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.