeq \mathrm{Proj}(k[x_{0}, \dots, x_{n}]/(f))$$ as $k$-schemes. Take any standard affine open subset $$U_{i} := D_{+}(x_{i}) \simeq \mathrm{Spec}(k[x_{0}/x_{i}, \dots, x_{n}/x_{i}])$$ such that $H \cap U_{i} \neq \emptyset.$ Since $H \cap U_{i}$ is of codimension $1$ in $U_{i},$ there must be a (monic) nonzero $f_{i} \in k[x_{0}/x_{i}, \dots, x_{n}/x_{i}]$ such that $$H \cap U_{i} = \mathrm{Spec}(k[x_{0}/x_{i}, \dots, x_{n}/x_{i}]/(f_{i})).$$ Since $H$ is the closure of $H \cap U_{i}$ in $\mathbb{P}^{n},$ it follows (e.g., from a previous posting) that topologically, we have $$V_{+}(f) = H,$$ where $f(x_{0}, \dots, x_{n}) = x_{i}^{\deg(f_{i})}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}),$ where $\deg(f_{i})$ denotes the maximum among the degrees of the monomials of $f_{i}$ in $x_{0}/x_{i}, \dots, x_{n}/x_{i},$ except $x_{i}/x_{i} = 1.$ However, to show that $$H \simeq \mathrm{Proj}(k[x_{0}, \dots, x_{n}]/(f)),$$ we also need to check that $$x_{i}^{\deg(f_{i})}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = x_{j}^{\deg(f_{j})}f_{j}(x_{0}/x_{j}, \dots, x_{n}/x_{j})$$ for all $i, j$ such that $U_{i}$ and $U_{j}$ both intersect $H,$ which holds when $f$ is irreducible.

Geometric argument. If $f$ is not irreducible, then $f = gh$ for some homogeneous polynomials $g, h$ with positive degrees. (To see this, decompose $g$ and $h$ into homogeneous pieces and argue by degree.) Thus, we would have $$V_{+}(f) = V_{+}(g) \cup V_{+}(h).$$ Since $V(f_{i})$ is irreducible, we know $V_{+}(f) = \overline{V(f_{i})}$ is an irreducible topological space, at least one of $V_{+}(g)$ and $V_{+}(h)$ is empty. This implies that one of $\sqrt{(g)}$ and $\sqrt{(h)}$ is equal to $(x_{0}, \dots, x_{n}).$ Without loss of generality, say $\sqrt{(g)} = (x_{0}, \dots, x_{n}),$ which is a maximal ideal. Then we have $\{(x_{0}, \dots, x_{n})\} = V(x_{0}, \dots, x_{n}) = V(g)$ in $\mathbb{A}^{n+1},$ although $V(g)$ has dimension $\geq n + 1 - 1 = n \geq 1,$ by Krull's principal ideal theorem. This is a contradiction as the singleton of a closed point has dimension $0.$

Combinatorial argument. If $f$ is not irreducible, then $f = gh$ for some homogeneous polynomials $g, h$ with positive degrees. We may assume that $g, h$ are monic since $k$ is a field. Then $$\begin{align*}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) &= x_{i}^{-\deg(f_{i})}f(x_{0}, \dots, x_{n}) \\ &= g(x_{0}/x_{i}, \dots, x_{n}/x_{i}) h(x_{0}/x_{i}, \dots, x_{n}/x_{i}).\end{align*}$$ Since $f_{i}$ is irreducible, at least one of the polynomials on the right-hand side must be constant in $k[x_{0}/x_{i}, \dots, x_{n}/x_{i}],$ which necessarily mean that it is $1$ (due to the monic assumption). Without loss of generality, say $g(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = 1.$ Then we have $g(x_{0}, \dots, x_{n}) = x_{i}^{\deg(g)},$ so $$f(x_{0}, \dots, x_{n}) = x_{i}^{\deg(g)}h(x_{0}, \dots, x_{n}).$$ Now, note from the definition of $f$ that it cannot vanish at the homogenizing variable $x_{i}.$ Thus, we must have $\deg(g) = 0,$ so $f$ must be irreducible.

Projective closure of affine hyper surface

I have been using the fact that if I have a single equation given in the affine space $\mathbb{A}^{n},$ then taking the closure in $\mathbb{P}^{n}$ can be given by homogenizing the equation. However, I have also realized that I never thought about why! We will only discuss for the case $n = 2,$ but the proof should work for any $n \geq 1.$ The references for this posting are Vakil's book and Kidwell's answer to this MSE question.

Computing projective closure. Let $g(x, y) \in k[x, y]$ be any polynomial and $d$ be the maximum total degree of the monomials of $g(x, y).$ Then $$\overline{V_{\mathbb{A}^{2}}(g(x, y))} = V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)),$$ where the closure is taken in $\mathbb{P}^{2}.$

Proof. Since $$V_{\mathbb{A}^{2}}(g(x, y)) = V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \cap D_{\mathbb{P}^{2}}(z) \subset V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)),$$ we have $$\overline{V_{\mathbb{A}^{2}}(g(x, y))} \subset V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)).$$ To show the reverse inclusion, first, note that there is a homogeneous ideal $J \subset k[x, y, z]$ such that $$V_{\mathbb{P}^{2}}(J) = \overline{V_{\mathbb{A}^{2}}(g(x, y))}.$$ We want to show that $V_{\mathbb{P}^{2}}(J) \supset V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)),$ and to do so, it is enough to show that $J \subset \sqrt{(z^{d}g(x/z, y/z))}$ in $k[x, y, z].$ Since $$V_{\mathbb{A}^{2}}(g(x, y)) \subset V_{\mathbb{P}^{2}}(J),$$ we see that $J$ must be contained in the homogeneous ideal $I(V_{\mathbb{A}^{2}}(g(x, y)))$ generated by all the homogeneous elements of $k[x, y, z]$ contained in homogeneous prime ideals in $$V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \cap D_{\mathbb{P}^{2}}(z) = V_{\mathbb{A}^{2}}(g(x, y)).$$ Thus, it is enough to show that $$I(V_{\mathbb{A}^{2}}(g(x, y))) \subset (z^{d}g(x/z, y/z))$$ in $k[x, y, z].$ Fix any homogenous $h(x, y, z) \in I(V_{\mathbb{A}^{2}}(g(x, y))).$ By definition, we have $$V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \cap D_{\mathbb{P}^{2}}(z) = V_{\mathbb{A}^{2}}(g(x, y)) \subset V_{\mathbb{P}^{2}}(h(x,y,z)),$$ so $$V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \subset V_{\mathbb{P}^{2}}(zh(x,y,z)).$$ This implies that (e.g., 4.5.H. (a) of Vakil) $$z^{N}h(x, y, z)^{N} \in (z^{d}g(x/z, y/z))$$ for some $N \gg 0.$ Due to the definition of $d,$ we see that $z^{d}g(x/z, y/z)$ is not divisible by $z$ in $k[x, y, z].$ Thus, if we denote by $$z^{d}g(x/z, y/z) = p_{1}(x, y, z)^{e_{1}} \cdots p_{r}(x, y, z)^{e_{r}}$$ the irreducible factorization in $k[x, y, z],$ then each $p_{i}(x, y, z)^{e_{i}}$ divides $h(x, y, z)^{N}.$ Hence, we must have $$h(x, y, z)^{N} \in (z^{d}g(x/z, y/z)).$$ This implies that $$h(x, y, z) \in \sqrt{(z^{d}g(x/z, y/z))},$$ as desired. $\Box$

What's going on? At first, I was wondering why Vakil's book is missing this important piece of information, but I realized that it was quite clear that this follows from 4.5.H. This is very general, say we have a graded ring $$S := S_{0} \oplus S_{1} \oplus S_{2} \oplus \cdots.$$ As usual, we write $S_{+} := S_{1} \oplus S_{2} \oplus \cdots$ the ideal generated by homogenous elements. Recall that we can consider $\mathrm{Proj}(S)$ as the set of homogeneous prime ideals that do not contain $S_{+}.$ We use the notation $$V_{+}(I) := \{\mathfrak{p} \in \mathrm{Proj}(S) : \mathfrak{p} \supset I\}$$ for a homogeneous ideal $I \subset S$ and $$D_{+}(f) := \mathrm{Proj}(S) \smallsetminus V_{+}(f).$$ Given any subset $T \subset \mathrm{Proj}(S),$ we denote by $I_{+}(T)$ to be the ideal generated by homogenous elements that are contained in any (homogeneous) prime ideals in $T.$ In Vakil, 4.5.H says:

(a) $V_{+}(I) \subset V_{+}(f)$ if and only if $f \in \sqrt{I}.$

(b) $V_{+}(I_{+}(T)) = \overline{T}$ in $\mathrm{Proj}(S)$ for any subset $T \subset \mathrm{Proj}(S).$

The statement (b) is basically proven in the proof above. To see (a), note that the only nontrivial direction is to assume $V_{+}(I) \subset V_{+}(f)$ and show $f \in \sqrt{I}.$ This easily follows once we show the following.

Lemma. Suppose that $V_{+}(I) \neq \emptyset.$ Then $\sqrt{I}$ is the intersection (in $S$) of all $\mathfrak{p} \in \mathrm{Proj}(S)$ that contains $I.$

Proof. We start with the fact that $\sqrt{I}$ is the intersection of all prime ideals of $S$ containing $I.$ Since we are assuming that $V_{+}(I)$ is nonempty, we do not need to consider the possibility of $S_{+}$ being the only prime ideal for this intersection. Hence, it remains to show that we can keep this intersection only using homogenous prime ideals of $S$ containing $I.$ Given any (not necessarily homogenous) ideal $\mathfrak{p}$ of $S,$ we can construct $$\mathfrak{p}^{h} := \bigoplus_{d \geq 0}(\mathfrak{p} \cap S_{d}),$$ which is a homogeneous ideal. What's great about this ideal is that it is prime if $\mathfrak{p}$ is prime, and because $I$ is homogenous, we have $I \subset \mathfrak{p}$ if and only if $I \subset \mathfrak{p}^{h}.$ This lets us get rid of all the primes $\mathfrak{p}$ in the intersection that are not homogeneous, which finishes the proof. $\Box$

Remark. Let $J$ be any ideal of a $\mathbb{Z}_{\geq 0}$-graded ring $S.$ We always have $J^{h} \subset J.$ and $J = J^{h}$ if and only if $J$ is homogeneous. We do not have this equality when $J$ is not homogenous. For instance, take $S = k[x, y]$ with the usual grading and $J = (x + y^{2}).$ Then $J^{h} = (0).$

More information. Having $V_{+}(g) \subset V_{+}(f)$ is equivalent to $D_{+}(f) \subset D_{+}(g),$ so we must have $$\mathrm{Spec}(S_{f})_{0} \hookrightarrow \mathrm{Spec}(S_{g})_{0},$$ which is given by $(S_{g})_{0} \rightarrow (S_{f})_{0}$ coming from the further localization may $S_{g} \rightarrow S_{f}.$ This seems to suggest that $D(f) \subset D(g),$ which is true, but this actually uses the argument in Lemma given above, so we could not finish the proof this way.

Remark. As an application, if $H \subset \mathbb{P}^{n}$ (over $k$) is a reduced closed subscheme of codimension $1,$ then we may find a homogeneous polynomial $F \in k[x_{0}, \dots, x_{n}]$ such that $$H \simeq \mathrm{Proj}\left(\frac{k[x_{0}, \dots, x_{n}]}{\sqrt{(F)}}\right).$$ I am not so sure if one can say more.

