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!