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
- 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
- $(f_{1}, \dots, f_{r})$ 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!
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!
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