A "very small" fiber gives the trivial residue field extension

Fix any scheme map $\pi : X \rightarrow Y$ that is affine. In this posting, I want to record the fact that given $y \in Y,$ if $$\dim_{\kappa(y)}\Gamma(\pi^{-1}(y), \mathscr{O}_{\pi^{-1}(y)}) = 1,$$ then we have $\pi^{-1}(y) = \{x\}$ with $\kappa(x) \simeq \kappa(y).$

Since the map is affine, we may assume that $X = \mathrm{Spec}(A)$ and $Y = \mathrm{Spec}(B)$ for some (commutative unital) rings $A$ and $B.$ We denote by $\phi : B \rightarrow A$ the ring map associated to $\pi.$ Write $\mathfrak{q} := y$ for the corresponding prime ideal of $B.$ 

Without the dimension $1$ hypothesis. We have $$\kappa(y) = B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$$ and $$\Gamma(\pi^{-1}(y), \mathscr{O}_{\pi^{-1}(y)}) = \frac{B_{\mathfrak{q}}}{\mathfrak{q}B_{\mathfrak{q}}} \otimes_{B} A \simeq \frac{B_{\mathfrak{q}}}{\mathfrak{q}B_{\mathfrak{q}}} \otimes_{B_{\mathfrak{q}}} A_{\mathfrak{q}} \simeq \frac{A_{\mathfrak{q}}}{\mathfrak{q}A_{\mathfrak{q}}}.$$ The first isomorphism is quite general, and the map looks like $$b \otimes a \mapsto b \otimes a.$$ For the second one, we consider the following general situation: given an ideal $I$ of a ring $R$ let $M$ be an $R$-module. Consider the exact sequence $$0 \rightarrow IM \rightarrow M \rightarrow M/IM\ \rightarrow 0.$$ We get the exact sequence $$0 = (R/I) \otimes_{R} IM \rightarrow (R/I) \otimes_{R} M \rightarrow (R/I) \otimes_{R} (M/IM)\ \rightarrow 0,$$ which gives $$\begin{align*}(R/I) \otimes_{R} M &\simeq (R/I) \otimes_{R} (M/IM) \\ &\simeq (R/I) \otimes_{R/I} (M/IM) \\ &\simeq M/IM \end{align*}$$ given by $\bar{r} \otimes m \mapsto \overline{rm}.$

Applying this for 

• $R = B_{\mathfrak{q}},$ 
• $I = \mathfrak{q}B_{\mathfrak{q}},$
• $M = A_{\mathfrak{q}} = (A \setminus \phi(\mathfrak{q}))^{-1}A,$

we have $$\frac{B_{\mathfrak{q}}}{\mathfrak{q}B_{\mathfrak{q}}} \otimes_{B_{\mathfrak{q}}} A_{\mathfrak{q}} \simeq \frac{A_{\mathfrak{q}}}{\mathfrak{q}A_{\mathfrak{q}}}$$ given by $$[b/t] \otimes (a/s) \mapsto [\phi(b)a/(\phi(t)s)].$$ Of course, this map is only an $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-linear map. Since the inverse map is a ring map $[a/s] \mapsto 1 \otimes (a/s),$ so is the original map. That is, this isomorphism is an isomorphism of $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-algebra.

With the dimension $1$ hypothesis. The hypothesis that $$\dim_{\kappa(y)}\Gamma(\pi^{-1}(y), \mathscr{O}_{\pi^{-1}(y)}) = \dim_{B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}}(A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}) = 1$$ implies that the map $$B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \rightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}$$ induced by $\phi$ is an isomorphism of rings. Why? First, note that since this is a ring map and $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$ is a field this map must be injective unless $A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} = 0.$ However, it is not, as it is a positive dimensional vector space, so the map must be an inejection. Next, note that this map is $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-linear, and since the target is $1$-dimensional over $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}},$ this injection must be a surjection. This also shows that $$\pi^{-1}(y) = \pi^{-1}(\mathfrak{q}) = \mathrm{Spec}(A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}})$$ has a unique point, which corresponds to a prime ideal $\mathfrak{p}$ of $A.$ What we need to show is that the field extension $$B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \rightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$ induced by $\phi$ is an isomorphism.

Proof. We only need to show the surjectivity. First, note that we have $$A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} \simeq B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \hookrightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$$ and this map literally looks like $a/s \mapsto a/s$ because it is given by $\phi(b)/\phi(t) \mapsto b/t \mapsto \phi(b)/\phi(t).$

Since $\phi^{-1}(\mathfrak{p}) = \mathfrak{q},$ we have $\phi(\mathfrak{q}) \subset \mathfrak{p},$ which implies that $$A \smallsetminus \mathfrak{p} \subset A \smallsetminus \phi(\mathfrak{q}).$$ This induces the ring map $$A_{\mathfrak{p}} \rightarrow A_{\mathfrak{q}}$$ that looks like $a/s \mapsto a/s.$ Denote by $J$ the kernel of the composition $$A_{\mathfrak{p}} \rightarrow A_{\mathfrak{q}} \twoheadrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}},$$ which induces the injection $A_{\mathfrak{p}}/J \hookrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}.$ We can only know that $J$ is a prime ideal, so we have $J \subset \mathfrak{p}A_{\mathfrak{p}}.$ We claim that this inclusion is actually an equality, which will finish the proof. To see this, note that we have $$A_{\mathfrak{p}}/J \hookrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} \hookrightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$$ and this map is given by $(a \mod J) \mapsto (a \mod \mathfrak{p}A_{\mathfrak{p}}).$ Because this is a ring injection, for any $a \in A_{\mathfrak{p}}$ if $a \in \mathfrak{p}A_{\mathfrak{p}},$ then the image of this map is $0,$ so $a$ must be $0$ modulo $J$ as well. This implies that $a \in J,$ so $\mathfrak{p}A_{\mathfrak{p}} \subset J,$ showing that $\mathfrak{p}A_{\mathfrak{p}} = J.$ $\Box$

Functor of points

Given a field $k$ and a polynomial $$f(x_{1}, \dots, x_{n}) \in k[x_{1}, \dots, x_{n}],$$ a $k$-point of the closed subscheme $$\mathrm{Spec}(k[x_{1}, \dots, x_{n}]/(f)) \simeq V(f) \subset \mathbb{A}^{n}_{k} = \mathrm{Spec}(k[x_{1}, \dots, x_{n}])$$ is a solution to the polynomial $f,$ which is $(a_{1}, \dots, a_{n}) \in k^{n}$ such that $$f(a_{1}, \dots, a_{n}) = 0.$$ A slogan I heard from my first algebraic geometry class is

"algebraic geometry is the study of solutions to polynomial equations."

That is, we are supposed to study $k$-points of polynomial equations.

References. This posting is a regurgitation of some parts of the following books:

• Eisenbud-Harris,  
• Olsson,  
• Poonen, and 
• Vakil.

Algebraic ways to think about $k$-points. First, note that a $k$-point $(a_{1}, \dots, a_{n}) \in k^{n}$ of $V(f)$ can be thought as the maximal ideal $$\mathfrak{m} = (x_{1} - a_{1}, \dots, x_{n} - a_{n})/(f)$$ of the ring $R = k[x_{1}, \dots, x_{n}]/(f).$ We have $$R/\mathfrak{m} \simeq \frac{k[x_{1}, \dots, x_{n}]}{(x_{1}-a_{1}, \dots, x_{n}-a_{n})} \simeq k,$$ where the second map is given by the evaluation at $(a_{1}, \dots, a_{n}) \in k^{n}.$ This illustrates that the $k$-points of $R$ are precisely maximal ideals of $R$ (i.e., closed points of $\mathrm{Spec}(R)$) whose residue field is $k.$ Such maximal ideals are precisely kernels of $k$-algebra maps $R \rightarrow k.$ They correspond to $k$-scheme maps $$\mathrm{Spec}(k) \rightarrow \mathrm{Spec}(R).$$ In general, a $k$-point of a $k$-scheme $X$ is a $k$-scheme map $\mathrm{Spec}(k) \rightarrow X,$ and one may check that $k$-points of $X$ correspond to the topological/set-theoretic points of $X$ whose residue fields are $k.$

Remark. Given a field $k,$ we know $\mathrm{Spec}(k)$ has one $k$-point. This is because the only $k$-algebra map $k \rightarrow k$ is the identity. If we just think of ring maps, there are many more in general. For instance, note that complex conjugation $\mathbb{C} \rightarrow \mathbb{C}$ gives a ring map that is not the identity.

Generalization. Fix a (commutative and unital) ring $A.$ Given any $A$-algebra $R$ and an $A$-scheme $X,$ an $R$-point of $X$ over $A$ is an $A$-scheme map $$\mathrm{Spec}(R) \rightarrow X.$$ Given any $A$-scheme $T,$ an $T$-point of $X$ over $A$ is an $A$-scheme map $$T \rightarrow X.$$ Note that an $R$-point on $X$ over $A$ is precisely $\mathrm{Spec}(R)$-point over $A.$ Note that we may generalize this notion further by replacing $A$ (or $\mathrm{Spec}(A)$) with another scheme.

Remark. Note that when we were discussing $k$-point earlier, we really meant $k$-point over $k.$ However, it seems that people do not really mention this out loud in general. That is, given an $S$-schemes $X$ and $T,$ we call an $S$-scheme map $T \rightarrow X$ a $T$-point, even though in principle, we should specify the scheme $S$ that we are over.

Functor of points. Given $S$-schemes $X$ and $T,$ we write $$X(T) := \mathrm{Hom}_{\textbf{Sch}_{S}}(T, X)$$ to mean the set of $T$-points. The functor $$h_{X} : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$$ given by $T \mapsto X(T)$ is called the functor of points on $X$ over $S$. By Yoneda lemma, we get an embedding $$\textbf{Sch}_{S}^{\mathrm{op}} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$$ given by $X \mapsto h_{X}.$

Restriction to affine schemes is enough. We can also have $$\textbf{Sch}_{S} \hookrightarrow \mathrm{Fun}(\textbf{Aff}_{S}^{\mathrm{op}}, \textbf{Set}),$$ where $\textbf{Aff}_{S}$ denotes the category of affine schemes over $S$. The functor is given by $$X \mapsto h_{X}|_{\textbf{Aff}_{S}^{\mathrm{op}}}$$ on the level of objects.

How come? Given any $S$-scheme $X,$ consider its functor of points $h_{X}.$ Given an $S$-scheme $T$, we recall that $T$-points on $X$ are precisely $S$-maps $T \rightarrow X.$ For any affine open subset $U \subset T,$ we have an $S$-map given by $$U \hookrightarrow T \rightarrow X,$$ a $U$-point of $X.$ If we denote by $\{U_{i}\}_{i \in I}$ the set of all affine open subsets of $T,$ then an $S$-map $T \rightarrow X$ is given by a set $\{U_{i} \rightarrow X\}_{i \in I}$ of $S$-maps that are compatible with inclusions among various $U_{i}.$ A clean way of saying this is to note that $X(T)$ is the limit (in $\textbf{Set}$) of the diagram $\{X(U_{i})\}_{i \in I},$ whose maps are induced by inclusions among $U_{i}.$ (This seems quite easy because we are dealing with the category of sets.) Thus, given any two $S$-schemes $X$ and $Y,$ if $$h_{X}|_{\textbf{Aff}_{S}^{\mathrm{op}}} \simeq h_{Y}|_{\textbf{Aff}_{S}^{\mathrm{op}}},$$ then $X \simeq Y,$ as $S$-schemes.

Moral. We may consider schemes as a very specific type of functors! Moreover, given any ring $R,$ since the opposite category $\textbf{Aff}_{R}^{\mathrm{op}}$ of the category affine schemes over $R$ is anti-equivalent to the category $\textbf{Alg}_{R}$ of $R$-algebras, so we actually have $$\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Alg}_{R}, \textbf{Set}).$$ In particular, taking $R = \mathbb{Z},$ we have $$\textbf{Sch} \hookrightarrow \mathrm{Fun}(\textbf{Ring}, \textbf{Set}),$$ where $\textbf{Ring}$ is the category of (commutative unital) rings.

