I am in trouble with Exercise 8.4 in Hartshorne's Chapter II; I am doing part (a). It is about (global) complete intersection in $\textbf P^n$. For those without Hartshorne' book at hand, I describe the problem:
$Y$ is a closed subscheme of $\textbf P^n$ of codimension $r<n$.
I assume that $Y$, as a scheme, is the same as $H_1\cap\dots\cap H_r$, where $H_i=V_+(f_i)$ are hypersurfaces. I have to show that $Y$ is a complete intersection (that is, its homogeneous ideal $I(Y)$ can be generated by $r$ elements).
Well, I know that the ring $S=k[x_0,\dots,x_n]$ is Cohen-Macauley, so we have the unmixedness theorem in $S$. That is, whenever an ideal $J\subset S$ of codimension $q$ can be generated by $q$ elements, it is unmixed: it shares its own height with all of its associated primes – the elements of $\textrm{Ass}_S(S/J)$. Thus what I can say (?) is
\begin{equation}
r=\textrm{ht}\,(f_1,\dots,f_r)=\textrm{ht}\,I(Y)
\end{equation}
and hence $L:=(f_1,\dots,f_r)$ is unmixed. So if $\mathfrak p$ is an associated (necessarily minimal) prime for $L$ then $\textrm{ht}\,\mathfrak p=r$. If I assume, by contradiction, that $I(Y)\neq L$, then what happens? I cannot find a contradiction.
Thank you so much!
Best Answer
Assume $Y$ is a scheme theoretic intersection $\mathcal{I}_Y=\mathcal{I}_{H_1}+...+\mathcal{I}_{H_r}$. I will use your notation, so that $S$ is the polynomial ring, $I(Y)$ is the (unique) saturated homogeneous ideal associated to $Y$, $(f_i)$ is the homogeneous ideal associated to $H_i$, and $L:=(f_1,...,f_r)$. I will also use the notation $U_j$ for the standard affine open sets of $\mathbb{P}^{n}$.
Claim: The ideals $I(Y)$ and $L$ define the same closed subscheme of $\mathbb{P}^{n}$.
Proof: Let $\mathcal{L}$ be the ideal sheaf associated to the closed subscheme defined by $L$. Observe that $\mathcal{I}_Y(U_j)=\sum{\mathcal{I}_{H_i}(U_j)}=(f_1/{x_j^{deg(f_1)}},...,f_r/{x_j^{deg(f_r)}})=\mathcal{L}(U_j)$. Since $I(Y)$ and $L$ are associated to closed subschemes which have the same ideal sheaves, the claim is proved.
It follows that $I(Y)$ is the saturation of $L$, and so it suffices to prove that $L$ is saturated. One can show that Hartshorne's definition of "saturation" of $L$ in Ex II.5.10 is equivalent to the definition $(L:m^\infty)=\cup(L:m^d)$ over all positive integers $d$, where $m$ denotes the irrelevant ideal.
Assume to the contrary that $L$ is not saturated. In a Noetherian ring at least, we have $(L:m^\infty)=\cap(q:m^\infty)$ where $q$ ranges over the ideals in some given minimal primary decomposition of $L$. Since $L$ is not saturated, we have for at least one of these $q$ that $(q:m^\infty)$ properly contains $q$. With some work, it follows that $q$ is $m$-primary. But then $L$ has an embedded component, which contradicts the unmixedness theorem. It must be that $L$ is saturated, hence $L=I(Y)$.