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(I am not a native English speaker hence there may be some mistakes.)

Recently I was working on the problem III.9.6 in Hartshorne, Algebraic Geometry. It states:


Let $Y \subset \mathbb{P}^n$ be a nonsingular variety of dimension $\ge2$ over an algebraically closed field $k$. Suppose $\mathbb{P}^{n-1}$ is a hyperplane in $\mathbb{P}^{n}$ which does not contain Y, and such that the scheme $Y'=Y \cap \mathbb{P}^{n-1}$ is also nonsingular. Prove that Y is a complete intersection in $\mathbb{P}^{n}$ if and only if $Y'$ is a complete intersection in $\mathbb{P}^{n-1}$.


However, I was stuck on the sufficiency. I have tried to search for this problem but found no results. Hartshorne's book has a hint: use (9.12) applied to the affine cone of $Y$ and $Y'$. I have tried this hint, however (9.12) is an algebraic result concerning about Noetherian local ring. For affine cone of $Y'$, although its coordinate ring is normal since $Y'$ is complete intersection hence projectively normal, but it seems to have no connection with the local ring. Also, I really do not understand how this problem is related to section 9, Flat Families.

For the completeness, I post (9.12) here:


Let $A$ be a local noetherian domain, which is a localization of an algebra of finite type over a field $k$. Let $t \in A$, and assume:

(1). $tA$ has only one minimal associated prime ideal $\mathfrak{p}$,

(2). $t$ generates the maximal ideal of $A_{\mathfrak{p}}$,

(3). $A/\mathfrak{p}$ is normal.

Then $\mathfrak{p}=tA$ and $A$ is normal.


Finally, I really appreciate any people who can offer help to me. Thank you very much!

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We use the following corollary of Lemma of Hironaka:

Let $A$ be a noetherian domain (not necessarily local!), which is excellent (e.g. a localization of an algebra of finite type over a field $k$).

Let $t \in A$, and assume:

(1). $tA$ has only one minimal associated prime ideal $\mathfrak{p}$,

(2). $t$ generates the maximal ideal of $A_{\mathfrak{p}}$,

(3). $A/\mathfrak{p}$ is normal.

Then $\mathfrak{p}=tA$.

Proof.

Let $\mathfrak{q} \subset A$ be any prime ideal.

Case 1. $t \notin \mathfrak{q}.$

$\mathfrak{p} A_{\mathfrak{q}} = A_{\mathfrak{q}} = tA_{\mathfrak{q}}.$

Case 2. $t \in \mathfrak{q}.$

In this case, the local ring $A_{\mathfrak{q}}$ satisfies the conditions of Lemma of Hironaka.

Therefore, $\mathfrak{p} A_{\mathfrak{q}} = t A_{\mathfrak{q}}.$

Summing up, at every prime $\mathfrak{q}$, $\mathfrak{p} A_{\mathfrak{q}} = t A_{\mathfrak{q}}.$

Hence $\mathfrak{p} = t A. \blacksquare$

Using this, we can conclude that $\operatorname{I}_{\mathbb{P}^n} (Y') = \operatorname{I}(Y) + (x_n).$

Since $\operatorname{I}_{\mathbb{P}^n} (Y') = \operatorname{I}_{\mathbb{P}^{n-1}} (Y') + (x_n)$, we can write $$\operatorname{I}_{\mathbb{P}^n} (Y') = (f_1, \cdots , f_r, x_n).$$

The rest is easy!

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    $\begingroup$ Excuse me. I can't figure out how to prove that $I(Y)$ is generated by $r$ elements. What should these generators be? I'd appreciate if you can make it clear. $\endgroup$ – Jerry.Li Jul 12 '20 at 13:20
  • $\begingroup$ By the argument above, we conclude that $I(Y) + ( x_n )=(f_1 , \cdots , f_r , x_n ).$ In particular, each $f_i$ can be written as a sum of an element of $I(Y)$ and that of $(x_n )$: $f_i = g_i + x_n \cdot (\mbox{some polynomial})$. Then $I(Y) = (g_1 , \cdots , g_r )$. (an easy(?) exercise!) $\endgroup$ – Hiro Wat Jul 16 '20 at 10:01
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    $\begingroup$ I have already thought of this way, but I couldn't get the equality. The most I can prove, is that these two ideals have the same height so that $Y$ is just a component of the closed subscheme defined by $(g_1, \cdots, g_r)$. $\endgroup$ – Jerry.Li Jul 17 '20 at 8:42
  • $\begingroup$ Put $J=(g_1, \cdots g_r )$. Let $V_+ (J) = Z_1 \cup \cdots \cup Z_s $ be the irreducible decomposition. By unmixedness, $codim(Z_i , \mathbb{P}^n) = r$ for each $i$. Now $Y' = V_+ (J) \cap V_+ (x_n ) = (Z_1 \cap V_+ (x_n ) ) \cup \cdots \cup (Z_r \cap V_+ (x_n ) )$, with $Y$ irreducible, $(Z_i \cap V_+ (x_n ) )$ having codim $\leq r+1$. Thus for any $i$, $Z_i \supset Y'$. On the other hand, by Jacobian criterion, at any closed point $P$ of $Y'$, $rk J(g_1 , \cdots , g_r , x_n ) = r+1$, and thus $rk J(g_1 , \cdots , g_r ) = r$. This shows $P$ is a regular point of $V_+ (J)$. $\endgroup$ – Hiro Wat Jul 22 '20 at 6:22
  • $\begingroup$ In particular, $V_+ (J)$ has only one irreducible component at $P$. Combining $Z_i \supset Y'$, $s=1$, i.e. $V_+ (J)$ is irreducible.The unique minimal prime over $J$ is clearly $I(Y)$. Again by unmixedness, $J$ has no embedded primes, i.e. $k[x_0 , \cdots , x_n ]/J$ satisfies $S_1$. Since $k[x_0 , \cdots , x_n ]/J$ is regular at $P$, it is so at the generic point (localization preserves regularity). Thus it is $R_0$. Therefore $k[x_0 , \cdots , x_n ]/J$ is reduced, and thus a domain, i.e. $J$ is prime, equal to $I(Y)$. $\endgroup$ – Hiro Wat Jul 22 '20 at 6:22

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