This is crazy! Isn't it? Schemes are just some functors from the category of rings to the category of sets, which sounds like scheme theory can be thought as a subfield of commutative algebra. Indeed, we often experience that to prove certain statement about schemes, we deduce something about affine opens and then glue together. Note that inducing the above embedding uses such an argument as well.

Remark. I think saying "scheme theory can be thought as a subfield of commutative algebra" is as philosophical as saying that "group theory can be thought of a subfield of topology" (just because every group is the fundamental group of some topological space).

As a terminology, the functors that are in the image of any of the following (Yoneda) embeddings are called representable by schemes (over the given base):

• $\textbf{Sch}_{S} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set});$
• $\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Aff}_{S}^{\mathrm{op}}, \textbf{Set});$
• $\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Alg}_{R}, \textbf{Set}).$

Geometry of functors. We are now going to think about the objects in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ as geometric spaces, which include schemes. This sounds quite ridiculous, but it seems that many people have thought about this for many years.

We will build the geometry of functors from the representable ones (i.e., $S$-schemes), so in some sense our notion of geometry will be still somewhat "concrete". For any representable functor $$h_{X} : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ with respect to an $S$-scheme $X,$ we called $$X(T) = h_{X}(T) = \mathrm{Hom}_{\textbf{Sch}_{S}}(T, X)$$ the set of $T$-points on $X.$ We have two takeaways here.

1. For any (not necessarily representable) functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$ the set $F(T)$ still makes sense. 
2. Each $T \rightarrow X$ can be thought as $h_{T} \rightarrow h_{X},$ using the Yoneda correspondence $\mathrm{Hom}_{\textbf{Sch}_{S}}(T, X) = h_{X}(T) \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{T}, h_{X}).$ This correspondence works even when $h_{X}$ is replaced by any other functor $F,$ so we have $F(T) \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{T}, F),$ as sets.

Hence, we may call the set $F(T)$ the set of $T$-points on $F$ (over $S$), while thinking of each element of $F(T)$ as a morphism $h_{T} \rightarrow F$, which we call a $T$-point of $F$ (over $S$).

Reading a bit more from Yoneda's lemma. Yoneda lemma is very simple, but it packs even more than the bijection $$F(T) \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{T}, F).$$ That is, this bijection is functorial both in $F$ and $T.$ The functoriality in $T$ says we may see the isomorphism as the isomorphism $$F \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{-}, F)$$ of functors $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ With $T$ fixed, the functoriality in $F$ does not seem to say much: it means that given any map $F \rightarrow G$ of functors, we have $$\mathrm{Hom}_{\textbf{Set}}(F(T), G(T)) \simeq \mathrm{Hom}_{\textbf{Set}}(\mathrm{Hom}(h_{T}, F), \mathrm{Hom}(h_{T}, G)),$$ where the hom-sets without subscripts were taken in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set}).$

Fiber product of functors. Let $\alpha : F \rightarrow H$ and $\beta : G \rightarrow H$ be morphisms in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ of functors. Then the fiber product $F \times_{H} G$ of $F$ and $G$ over $H$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ can be constructed as follows. For each $S$-scheme $T,$ we construct $$(F \times_{H} G)(T) := F(T) \times_{H(T)} G(T),$$ which consists of $(x, y) \in F(T) \times G(T)$ such that $\alpha_{T}(x) = \beta_{T}(y) \in H(T).$ Given any $S$-scheme morphism $\phi : T' \rightarrow T,$ we have $$F(T) \times_{H(T)} G(T) \rightarrow F(T') \times_{H(T')} G(T')$$ given by $(x, y) \mapsto (\phi^{F}(x), \phi^{G}(y)).$ When $\phi$ is the identity map of an object, so is this map. This map also respects the composition with another $S$-scheme morphism, so $$F \times_{H} G : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ is indeed a functor. One may easily check that the set projections $$F(T) \leftarrow F(T) \times_{H(T)} G(T) \rightarrow G(T)$$ for $S$-schemes $T$ give rise to morphisms  $$F \leftarrow F \times_{H} G \rightarrow G$$ of functors such that the following two compositions are identical: $$F \times_{H} G \rightarrow F \overset{\alpha}{\longrightarrow} H$$ and $$F \times_{H} G \rightarrow G \overset{\beta}{\longrightarrow} H.$$ It requires a small check to notice that $F \times_{H} G$ is indeed the fiber product of $F$ and $G$ over $H$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set}).$

Open subfunctors. We are still thinking about geometry of functors. How do we think about "open subsets" of a functor $$F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ even when $F$ is not a $S$-scheme (i.e., not representable)?

Given any open subset $U \hookrightarrow X,$ one can easily check that any $S$-scheme map $f : Y \rightarrow X$ yields $$h_{f^{-1}(U)} \simeq h_{U} \times_{h_{X}} h_{Y}$$ in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ with respect to $$h_{U} \leftarrow h_{f^{-1}(U)} \rightarrow h_{Y},$$ induced by the restriction $f^{-1}(U) \rightarrow U$ of $f$ and the open embedding $f^{-1}(U) \hookrightarrow Y.$ In particular $$f^{-1}(U) \simeq U \times_{X} Y$$ in the category $\textbf{Sch}_{S}.$ We mimic this situation for the case when $h_{X}$ is replaced by more general functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$

A morphism $G \rightarrow F$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ is called an open subfunctor if for every $S$-scheme $Y$ and a morphism $h_{Y} \rightarrow F$ of functors, the induced map $$G \times_{F} h_{Y} \rightarrow h_{Y}$$ is given by an open embedding $V \hookrightarrow Y.$ This means that $$h_{V} \simeq G \times_{F} h_{Y} \rightarrow h_{Y},$$ with another structure map $h_{V} \rightarrow G$ to define the fiber product.

Remark. Note that an $S$-scheme map $U \rightarrow X$ is an open embedding if and only if $h_{U} \rightarrow h_{X}$ is an open subfunctor. Since two Cartesian squares make a Cartesian rectangle (1.3.G in Vakil), composition of two open subfunctors is an open subfunctor. Eisenbud-Harris (Definition VI-5) gives a weaker definition of an open subfunctor where $Y$ is only allowed to be affine schemes. This is not the same as the definition we gave (from Section 9.1 of Vakil), but the upshot is that they only used affine schemes instead of using all schemes. (I thank Sridhar Venkatesh for explaining this to me.)

Open cover. Given an $S$-scheme, an open cover $\{U_{i}\}_{i \in I}$ of $X$ can be realized as the family of corresponding open subfunctors $\{h_{U_{i}} \hookrightarrow h_{X}\}_{i \in I}.$ Note that for any $S$-scheme map $f : Y \rightarrow X,$ we have an open cover $\{f^{-1}(U_{i}) \hookrightarrow Y\}_{i \in I}$ of $Y.$

More generally, an open cover of a functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ is a family $\{G_{i} \hookrightarrow F\}_{i \in I}$ of open subfunctors such that for every $S$-scheme $Y$ and $h_{Y} \rightarrow F,$ the fiber products $$G_{i} \times_{F} h_{Y} \rightarrow h_{Y}$$ gives rise to an open covering of $Y.$

Sheaves as functors. Given any topological space $X,$ denote by $\mathscr{C}_{X}$ the category of open subsets of $X,$ whose morphisms are given by inclusions. Recall that a sheaf of sets is a functor $F : \mathscr{C}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ such that for any open subset $U$ of $X$ (i.e., an object of $\mathscr{C}_{X}$) and an open cover $U = \bigcup_{i \in I}U_{i}$, we have the following equalizer sequence $$F(U) \rightarrow \prod_{i \in I}F(U_{i}) \rightrightarrows \prod_{i, j \in I}F(U_{i} \cap U_{j}),$$ where the first map is given by $s \mapsto (s|_{U_{i}})_{i \in I}$ and the second (double) map is given as follows: for each ordered pair $(i', j') \in I^{2},$ we have two maps $$\prod_{i \in I}F(U_{i}) \rightrightarrows F(U_{i'} \cap U_{j'}),$$ the first of which is $$(s_{i})_{i \in I} \mapsto (s_{i'}|_{U_{i'} \cap U_{j'}}),$$ and the second of which is $$(s_{i})_{i \in I} \mapsto (s_{j'}|_{U_{i'} \cap U_{j'}}),$$ where we note that that these two maps can be equal. Since there are two maps per $F(U_{i} \cap U_{j}),$ we get the desired double map.

Functorial construction of the equalizer sequence. For our purpose it is important that the maps in the equalizer sequence (i.e., the sequence we check to be an equalizer for the definition of sheaves) can be functorially induced. For the first map, the open embeddings $U_{i} \hookrightarrow U$ gives $F(U) \rightarrow F(U_{i}),$ which gives the first map by the property of the product in $\textbf{Set}$. For the second map, we the open embeddings $U_{i'} \cap U_{j'} \hookrightarrow U_{i'}$ and $U_{i'} \cap U_{j'} \hookrightarrow U_{j'}$ to get $F(U_{i'}) \rightarrow F(U_{i'} \cap U_{j'})$ and $F(U_{j'}) \rightarrow F(U_{i'} \cap U_{j'}),$ and using these two maps we have gotten the following two maps: $$\prod_{i \in I}F(U_{i}) \rightarrow F(U_{i'}) \rightarrow F(U_{i'} \cap U_{j'})$$ and $$\prod_{i \in I}F(U_{i}) \rightarrow F(U_{j'}) \rightarrow F(U_{i'} \cap U_{j'}).$$ We can thus use the equalizer sequence that defines a functor as a sheaf works in more generality, even when we replace $\mathscr{C}_{X}$ with another categoty (with some extra structure).

If we assumed that $X$ is a scheme, then we could consider a sheaf on $X$ valued in the category $\textbf{Ring}$ of rings or the category $\textbf{Ab}$ of abelian groups (or that of $R$-modules with a fixed ring $R$, etc.) However, all such categories we take are in the category $\textbf{Set}$ of sets, so we take the equalizer sequence there to give a definition of a sheaf. The bottom line is that a sheaf is a functor $F : \mathscr{C}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}.$

Remark. A bold idea to replace $\mathscr{C}_{X}$ with $\textbf{Sch}_{X}.$ Objects of either category have the notion of open covers, so we call them "sites" (which we formally define below). More formally, we call $\mathscr{C}_{X}$ the small Zariski site on $X,$ while $\textbf{Sch}_{X}$ the big Zariski site on $X.$ The word "small" and "big" are due to the fact that the objects of $\mathscr{C}_{X}$ are open subsets $U \hookrightarrow X,$ while the objects of $\textbf{Sch}_{X}$ are all morphisms $U \rightarrow X,$ not just open embeddings. The word "Zariski" is due to the fact that the notion of "open cover" is given by the Zariski topology on $X.$

A functor $F : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ is called a Zariski sheaf if for any $X$-scheme map $U \rightarrow X$ an open cover $U = \bigcup_{i \in I}U_{i},$ the sequence $$F(U) \rightarrow \prod_{i \in I}F(U_{i}) \rightrightarrows \prod_{i, j \in I}F(U_{i} \cap U_{j})$$ induced by the open inclusions is an equalizer in $\textbf{Set}$.

Remark. Note that any representable functor (i.e., the objects in the image of $\textbf{Sch}_{X} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set})$) is a Zariski sheaf. This is just a fancier way of saying a morphism $U \rightarrow X$ of schemes can be glued from morphisms $U_{i} \rightarrow X$ that are compatible with an open covering $U = \bigcup_{i \in I} U.$

Theorem (Which functors are representable?). A functor $F : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ is representable if and only if it is a Zariski sheaf and has an open cover of open subfunctors represented by schemes over $X$.

Proof. As in the previous remark any representable functor $h_{U} : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ is a Zariski sheaf. Moreover, taking any open cover $U = \bigcup_{i \in I}U_{i}$ gives rise to the desired open cover $\{h_{U_{i}} \rightarrow h_{U}\}_{i \in I}$ of the functor $h_{U}.$

Conversely, let $F : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ be a Zarisk sheaf with an open cover $$\{h_{U_{i}} \rightarrow F\}_{i \in I}$$ such that each $U_{i}$ is a scheme over $X.$ By definition of open subfunctors for any $i, j \in I$, we have an scheme $U_{ij}$ over $X$ such that $$h_{U_{ij}} = h_{U_{i}} \times_{F} h_{U_{j}} \rightarrow h_{U_{i}}$$ is an open subfunctor in $\mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set}),$ which gives rise to an open embedding $\varphi_{i} : U_{ij} \hookrightarrow U_{i},$ which we may think as an actual inclusion by taking its image. Note that $U_{ii} = U_{i}$ because $h_{U_{i}} = h_{U_{i}} \times_{F} h_{U_{i}}.$ The isomorphim $$h_{U_{ij}} = h_{U_{i}} \times_{F} h_{U_{j}} \simeq h_{U_{j}} \times_{F} h_{U_{i}} = h_{U_{ji}}$$ gives rise to an isomorphism $\varphi_{ij} : U_{ij} \overset{\sim}{\longrightarrow} U_{ji}$ of schemes over $X.$ One may check that $$h_{U_{ij} \cap U_{ik}} = h_{U_{ij}} \times_{h_{U_{i}}} h_{U_{ik}}.$$ We continue to note that $$\begin{align*}h_{U_{ij} \cap U_{ik}} &= h_{U_{ij}} \times_{h_{U_{i}}} h_{U_{ik}} \\ &\simeq h_{U_{ij}} \times_{h_{U_{i}}} (h_{U_{i}} \times_{F} h_{U_{k}}) \\ &\simeq (h_{U_{ij}} \times_{h_{U_{i}}} h_{U_{i}}) \times_{F} h_{U_{k}} \\ &\simeq h_{U_{ij}} \times_{F} h_{U_{k}} \\ &\simeq (h_{U_{ij}} \times_{h_{U_{j}}} h_{U_{j}}) \times_{F} h_{U_{k}} \\ &\simeq h_{U_{ij}} \times_{h_{U_{j}}} (h_{U_{j}} \times_{F} h_{U_{k}}) \\ &= h_{U_{ij}} \times_{h_{U_{j}}} h_{U_{jk}} \\ &\simeq h_{U_{ji}} \times_{h_{U_{j}}} h_{U_{jk}} \\ &= h_{U_{ji} \cap U_{jk}},\end{align*}$$ and when we follow the isomorphisms, we notice that all of them are compatible with $h_{U_{ij}} \simeq h_{U_{ji}}$ we fixed before, so this shows that $\varphi_{ij} : U_{ij} \overset{\sim}{\longrightarrow} U_{ji}$ restircts to $$U_{ij} \cap U_{ik} \simeq U_{ji} \cap U_{jk}.$$ Likewise, the isomorphism $\varphi_{jk} : U_{jk} \overset{\sim}{\longrightarrow} U_{kj}$ restricts to $$U_{ji} \cap U_{jk} \simeq U_{ki} \cap U_{kj}.$$ When we combine the isomorphisms $$h_{U_{ij} \cap U_{ik}} \simeq h_{U_{ji} \cap U_{jk}} \simeq h_{U_{ki} \cap U_{kj}},$$ we may note that all the isomorphisms involved are compatible with $h_{U_{ik}} \simeq h_{U_{ki}}.$ Thus, we have $$\varphi_{ik} = \varphi_{jk} \circ \varphi_{ij},$$ with appropriate restrictions. Hence, we may glue the schemes $U_{i}$ to get a scheme $U$ (4.4.A in Vakil), where we have open embeddings $\phi_{i} : U_{i} \hookrightarrow U,$ such that $U = \bigcup_{i \in I}\phi_{i}(U_{i})$ and $$\phi_{i}(U_{i}) \cap \phi_{j}(U_{j}) = \phi_{i}(U_{ij}) = \phi_{j}(U_{ji})$$ so that we may consider $U_{i}$ as an actual open subset of $U.$

Remark. The proof is not over, but so far we have not used either of the hypotheses that the open subfunctors $h_{U_{i}} \rightarrow F$ form a cover nor the functor $F$ is a Zariski sheaf!

Now, let's use the hypothesis that $F$ is a Zariski sheaf. Write $s_{i} : h_{U_{i}} \rightarrow F$ to mean the open subfunctors. We may use the Yoneda correspondence $$\mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{U_{i}}, F) \simeq F(U_{i}),$$ given by $$\phi \mapsto \phi_{U_{i}}(\mathrm{id}_{U_{i}}),$$ (functorial in $U_{i}$) to think of $s_{i} \in F(U_{i}).$ Thinking all $U_{i}$ in $U,$ we have $$s_{i}|_{U_{i} \cap U_{j}} = s_{j}|_{U_{i} \cap U_{j}}$$ for all $i, j \in I$, so we must have a unique $s \in F(U)$ such that $$s|_{U_{i}} = s_{i}$$ for all $i \in I$ because $F$ is a Zariski sheaf. We may interpret $s : h_{U} \rightarrow F$ as a map of functors. One may check that this is an isomorphism of functors by constructing its inverse as follows: for any scheme $W,$ an element of $F(W)$ correponds to a map $h_{W} \rightarrow F$ of functors. Since $h_{U_{i}} \rightarrow F$ is an open subfunctor, we have $h_{U_{i}} \times_{F} h_{W} = h_{W_{i}} \rightarrow h_{W},$ given by an open embedding $W_{i} \hookrightarrow W.$ We have projections $h_{W_{i}} \rightarrow h_{U_{i}},$ which can be composed with $h_{U_{i}} \rightarrow h_{U}$ given by previously constructed open embeddings $U_{i} \hookrightarrow U$ to give maps $$h_{W_{i}} \rightarrow h_{U_{i}} \rightarrow h_{U}.$$ We are not ready to use the hypothesis that $\{h_{U_{i}} \rightarrow F\}_{i \in I}$ is an open cover, which lets us have $$W = \bigcup_{i} W_{i}.$$ Since $F$ is a Zariski sheaf, the maps $h_{W_{i}} \rightarrow h_{U}$ with all $i \in I$ gives rise to $h_{W} \rightarrow h_{U},$ which can be thought as an element of $h_{U}(W).$ Thus, we have constructed a set map $$F(W) \rightarrow h_{U}(W).$$ Using this map, one can construct a map $F \rightarrow h_{U}$ of functors and can check that it is inverse to the map $h_{U} \rightarrow F$ we have constructed earlier. Therefore, we get $F \simeq h_{U},$ so $F$ is representable, as desired. $\Box$

Language of sites. We revisited the usual notion of sheaves on a scheme $X$ as a functor $\mathscr{C}_{X}^{\mathrm{op}} \rightarrow \textbf{Set},$ with an equalizer sequence attached to every open cover of an object of $\mathscr{C}_{X}^{\mathrm{op}},$ where $\mathscr{C}_{X}$ is the category of open subsets of $X,$ whose morphisms are given by inclusions. Then we replaced $\mathscr{C}_{X}$ with $\textbf{Sch}_{X},$ the category of schemes over $X,$ and called the functors $\textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set},$ with an equalizer sequence attached to every open cover of an object of $\textbf{Sch}_{X}.$ We now introduce a general language that combines these two situations.

Given a category $\mathscr{C},$ a Grothendieck topology on $\mathscr{C}$ is given by the following data: for every object $U$ of $\mathscr{C},$ we have a set $\mathrm{Cov}(U),$ each of whose element is a collection of morphisms into $U,$ called the set of coverings of $U$, with the following properties.

1. Any isomorphism into $U$ is in $\mathrm{Cov}(U).$
2. Coverings respect base change: if $\{U_{i} \rightarrow U\}_{i \in I} \in \mathrm{Cov}(U),$ for any morphism $V \rightarrow U,$ we have $\{U_{i} \times_{U} V \rightarrow V\} \in \mathrm{Cov}(V),$ where we also require the fiber products exist in $\mathscr{C}.$
3. We can take subcoverings: if $\{U_{i} \rightarrow U\}_{i \in I} \in \mathrm{Cov}(U)$ and $\{U_{ij} \rightarrow U_{i}\}_{j \in J_{i}} \in \mathrm{Cov}(U_{i})$ for each $i \in I,$ then $\{U_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}} \in \mathrm{Cov}(U).$

Another word. If a category $\mathscr{C}$ is equipped with a Grothendieck topology, then we call it a site.

Example. Note that $\mathscr{C}_{X}$ can get a Grothendieck topology and be made into a site. Given any object $U$ (i.e., an open subset $U \hookrightarrow X$), the set $\mathrm{Cov}(U)$ consists of all open covers of $U.$ As we mentioned earlier, this site is called the small Zariski site on $X.$

Example. The category $\textbf{Sch}_{X}$ can also be equipped with a Grothendieck topology and be made into a site. Given any object $U$ (i.e., a scheme map $U \rightarrow X$), the set $\mathrm{Cov}(U)$ consists of all open covers of $U.$ As we mentioned earlier, this site is called the big Zariski site on $X.$

Given a site $\mathscr{C},$ a functor $F : \mathscr{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ is called a presheaf. A presheaf $F$ is a called a sheaf if given every object $U$ of $\mathscr{C}$ and $\{U_{i} \rightarrow U\}_{i \in I},$ the sequence (given similarly from the case of the small/big Zariski site on a scheme) $$F(U) \rightarrow \prod_{i \in I}F(U_{i}) \rightrightarrows \prod_{i \in I}F(U_{i} \times_{U} U_{j})$$ is an equalizer in $\textbf{Set}.$

Remark. We have observed before that given $X,$ all the representable functors in $\mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set})$ are sheaves on the big Zariski site of $X.$ Moreover, we learned that a functor in $\mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set})$ is representable if and only if

• it is a sheaf on Zariski site of $X$ and
• it can be covered by representable open subfunctors.

Construction of fiber product of schemes. We finish the posting by giving a basic application: construction of the fiber product of two schemes over a ring $R.$ I am not sure whether we can do this for more general base instead of $\mathrm{Spec}(R)$, because generally structure maps over a base scheme are not necessarily affine, which we would need in the arguments below.

Consider $R$-schemes $X, Y, Z$ with $R$-scheme maps $X \rightarrow Z$ and $Y \rightarrow Z.$ Using the Yoneda embedding $$\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{R}^{\mathrm{op}}, \textbf{Set}),$$ we note that if the functor $h_{X} \times_{h_{Z}} h_{Y}$ is representable, then the scheme that represents this functor must be the fiber product $X \times_{Z} Y$ in $\textbf{Sch}_{R}.$ Let $\mathrm{Spec}(A) \subset X, \mathrm{Spec}(B) \subset Y, \mathrm{Spec}(C) \subset Z$ be affine open subsets. We have a canonical map $$h_{\mathrm{Spec}(A)} \times_{h_{\mathrm{Spec}(C)}} h_{\mathrm{Spec}(B)} \rightarrow h_{X} \times_{h_{Z}} h_{Y}$$ given by the fiber product in $\mathrm{Fun}(\textbf{Sch}_{R}^{\mathrm{op}}, \textbf{Set}).$ We claim that these are representable open subfunctors that cover $h_{X} \times_{h_{Z}} h_{Y}$ and the functor $h_{X} \times_{h_{Z}} h_{Y}$ is a sheaf on the big Zariski site on $X.$ This will finish showing that $h_{X} \times_{h_{Z}} h_{Y}$ is representable.

Proof. For representability, one can check that $$h_{\mathrm{Spec}(A \otimes_{C} B)} = h_{\mathrm{Spec}(A)} \times_{h_{\mathrm{Spec}(C)}} h_{\mathrm{Spec}(B)}.$$ To check the functor maps into $h_{X} \times_{h_{Z}} h_{Y},$ we can do a bit more generally. Given any open subsets $U \hookrightarrow X, V \hookrightarrow Y, W \hookrightarrow Z,$ where $U$ and $V$ map into $W$ using the maps $X \rightarrow Z$ and $Y \rightarrow Z$ we fixed in the beginning. Given any $h_{T} \rightarrow h_{X} \times_{h_{Z}} h_{Y},$ we have $R$-scheme maps $\alpha : T \rightarrow X$ and $\beta : T \rightarrow Y,$ that are also compatible with the maps from $X, Y$ to $Z.$ We have an open subset $\alpha^{-1}(U) \cap \beta^{-1}(V) \hookrightarrow T,$ and one may check that the corresponding map $$h_{\alpha^{-1}(U) \cap \beta^{-1}(V)} \rightarrow h_{T}$$ gives the fiber prduct $$h_{\alpha^{-1}(U) \cap \beta^{-1}(V)} = (h_{U} \times_{h_{W}} h_{V}) \times_{h_{X} \times_{h_{Z}} h_{Y}} h_{T}.$$ To show the covering condition, take any affine open $\mathrm{Spec}(C) \subset Z$ and cover its preimage in $X$ by affine opens and the same in $Y.$ Then we may cover $T$ with open subsets of the form $\alpha^{-1}(\mathrm{Spec}(A)) \cap \beta^{-1}(\mathrm{Spec}(B)),$ where $\mathrm{Spec}(A), \mathrm{Spec}(B)$ are affine opens of $X$ and $Y,$ respectively.

Finally, we need to show that $h_{X} \times_{h_{Z}} h_{Y}$ is a sheaf (on the big Zariski site). Let $U$ be an $R$-scheme and say $U = \bigcup_{i \in I}U_{i}$ is an open cover. We use the construction of the fiber product of the functors to show that the following sequence is an equalizer in $\textbf{Set}$: $$h_{X}(U) \times_{h_{Z}(U)} h_{Y}(U) \rightarrow \prod_{i \in I} h_{X}(U_{i}) \times_{h_{Z}(U_{i})} h_{Y}(U_{i}) \rightrightarrows \prod_{i, j \in I} h_{X}(U_{ij}) \times_{h_{Z}(U_{ij})} h_{Y}(U_{ij}),$$ where $U_{ij} := U_{i} \cap U_{j}.$ Given $(s_{i}, t_{i}) \in h_{X}(U_{i}) \times_{h_{Z}(U_{i})} h_{Y}(U_{i})$ for $i \in I,$ if $$(s_{i}, t_{i})|_{U_{ij}} = (s_{j}, t_{j})|_{U_{ij}}$$ for all $i, j \in I,$ then $s_{i}|_{U_{ij}} = s_{j}|_{U_{ij}}$ and $t_{i}|_{U_{ij}} = t_{j}|_{U_{ij}},$ for all $i, j \in I.$ Since $h_{X}$ and $h_{Y}$ are sheaves, we have a unique $s \in h_{X}(U)$ such that $s|_{U_{i}} = s_{i}$ for all $i \in I$ and a unique $t \in h_{Y}(U)$ such that $t|_{U_{i}} = t_{i}$ for all $i \in I.$ The compositions $U \overset{s}{\longrightarrow} X \rightarrow Z$ and $U \overset{t}{\longrightarrow} Y \rightarrow Z$ must be identical because their restrictions on $U_{i} \hookrightarrow U$ are for every $i \in I.$ Thus, we have found a unique $(s, t) \in h_{X}(U) \times_{h_{Z}(U)} h_{Y}(U)$ such that $$(s, t)|_{U_{i}} = (s_{i}, t_{i})$$ for all $i \in I.$ This finishes the proof, so in particular we proved that $h_{X} \times_{h_{Z}} h_{Y}$ is representable by an $R$-scheme $\Box$

Taking cohomology commutes with any exact functor

In a previous positing, we have talked about why taking Čech cohomology of a quasi-compact separated scheme $X$ over a field $k$ commutes with base change along any field extension $k \hookrightarrow K.$ If one carefully reads the proof,  it is not difficult to note that we have proved that given any cochain $$C : 0 \rightarrow C^{0} \rightarrow C^{1} \rightarrow C^{2} \rightarrow \cdots$$ of $k$-vector spaces, if $$C \otimes_{k} K : 0 \rightarrow C^{0} \otimes_{k} K \rightarrow C^{1} \otimes_{k} K \rightarrow C^{2} \otimes_{k} K \rightarrow \cdots$$ denotes the induced cochain of $K$-vector spaces by the tensor product, then we have $$H^{i}(C \otimes_{k} K) \simeq H^{i}(C) \otimes_{k} K$$ for all $i \geq 0.$ There are more things to keep in mind from the proof (in the previous positing):

• we don't have to assume that $C^{i} = 0$ for $i < 0$;
• the fact that $H^{i}(C) \otimes_{k} K = (\ker(d^{i}) \otimes_{k} K)/\mathrm{im}(d^{i-1} \otimes_{k} K)$ only used the fact that $(-) \otimes_{k} K$ is a right-exact functor;
• exactness of $(-) \otimes_{k} K$ (as $k$ is a field) is used to ensure that the canonical map $\ker(d^{i}) \otimes_{k} K \rightarrow \ker(d^{i} \otimes_{k} K)$ is an isomorphism, which resulted in $$\begin{align*}H^{i}(C) \otimes_{k} K &= (\ker(d^{i}) \otimes_{k} K)/\mathrm{im}(d^{i-1} \otimes_{k} K) \\ &\simeq \ker(d^{i} \otimes_{k} K)/\mathrm{im}(d^{i-1} \otimes_{k} K) \\ &= H^{i}(C \otimes_{k} K).\end{align*}$$

In particular, our observations immediately show that if $C$ is any cochain complex of $A$-modules over a (commutative unital) ring $A$ and $B$ is any $A$-algebra, then for each $i \in \mathbb{Z},$ we always have a map $$H^{i}(C) \otimes_{A} B \rightarrow H^{i}(C \otimes_{A} B),$$ and this map is an ismorphism if $B$ is flat over $A$ (i.e., if the functor $(-) \otimes_{A} B : \textbf{Mod}_{A} \rightarrow \textbf{Mod}_{B}$ is exact). Let's prove a more general statement.

Theorem. Given two rings $A, B,$ let $F : \textbf{Mod}_{A} \rightarrow \textbf{Mod}_{B}$ be any additive functor and fix any cochain complex $$C : \cdots \overset{d^{i-2}}{\longrightarrow} C^{i-1} \overset{d^{i-1}}{\longrightarrow} C^{i} \overset{d^{i}}{\longrightarrow} C^{i+1} \overset{d^{i+1}}{\longrightarrow} \cdots$$ in the category $\textbf{Mod}_{A}$ of $A$-modules. Denote by $F(C)$ the cochain complex of $B$-modules given by applying the additive functor $F$ to the cochain compelx $C$ of $A$-modules.

1. If $F$ is right-exact, then we have a canonical $B$-linear map $F(H^{i}(C)) \rightarrow H^{i}F(C)$ for each $i \in \mathbb{Z}.$
2. If $F$ is left-exact, then we have a canonical $B$-linear map $F(H^{i}(C)) \leftarrow H^{i}F(C)$ for each $i \in \mathbb{Z}.$
3. If $F$ is exact, then for each $i \in \mathbb{Z},$ the maps given above are inverses to each other so that $$F(H^{i}(C)) \simeq H^{i}(F(C)).$$

Remark. Note that we did not even mention what a "canonical map" means. In this case, one needs to look at the proof to see how such a map is induced and figure out in what sense it is canonical.

Before proving the theorem, let us not assume left-exactness nor right-exactness for $F.$

1st map. We may apply $F$ to the exact sequence $$C^{i-1} \overset{d^{i-1}}{\longrightarrow} \ker(d^{i}) \overset{\pi_{i}}{\longrightarrow} \ker(d^{i})/\mathrm{im}(d^{i-1}) = H^{i}(C) \rightarrow 0$$ to get the following cochain: $$F(C^{i-1}) \overset{F(d^{i-1})}{\longrightarrow} F(\ker(d^{i})) \overset{F(\pi_{i})}{\longrightarrow} F(\ker(d^{i})/\mathrm{im}(d^{i-1})) = F(H^{i}(C)).$$ By the universal property of the cokernel map $$F(\ker(d^{i}))\overset{\rho_{i}}{\longrightarrow} \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \rightarrow 0,$$ there is a unique map $$ \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\psi_{i}}{\longrightarrow} F(H^{i}(C))$$ such that $$F(\pi_{i}) = \psi_{i} \circ \rho_{i}.$$ In this sense, the map $\psi_{i}$ is canonical.

Remark. If $F$ is right-exact, the map $\psi_{i}$ is an isomorphism.

2nd map. We are still not assuming any exactness of $F.$ The exact sequence $$0 \rightarrow \ker(d^{i}) \overset{\iota_{i}}{\longrightarrow} C^{i} \overset{d^{i}}{\longrightarrow} C^{i+1}$$ gives rise to the following cochain: $$F(\ker(d^{i})) \overset{F(\iota_{i})}{\longrightarrow} F(C^{i}) \overset{F(d^{i})}{\longrightarrow} F(C^{i+1}).$$ This induces the cochian $$\frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\overline{F(\iota_{i})}}{\longrightarrow} \frac{F(C^{i})}{{\mathrm{im}(F(d^{i-1}))}} \overset{\overline{F(d^{i})}}{\longrightarrow} F(C^{i+1}).$$ We note that $$0 \rightarrow H^{i}(F(C)) = \frac{\ker(F(d^{i}))}{{\mathrm{im}(F(d^{i-1}))}} \overset{\omega_{i}}{\longrightarrow} \frac{F(C^{i})}{{\mathrm{im}(F(d^{i-1}))}} \overset{\overline{F(d^{i})}}{\longrightarrow} F(C^{i+1})$$ is exact, so by the universal property of the kernel, there exists a unique map $$\frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\phi_{i}}{\longrightarrow} H^{i}(F(C))$$ such that $$\overline{F(\iota_{i})} = \omega_{i} \circ \phi_{i}.$$ This uniqueness is why we say the map $\phi_{i}$ is canonical.

Remark. If $F$ is left-exact, the map $\phi_{i}$ is an isomorphism.

Proof. For the first statement, assume that $F$ is right-exact. Then the map $\psi_{i}$ is an isomorphism, so we get $$F(H^{i}(C)) \overset{\psi_{i}^{-1}}{\longrightarrow} \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\phi_{i}}{\longrightarrow} H^{i}(F(C)),$$ as desired.

For the second statement, assume that $F$ is left-exact. Then $\phi_{i}$ is left-exact, so we get $$H^{i}(F(C)) \overset{\phi_{i}^{-1}}{\longrightarrow} \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\psi_{i}}{\longrightarrow} F(H^{i}(C)),$$ as desired.

The third statement follows because if $F$ is exact, then both $\psi_{i}$ and $\phi_{i}$ are invertible and $$(\psi_{i}^{-1} \circ \phi_{i})^{-1} = \phi_{i}^{-1} \circ \psi_{i},$$ as desired. $\Box$

Non-mathematical remarks. Due to the formula $$F(H^{i}(C)) \simeq H^{i}(F(C)),$$ Ravi Vakil called this "FHHF Theorem" (Vakil, 1.6.H), and I believe in an earlier draft of his book, he called it "From Here Hop Far," which perhaps did not catch on. He also calls this Fernbahnhof(FernbaHnHoF), which seems like a German word, but I did not get it even after looking up what it meant.

More general situations. In fact, Vakil's book (1.6.H) formulates this in a much more general situation where an additive functor given between any two abelian categories.

Theorem. Given two abelian categories $\mathscr{A}, \mathscr{B},$ let $F : \mathscr{A} \rightarrow \mathscr{B}$ be any additive functor and fix any cochain complex $$C : \cdots \overset{d^{i-2}}{\longrightarrow} C^{i-1} \overset{d^{i-1}}{\longrightarrow} C^{i} \overset{d^{i}}{\longrightarrow} C^{i+1} \overset{d^{i+1}}{\longrightarrow} \cdots$$ in $\mathscr{A}$. Denote by $F(C)$ in $\mathscr{B}$ given by applying the additive functor $F$ to the cochain compelx $C$ in $\mathscr{A}.$

1. If $F$ is right-exact, then we have a canonical morphism $F(H^{i}(C)) \rightarrow H^{i}F(C)$ in $\mathscr{B}$ for each $i \in \mathbb{Z}.$
2. If $F$ is left-exact, then we have a canonical morphism $F(H^{i}(C)) \leftarrow H^{i}F(C)$ in $\mathscr{B}$ for each $i \in \mathbb{Z}.$
3. If $F$ is exact, then for each $i \in \mathbb{Z},$ the maps given above are inverses to each other so that $$F(H^{i}(C)) \simeq H^{i}(F(C)).$$

Proof. When we gave the proof for the case $\mathscr{A} = \textbf{Mod}_{A}$ and $\mathscr{B} = \textbf{Mod}_{B},$ all the maps are induced by certain universal properties, so this should work in general as well. $\Box$

Complete intersection

I am scared about the phrase "complete intersection", so I decided to write about it to get rid of this phobia. In Vakil's book, complete intersections are defined for the first time in 8.4.8 as follows.

Given a scheme $Y,$ a complete intersection is the (scheme-theoretic) intersection in $Y$ is the intersection of some finitely many effective Cartier divisors $D_{1}, \dots, D_{r}$ giving a regular sequence at every point in $D_{1} \cap \cdots \cap D_{r}.$

What do these words mean? An effective Cartier divisor $D \hookrightarrow Y$ is a closed subscheme that is (not necessarily arbitrary) affine locally given by single nzd (non-zerodivisor). This means that there is an affine open cover $Y = \bigcup_{i \in I} \mathrm{Spec}(A_{i})$ such that $$D|_{\mathrm{Spec}(A_{i})} = \mathrm{Spec}(A_{i}/(f_{i}))$$ for some nzds $f_{i} \in A_{i}.$ If $Y$ is locally Noetherian (so that $A_{i}$ are Noetherian), this amounts to say that the local sections $f_{i}$ do not vanish at any associated points of $Y$ (or equivalently, any associated primes of $A_{i}$).

A regular sequence of a ring $A,$ is a finite sequence $f_{1}, \dots, f_{r}$ in $A$ such that

1. the image of $f_{i}$ in $A/(f_{1}, \dots, f_{i-1})$ is not an nzd (where we denote $f_{0} := 0$) for all $1 \leq i \leq r,$ and
2. $(f_{1}, \dots, f_{r})$ is not the unit ideal $(1) = A.$

Why would anyone need to remember a definition like this? This is because with the conditions given above, the locus of $f_{1}, \dots f_{r}$, or more precisely $$\mathrm{Spec}(A/(f_{1}, \dots, f_{r})) \simeq V(f_{1}, \dots, f_{r}) \subsetneq \mathrm{Spec}(A),$$ has codimension $r$ in $\mathrm{Spec}(A).$ In more plain algebraic words, this means that any minimal prime of $A$ containing $f_{1}, \dots, f_{r}$ has height $r.$ This is an application of Krull's Principal Ideal Theorem (11.3.3 in Vakil).

Remark. One needs to be careful about the order of a regular sequence unless $A$ is Noetherain and local. (See 8.4.5 and 8.4.6 in Vakil.)

Now, let's unpack the definition of a complete intersection of effective Cartier divisors $D_{1}, \dots, D_{r}$ in $Y.$ By definition of effective Cartier divisors, we have an affine open cover $\mathscr{U}$ of $Y$ such that for any $\mathrm{Spec}(A) \in \mathscr{U}$ we have $D_{i} = \mathrm{Spec}(A/(f_{i}))$ with an nzd $f_{i} \in A.$ We may write $$D_{1} \cap \cdots \cap D_{r} \cap \mathrm{Spec}(A) = \mathrm{Spec}(A/(f_{1}, \dots, f_{r})).$$ The condition that $D_{1}, \dots, D_{r}$ form a regular sequence at any point in $D_{1} \cap \cdots \cap D_{r}$ means that for any prime ideal $\mathfrak{p}$ of $A$ containing $f_{1}, \dots, f_{r},$ the images of $f_{1}, \dots, f_{r}$ in $A_{\mathfrak{p}}$ (or in the unique maximal ideal $\mathfrak{p}A_{\mathfrak{p}}$) form a regular sequence. It is enough to ask this condition to hold for minimal primes $\mathfrak{p}$ containing $f_{1}, \dots, f_{r}.$ Such primes correspond to the irreducible component $V(\mathfrak{p})$ of $V(f_{1}, \dots, f_{r}) \subset \mathrm{Spec}(A).$ This implies that $\mathfrak{p}A_{\mathfrak{p}}$ is minimal among the primes in $A_{\mathfrak{p}}$ containing (the images of) $f_{1}, \dots, f_{r},$ but it is already a (unique) maximal ideal of $A_{\mathfrak{p}}.$ Hence, we see that $\mathfrak{p}A_{\mathfrak{p}}$ is the only prime containing $f_{1}, \dots, f_{r}.$ This implies that $$\mathfrak{p}A_{\mathfrak{p}} = \sqrt{(f_{1}, \dots, f_{r})} \subset A_{\mathfrak{p}}.$$ Thus, we see that $$\begin{align*}\mathrm{ht}(\mathfrak{p}) &= \dim(A_{\mathfrak{p}}) \\ &= \mathrm{ht}(\mathfrak{p}A_{\mathfrak{p}}) \\ &= r,\end{align*}$$ by Krull's Principal Ideal Theorem. This implies that the codimension of $D_{1} \cap \cdots \cap D_{r}$ in $Y$ is equal to $r.$

Twisted cubic is not a complete intersection of hypersurfaces in $\mathbb{P}^{3}.$ Let us work over a field $k.$ Recall that the twisted cubic is the image Veronese embedding $\mathbb{P}^{1} \hookrightarrow \mathbb{P}^{3},$ which looks like $[x : y] \mapsto [x^{3} : x^{2}y : xy^{2} : y^{3}]$ on the sets of $k$-points. It is indeed given by the line bundle $\mathscr{O}_{\mathbb{P}^{1}}(3)$ on $\mathbb{P}^{1},$ so the degree of this curve as a closed subschme is $3$ (and hence the name "cubic"). The name "twisted" suggests that the curve should not be in any hyperplane (which is a copy of $\mathbb{P}^{2}$) in $\mathbb{P}^{3}.$

Indeed, if there is such a hyperplane $H$ so that we have $\mathbb{P}^{1} \hookrightarrow H \subset \mathbb{P}^{3},$ then denote by $a_{0}y_{0} + a_{1}y_{1} + a_{2}y_{2} + a_{3}y_{3}$ the (nonzero) linear form defining $H.$ The graded ring maps defining the twisted cubic are given by $$k[y_{0}, y_{1}, y_{2}, y_{3}] \twoheadrightarrow k[x_{0}^{3}, x_{0}^{2}x_{1}, x_{0}x_{1}^{2}, x_{1}^{3}] \subset k[x_{0}, x_{1}],$$ where the surjection is given by $y_{i} \mapsto x_{0}^{3-i}x_{1}^{i},$ and the image subring is the Veronese subring of degree $3.$ Under this map, we will have $$a_{0}y_{0} + a_{1}y_{1} + a_{2}y_{2} + a_{3}y_{3} \mapsto a_{0}x_{0}^{3} + a_{1}x_{0}^{2}x_{1} + a_{2}x_{0}x_{1}^{2} + a_{3}x_{1}^{3}.$$ Since the monomials in $x_{0}, x_{1}$ are $k$-linearly independent, the degree three homogenous polynomial in the image is not zero. This implies that $\mathbb{P}^{1} \cap H$ (in $\mathbb{P}^{2}$) has dimension $0,$ so $H$ cannot contain $\mathbb{P}^{1}.$ This proves that the twisted cubic is "twisted" in $\mathbb{P}^{3}.$

Finally, since the twisted cubic $C$ has dimension $1$ (as it is isomorphic to $\mathbb{P}^{1}$), if it were possible that we can write it as a complete intersection of hypersurfaces in $\mathbb{P}^{3},$ it must be given by two homogeneous polynomials $f_{1}$ and $f_{2}$ due to the (co)dimensional reason. By Bézout's theorem, we must have $$3 = \deg(C) = \deg(f_{1})\deg(f_{2}),$$ but we have already seen that $\deg(f_{1}), \deg(f_{2}) \neq 1,$ because $C$ is "twisted" in $\mathbb{P}^{3}.$ Thus, we have $$3 = \deg(f_{1})\deg(f_{2}) \geq 2 \cdot 2 = 4,$$ which is a contradiction. This shows that $C$ cannot be written as a complete intersection of hypersufaces in $\mathbb{P}^{3}.$

Remark. Recall that in $\mathbb{P}^{3},$ the twisted cubic is precisely given as $$\mathrm{Proj}\left(\frac{k[y_{0}, y_{1}, y_{2}, y_{3}]}{(y_{0}y_{1} - y_{2}^{2}, y_{0}y_{2} - y_{1}^{2}, y_{0}y_{3} - y_{1}y_{2})}\right).$$ What we have shown is that removing any one of the three given equations will not cut out the twisted cubic. That is, it is VERY TWISTED!

WLOG: algebraically closed

I have seen many algebraic geometors saying, "Without loss of generality, we may assume $k$ is an algebraically closed field." This does not necessarily mean that they don't care about non-algebraically closed fields like $\mathbb{Q}, \mathbb{R},$ or $\mathbb{F}_{q}.$ It's just that there are neat tricks out there that we can use in order to get help from statements over an algebraically closed field. (My main reference for algebraic geometry is Vakil's book, so you may consider this posting as a regurgitation of it.)

Cohomology is insensitive to field extensions. Let $X$ be a quasicompact separated scheme over a field $k$ so that we have a well-defined Čech cohomology for any coherent sheaf $\mathscr{F}$ on $X.$ For any field extension $k \hookrightarrow K,$ we have $$H^{i}(X, \mathscr{F}) \otimes_{k} K \simeq H^{i}(X_{K}, \mathscr{F}_{K}),$$ where $X_{K} := X \times_{\mathrm{Spec}(k)} \mathrm{Spec}(K)$ and $\mathscr{F}_{K}$ means the pullback of $\mathscr{F}$ under the map $X_{K} \rightarrow X.$

Moral. Taking $K = \bar{k}$ in the above fact, we see that the cohomology (or more precisely, the set of Betti numbers) does not change when we replace $X$ with $X_{\bar{k}}.$ To see why this fact is useful in practice, we consider the following exercise.

Exercise. Let $C$ be an one-dimensional closed subscheme of $\mathbb{P}^{3}_{k}$ (for a given field $k$). If $\deg(C) = 1,$ then $C \simeq \mathbb{P}^{1}_{k}.$

How to reduce to algebraically closed case. Suppose that we know that the statement in the above exercise is true when $k$ is algebraically closed. We will see below that $\deg(C) = \deg(C_{\bar{k}}).$ Since dimension does not change when we base change to $\bar{k}$ (as we can see it affine-locally as transcendence degree), we will then apply the exercise for the algebraically closed field case to prove $C_{\bar{k}} \simeq \mathbb{P}^{1}_{\bar{k}}.$ This does not prove that $C$ is isomorphic to $\mathbb{P}^{1}_{k}$ just yet, but this is great amount of information to deduce something like it.

Base changing the inclusion $i : C \hookrightarrow \mathbb{P}^{3}_{k}$ over $\bar{k}$ gives the inclusion $\bar{i} : C_{\bar{k}} \hookrightarrow \mathbb{P}^{3}_{\bar{k}}$. (Probably, the most painless way to see this is to check this affine locally.) In particular, denoting $\pi_{1} : C_{\bar{k}} \rightarrow C$ and $\pi_{2} : \mathbb{P}^{1}_{\bar{k}} \rightarrow \mathbb{P}^{1}_{k}$ for projection maps, we have $$\pi_{2} \circ \bar{i} = i \circ \pi_{1}.$$ From explicit descriptions of $\mathscr{O}(m)$ of $\mathbb{P}^{3}_{k}$ and $\mathbb{P}^{3}_{\bar{k}},$ we note that $$\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m) = \pi_{2}^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m).$$ This implies that $$\bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m) = \pi_{1}^{*}i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m),$$ which are line bundles over $C_{\bar{k}}.$ Thus, we have (for all $i \geq 0$) $$H^{i}(C_{\bar{k}}, \bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m)) = H^{i}(C_{\bar{k}}, \pi_{1}^{*}i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m)) \simeq H^{i}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m)) \otimes_{k} \bar{k}.$$ In particular, we have $$h^{i}(C_{\bar{k}}, \bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m)) = h^{i}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m)).$$ Thus, the Hilbert polynomial of $C$ in $\mathbb{P}^{3}_{k}$ is equal to that of $C_{\bar{k}}$ in $\mathbb{P}^{3}_{\bar{k}}.$ In particular, we have $\deg(C) = \deg(C_{\bar{k}}),$ so applying the statement for the base field $\bar{k},$ we have $C_{\bar{k}} \simeq \mathbb{P}^{1}_{\bar{k}}.$ Thus, we have $$\begin{align*}1 &= \deg(C) \\ &= \deg(C_{\bar{k}}) \\ &= \deg(\bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) \\ &= \chi(C_{\bar{k}}, \bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - \chi(C_{\bar{k}}, \mathscr{O}_{C_{\bar{k}}}) \\ &= \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - \chi(\mathbb{P}^{1}_{\bar{k}}, \mathscr{O}_{\mathbb{P}^{1}_{\bar{k}}}) \\ &= \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - 1. \end{align*}$$ That is, we have $$h^{0}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - h^{1}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = 2.$$ Since $C_{\bar{k}} \simeq \mathbb{P}^{1}_{\bar{k}},$ so the genus of $C$ is $0.$ By Serre duality, we have a line bundle $\omega$ on $C_{\bar{k}}$ with degree $-2$ (18.5.A (b) in Vakil) such that $$h^{1}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = h^{0}(C_{\bar{k}}, \omega \otimes i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) = 0$$ because we have the negative degree (19.2.3 in Vakil) $$\begin{align*}\deg(\omega \otimes i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) &= \deg(\omega) + \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) \\ &= -2 + \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) \\ &= -3,\end{align*}$$ where we know the last equality holds because $$\begin{align*}\deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) - 1 &= \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) + \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) \\ &= \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}) \\ &= \deg(\mathscr{O}_{C_{\bar{k}}}) \\ &= 0.\end{align*}.$$ Thus, we have $$h^{0}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(1)) = h^{0}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = 2.$$ This may not look like much, but it involves quite a bit. Recall that the line bundle $i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)$ has four global sections $s_{0}, s_{1}, s_{2}, s_{3} \in \Gamma(C, \mathscr{O}_{C})$ that do not have common have common zeros, which corresponds to the closed embedding: $$i : C \hookrightarrow \mathbb{P}^{3}_{k} = \mathrm{Proj}(k[x_{0}, x_{1}, x_{2}, x_{3}]).$$ Explicitly, for any nonempty affine open $\mathrm{Spec}(A) \subset C,$ this map is associated to the ring surjections (for $i = 0, 1, 2, 3$) $$k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}, x_{3}/x_{i}] \twoheadrightarrow A$$ given by $x_{j}/x_{i} \mapsto s_{j}/s_{i}.$ Since $h^{0}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = 2,$ it must be the case that 

$$s_{0}, s_{1}, s_{2}, s_{3} \in \Gamma(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1))$$ are $k$-linearly dependent. Thus, we may write $$s_{3} = b_{0}s_{0} + b_{1}s_{1} + b_{2}s_{2}$$ for some $b_{0}, b_{1}, b_{2} \in k.$ Similarly, we see that $s_{0}, s_{1}, s_{2}$ must be $k$-linearly dependent, so we may write $$s_{2} = a_{0}s_{0} + a_{1}s_{1}$$ for some $a_{0}, a_{1} \in k$ as well. This implies that the ring maps $$k[x_{0}/x_{i}, x_{1}/x_{i}] \rightarrow A$$ given by $x_{j}/x_{i} \mapsto s_{j}/s_{i}$ are surjective as well, which induces the closed embedding $$C \hookrightarrow \mathbb{P}^{1}_{k} \subset \mathbb{P}^{3}_{k}.$$ Since $\mathbb{P}^{1}_{k}$ is one-dimensional and irreducible, this closed embedding must be an isomorphism, so we get $C \simeq \mathbb{P}^{1}_{k}.$

Thus, we only need to do the exercise for the case where $k$ is algebraically closed.

Algebraically closed case. If $k$ is algebraically closed, then the above exercise can be done as follows. Pick any two (distinct) closed points $p, q$ on $C,$ which exist because $\dim(C) = 1 > 0.$ Since $k$ is algebriacally closed, by Nullstellensatz, we know $p, q$ are $k$-points. Thus, we may write $$p = [p_{0} : p_{1} : p_{2} : p_{3}] \in \mathbb{P}^{3}(k)$$ and $$q = [q_{0} : q_{1} : q_{2} : q_{3}] \in \mathbb{P}^{3}(k).$$ Then there is a hyperplane $H$ of $\mathbb{P}^{3}_{k}$ that contains $p$ and $q.$ 

Why does such a hyperplane exist? We must find a $k$-point $a = [a_{0} : a_{1} : a_{2} : a_{3}] \in \mathbb{P}^{3}(k)$ such that $$a_{0}p_{0} + a_{1}p_{1} + a_{2}p_{2} + a_{3}p_{3} = a_{0}q_{0} + a_{1}q_{1} + a_{2}q_{2} + a_{3}q_{3} = 0.$$ These are precisely $k$-points of the intersection of the hyperplanes defined by the two linear equations $$p_{0}x_{0} + p_{1}x_{1} + p_{2}x_{2} + p_{3}x_{3} = 0$$ and $$q_{0}x_{0} + q_{1}x_{1} + q_{2}x_{2} + q_{3}x_{3} = 0.$$ Since $p \neq q,$ the intersection of these two hyperplanes is isomorphic to $\mathbb{P}^{1}_{k},$ which has many $k$-points (i.e., as many as $k \sqcup \{\infty\}$). In particular, there is such an $a = [a_{0} : a_{1} : a_{2} : a_{3}].$ This establishes that there is a hyperplane $H$ in $\mathbb{P}^{3}_{k}$ containing $p, q.$

Remark. The above argument shows that (for an algebraically closed field $k$) there are infinitely many hyperplanes in $\mathbb{P}^{3}_{k}$ passing through $p$ and $q.$ We will use this remark soon.

If $C$ is reduced and irreducible, by Bézout's theorem, we must have $C \subset H$ because otherwise $$\deg(C \cap H) = \deg(C)\deg(H) = 1$$ so that there is one point in the intersection $C \cap H,$ contradicting the fact that there are at least two points in the intersection, namely $p, q \in C \cap H.$

If $C$ is irreducible but not necessarily reduced, then its associated reduced scheme $C'$ is in $H,$ but the underlying topological spaces for $C$ and $C'$ are the same, so $C \subset H$ as well. Since there are infinitely many such $H$ (and they all contain $C$), we must have $C \hookrightarrow \mathbb{P}^{1}_{k}$ as a closed embedding. Thus, we must have $C \simeq \mathbb{P}^{1}_{k}.$

If $C$ is not irreducible, write $$C = C_{1} \cup \cdots \cup C_{r},$$ where each $C_{i}$ is an irreducible component of $C.$ Then (by 18.6.F in Vakil) $$\deg(C_{i}) \leq \deg(C) = 1,$$ and since the Hilbert polynomial of $C_{i}$ has degree equal to $\dim(C_{i}) = 1$ (Theorem 18.6.2 in Vakil), we cannot have $\deg(C_{i}) = 0.$ Since any degree is an integer (18.6.G in Vakil), this implies that $$\deg(C_{i}) = 1.$$ Similarly, any finite union of them has degree $1.$

Again, by Bézout's theorem, we either have $|C_{i} \cap H| = 1$ or $C_{i} \subset H.$ There cannot be more than one $C_{i}$ with $|C_{i} \cap H| = 1,$ because otherwise their union will violate Bézout's theorem. Thus, we have at most one such $C_{i},$ so let us assume that $C_{1}$ is such irreducible component so that $C_{2}, \dots, C_{r} \subset H.$ Writing $C_{1} \cap H = \{p_{1}\},$ we may assume that $p_{1} \neq q.$ Take any hyperplane $H_{1}$ in $\mathbb{P}^{3}_{k}$ that contains $p_{1}$ and $q.$ We may assume $q \in C_{2}.$ Applying Bézout's theorem in $H \simeq \mathbb{P}^{2}_{k},$ we see that $H \cap H_{1}$ meets $C_{1}$ at one point, namely $p_{1},$ and $C_{2}$ at one point, namely $q.$ Since $C_{2} \subset H,$ this means that $H_{1}$ meets $C_{2}$ only at $q,$ which implies that $C_{1} \subset H_{1}.$ We may take another hyperplane $H'_{1}$ that passes $p_{1}$ and $q$ so that $C_{1} \subset H_{1} \cap H'_{1} = H_{1} \cap H \subset H.$ Thus, we have shown that $$C = C_{1} \cup C_{2} \cup \cdots \cup C_{r} \subset H,$$ for any hyperplane $H$ containing $p$ and $q.$ Now, taking another such $H,$ we get $C \hookrightarrow \mathbb{P}^{1}_{k},$ which implies $C \simeq \mathbb{P}^{1}_{k},$ as desired. $\Box$

Why is cohomology insensitive to field extension? Denote by $$\pi : X_{K} = X \times_{\mathrm{Spec}(k)} \mathrm{Spec}(K) \rightarrow X$$ the projection map. For any affine open subset $\mathrm{Spec}(A) \subset X,$ we have $$\begin{align*}\Gamma(\mathrm{Spec}(A) \times_{\mathrm{Spec}(k)} \mathrm{Spec}(K), \pi^{*}\mathscr{F}) &= \Gamma(\mathrm{Spec}(A \otimes_{k} K), \pi^{*}\mathscr{F}) \\ &= \Gamma(\mathrm{Spec}(A) , \mathscr{F}) \otimes_{k} K. \end{align*}$$ Thus, we may construct a Čech cochain complex of $X_{K}$ by applying $(-) \otimes_{k} K$ to a Čech cochain complex of $X.$

Fix any cochain complex $$C : 0 \rightarrow C^{0} \rightarrow C^{1} \rightarrow C^{2} \rightarrow \cdots,$$ where each term is a $k$-vector space. Given a field extension $k \hookrightarrow K,$ we get $$C \otimes_{k} K: 0 \rightarrow C^{0} \otimes_{k} K \rightarrow C^{1} \otimes_{k} K \rightarrow C^{2} \otimes_{k} K \rightarrow \cdots,$$ which is a cochain complex of $K$-vector spaces. Our goal is to show that $$H^{i}(C) \otimes_{k} K \simeq H^{i}(C \otimes_{k} K).$$ The exact sequence $$C^{i-1} \overset{d^{i-1}}{\longrightarrow} \ker(d^{i}) \rightarrow H^{i}(C) \rightarrow 0$$ will stay exact if we apply $(-) \otimes_{k} K$ as it is a right exact functor, so we get the following exact sequence: $$C^{i-1} \otimes_{k} K \overset{d^{i-1} \otimes_{k} K}{\longrightarrow} \ker(d^{i}) \otimes_{k} K \rightarrow H^{i}(C) \otimes_{k} K  \rightarrow 0.$$ This implies that $$H^{i}(C) \otimes_{k} K = \frac{\ker(d^{i}) \otimes_{k} K}{\mathrm{im}(d^{i-1} \otimes_{k} K)}.$$ Since $(-) \otimes_{k} K$ is left exact (because $k$ is a field), we can identify $\ker(d^{i}) \otimes_{k} K$ with $\ker(d^{i} \otimes_{k} K).$ Thus, we have $$H^{i}(C) \otimes_{k} K = \frac{\ker(d^{i}) \otimes_{k} K}{\mathrm{im}(d^{i-1} \otimes_{k} K)} \simeq \frac{\ker(d^{i} \otimes_{k} K)}{\mathrm{im}(d^{i-1} \otimes_{k} K)} = H^{i}(C \otimes_{k} K),$$ as desired. $\Box$

Remark. In general, it seems to be true that "cohomology commutes with any exact functor" (1.6.H (c) in Vakil). However, I am not in the mood for chasing diagrams tonight.

Bézout's theorem

Bézout's theorem is one of the theorems I was scared of. Let alone its proof, I was always uncomfortable about the word "multiplicity" because I was not sure what it really meant. What I am writing here is more or less a poor regurgitation of a good exposition in Vakil's book, which made me realize that this theorem was not so bad after all.

Bézout's theorem. Over any field $k,$ any two curves $C_{1}$ and $C_{2}$ in $\mathbb{P}^{2}$ such that $\dim(C_{1} \cap C_{2}) = 0$ meet at precisely at $\deg(C_{1})\deg(C_{2})$ points counting with multiplicity.

Really, most of the effort in understanding this theorem goes into understanding what some of the words mean in the statement: "curve", "degree", and "multiplicity". For the rest, we will be working over a fixed field $k.$

Degree. Let $i : X \hookrightarrow \mathbb{P}^{n}$ be a closed subscheme. Then $p_{X}(m) := h^{0}(X, \mathcal{O}_{X}(m))$ is a polynomial in $m$ when $m$ is large enough. This is called the Hilbert polynomial of $X$ in $\mathbb{P}^{n}.$ The coefficients of this polynomial are in $\mathbb{Q}.$ It turns out that the leading coefficient of the Hilbert polynomial $p_{X}(m)$ of $X \hookrightarrow \mathbb{P}^{n}$ times $d!$ is always an integer (18.6.G in Vakil), where $d = \deg(p_{X}).$ The degree of the closed subscheme $X \hookrightarrow \mathbb{P}^{n}$ is the leading coefficient of its Hilbert polynomial $p_{X}(m)$ times $d!,$ which is a non-negative integer.

Dependence on the embedding. Note that the Hilbert polynomial and the degree of a closed subscheme $i : X \hookrightarrow \mathbb{P}^{n}$ depends on the embedding $i$, not just the scheme $X.$ For instance, the Hilbert polynomial of $\mathbb{P}^{1}$ in itself is equal to $m + 1$ with the degree $1,$ but when $\mathbb{P}^{1}$ is considered as a closed subscheme of $\mathbb{P}^{3}$ given by the Veronese embedding, its Hilbert polynomial is $3m + 1$ with degree $3.$ (See Example 1 on p.490 and 18.6.D in Vakil.) After all, the dependence on $i$ should not be so surprising because $$i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m) \simeq \mathscr{O}_{X}(m).$$ This isomorphism is more or less a definition.

Example. The notion of degree will be used to express what it means to the "multiplicity" of a point at the intersection of two curves. For instance, consider the intersection of the parabola $y = x^{2}$ and the $x$-axis $y = 0$ in the affine plane in $(x, y)$-coordinates. When we intersect these two curves, we have $(x^{2}, y) = (0,0),$ so $(x, y) = (0, 0)$ is the only point at the intersection. Well, this is correct, except the fact that when we work with schemes, we need to remember $(x^{2}, y) = (0, 0).$ That is, we are intersecting two affine schemes whose data are equivalent to the following two rings: $k[x, y]/(y-x^{2})$ and $k[x, y]/(y).$ Hence, their intersection corresponds to $k[x]/(x^{2})$ not $k[x]/(x) \simeq k,$ although as topological spaces they are equal $$\mathrm{Spec}(k[x]/(x^{2})) \simeq V(x^{2}) = V(x) = \{(x)\} \simeq \mathrm{Spec}(k[x]/(x)),$$ a single closed point of the affine line $\mathrm{Spec}(k[x]) = \mathbb{A}^{1}.$ Note that this issue comes in even if one only considers the most classical algebraic geometry, namely when $k = \mathbb{C}.$ In this case, the degree of the point at the intersection is equal to the dimension of the ring $k[x]/(x^{2})$ over $k,$ which is $2$.

To fit this example to our situation, we need to projectivize these curves. Namely, the parabola $y = x^{2}$ becomes $yz = x^{2}$ in $\mathbb{P}^{2}$ and the line $y = 0$ becomes $y = 0$ in $\mathbb{P}^{2}.$ More technically the parabola is given by $$\mathrm{Proj}(k[x,y,z]/(yz - x^{2})) \hookrightarrow \mathrm{Proj}(k[x,y,z]) = \mathbb{P}^{2},$$ and the line is given by $$\mathbb{P}^{1} = \mathrm{Proj}(k[x,y,z]/(y)) \hookrightarrow \mathrm{Proj}(k[x,y,z]) = \mathbb{P}^{2}.$$ The intersection of these two curves is given by $$\mathrm{Proj}(k[x,z]/(x^{2})) \hookrightarrow \mathrm{Proj}(k[x,z]) = \mathbb{P}^{2}.$$ If we just look at this as a topological space, then this is identical to $V_{\mathbb{P}^{2}}(x),$ which corresponds to the $k$-point $[x : z] = [0 : 1]$ in $\mathbb{P}^{2}.$ However, the point $[0 : 1]$ also corresponds to $$\mathrm{Proj}(k[x,z]/(x)) \hookrightarrow \mathrm{Proj}(k[x,z]) = \mathbb{P}^{2},$$ but the former point has degree $2,$ and the latter will have degree $1$ (in the sense of the degree defined with the Hilbert polynomial).

More generally, any degree $d$ hypersurface of $\mathbb{P}^{n}$ (i.e., the locus given by a single degree $d$ homogeneous polynomial) has degree $d$ in the sense of Hilbert polynomial, and this is great because the latter notion of degree is more intrinsic to the scheme That is, it does not require explicit description of polynomials. (See 18.6.H in Vakil's book.)

As this example illustrates, when the closed subscheme $X$ we are dealing with consists of finitely many closed points on a projective curve on $C$, then we may consider $X$ as a divisor on $C,$ namely $$X = n_{1}x_{1} + \cdots + n_{r}x_{r},$$ where $x_{i}$ are closed points of $C,$ and the degree of $X$ is precisely the degree of $X$ as a divisor: $$\deg(X) = \sum_{i=1}^{r}n_{i}[\kappa(x_{i}) : k].$$ Thus, in this case, the degree can be computed by local information even though we needed somewhat global setting (i.e., projective situation) to define it. This is because for any closed point $x \in C,$ if we consider $\{x\} \hookrightarrow C$ as a closed subscheme (and hence a closed subscheme of some $\mathbb{P}^{n}$), then its degree is precisely the degree $[\kappa(x) : k]$ of the residue field $\kappa(x)$ of $x$ in $\mathbb{P}^{n}$ (or in $C$).

Multiplicity at intersection. Due to the discussion that follows from the previous example, if we have two curves $C_{1}$ and $C_{2}$ in $\mathbb{P}^{2}$ and $C_{1} \cap C_{2}$ is a finite set of closed points, then $\deg(C_{1} \cap C_{2})$ is equal to the sum of degrees of the residue fields of the points in $C_{1} \cap C_{2}$ over $k,$ and this is really what we should mean when we say the "number of points with multiplicity". Sorting this out lets us formulate our statement more rigorously as follows.

Restating Bézout's theorem. Let $C_{1}$ and $C_{2}$ be curves in $\mathbb{P}^{2}.$ If $\dim(C_{1} \cap C_{2}) = 0,$ then $$\deg(C_{1} \cap C_{2}) = \deg(C_{1}) \deg(C_{2}),$$ where the degrees are given as the closed subschemes of $\mathbb{P}^{2}.$

Remark. By saying that $C$ is a curve in $\mathbb{P}^{2},$ we mean that $C \hookrightarrow \mathbb{P}^{2}$ is an $1$-dimensional closed subscheme. Requiring that $\dim(C_{1} \cap C_{2}) = 0$ means that $C_{1}$ does not pass through none of the generic points of irreducible components of $C_{2},$ and vice versa. This is equivalent to saying that these two curves do not share any irreducible components or that they intersect at finitely many points. Under any of these equivalent hypotheses, it is necessarily true that $C_{1} \cap C_{2}$ consists of finitely many closed points. If $C_{1}$ is irreducible, then these conditions are equivalent to saying that $C_{1}$ is not contained in $C_{2}.$

Reduction. Say we have an $1$-dimensional irreducible closed subscheme $C \hookrightarrow \mathbb{P}^{2}.$ Considering the standard affine open cover of $\mathbb{P}^{2},$ we have a standard affine open subset $U \subset \mathbb{P}^{2}$ such that $U \simeq \mathbb{A}^{2}$ and $C \cap U \neq \emptyset.$ Note that $C \cap U$ has codimension $1$ in $U \simeq \mathbb{A}^{2}.$ Since every height $1$ prime in a UFD is principal, there is a single polynomial in two variables that defines $C \cap U.$ The vanishing locus of the homogenization of this polynomial is precisely $C.$

Using this argument, we see that $C_{1}$ and $C_{2}$ are hypersurfaces of $\mathbb{P}^{2}.$ What's great about hypersurfaces of $\mathbb{P}^{2}$ is that they don't have embedded points. Thus, our hypothesis says that $C_{1}$ and $C_{2}$ do not intersect at any associated points. (The language of associated points may be found Section 5.5 in Vakil's book.) This lets us see that the following is a more general statement of Bézout's theorem.

More general version of Bézout's theorem. Let $X \hookrightarrow \mathbb{P}^{n}$ be a projective scheme of dimension $\geq 1$ and $H$ a hypersurface of $\mathbb{P}^{n}$ that do not pass through any associated points of $X.$ Then $$\deg(H \cap X) = \deg(H)\deg(X),$$ where degrees are given by the relevant closed subscheme structures in $\mathbb{P}^{n}.$

Proof. Let $d := \deg(H)$ and $f \in \Gamma(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(d))$ the defining homogeneous polynomial of $H.$ Denoting $i : X \hookrightarrow \mathbb{P}^{n}$ for the given closed embedding, we see that $H \cap X$ is given by the vanishing locus of the section $i^{*}f \in \Gamma(X, \mathscr{O}_{X}(d)).$ The hypothesis that $H$ avoids all the associated points of $X$ implies that affine locally the section $i^{*}f$ is a non-zerodivisor. Hence, we see that $H \cap X$ defines an effective Cartier divisor of $X,$ which gives the following exact sequence: $$0 \rightarrow \mathscr{O}_{X}(-d) \rightarrow \mathscr{O}_{X} \rightarrow i_{*}\mathscr{O}_{H} \rightarrow 0.$$ Applying $(-) \otimes_{\mathscr{O}_{X}} \mathscr{O}_{X}(m),$ we get $$0 \rightarrow \mathscr{O}_{X}(m-d) \rightarrow \mathscr{O}_{X}(m) \rightarrow i_{*}\mathscr{O}_{H}(m) \rightarrow 0,$$ so we have $$p_{H}(m) = p_{X}(m) - p_{X}(m-d).$$ Computing the leading coefficient of the right-hand side gives the result. $\Box$