Relation between intersection and product of ideals

Question

Let $C$ be a smooth projective (irreducible) curve in $\mathbb{P}^n$ for some $n$. Denote by $I_C$ the ideal of $C$. Let $g \in I_C\backslash I_{C}^2$, an irreducible element. Is it true that for any positive integer $m$, $I_C^m \cap (g) \cong (g)I_C^{m-1}$?

Answer 1

Since you asked, here is a translation of the answer from MathOverflow so that it does not use the language of sheaves:

First, a short summary of this proof seems to be that "In this situation (a smooth integral subvariety of a smooth variety) we have a notion of $I$-valuation that is strictly additive; but the $I$-valuation of $g$ is 1, and we produce $f$ so that the valuation of $f/g$ is $m_0$, and the valuation of $f$ is strictly greater than $m_0+1$. This is a contradiction."

Careful proof (from MO): Let $m$ be the first value where the equality fails, so there exists $f \in I^m \cap (g)$ but not in $I^{m-1}(g)$.

Write $f = gh$. Note that it must be true that $h \notin I^{m-1}$, so let $m_0$ be the largest value so that $h \in I^{m_0}$ but not $I^{m_0+1}$. There are two relevant inequalities here: $$m_0 < m-1\ \ \text{ and }\ \ m_0+2 \leq m.$$

Now we consider the associated graded ring $$\mathrm{gr}_I R = R/I \oplus I/I^2 \oplus I^2/I^3 \oplus \cdots,$$ which by the theorem from Hartshorne is isomorphic to $$R/I[t_1, ..., t_{n-1}],$$ where the $t_i$'s are a regular sequence generating $I$. (The isomorphism sends $t_i$ to the $i$-th generator; this works because $\mathbb{P}^n$ and $C$ are smooth, in particular $\mathbb{P}^n$ is Cohen-Macaulay and $I_C$ is generated by a regular sequence.)

Look at the image of $h \in I^{m_0} / I^{m_0+1}$. This is nonzero by the choice of $h$. Likewise the image of $g \in I/I^2$ is nonzero by hypothesis.

In the graded ring, multiply them to obtain $h\cdot g \in I^{m_0+1} / I^{m_0+2}.$ Now note that $h\cdot g = f$ is actually in $I^m \subseteq I^{m_0+2}$ by the second inequality above, so the product is zero in the graded ring.

This is a contradiction because $R/I[t_1, \ldots, t_n]$ is an integral domain.

Answer 2

Edit: This proof has an error. :-( I will leave it up for now to see if it can be fixed (comments welcome).

We have an obvious injection $I^{m-1}_C \cdot(g) \hookrightarrow I^m_C \cap (g)$, which becomes an isomorphism locally at any point $p \notin C$, since then $I_C$ localizes to $(1)$.

Now, let $p \in C$ and let $R_p$ be the local ring. Since $\mathbb{P}^n$ and $C$ are smooth, $C$ is locally a complete intersection, hence $I_C$ is generated (in $R_p$) by $n-1$ elements. By Nakayama's Lemma, your hypothesis on $g$ says that we can extend $g$ to a minimal generating set [edit: this is not true!], say $g, f_1, \ldots, f_{n-2}$. Let $J = (f_1, \ldots, f_{n-2})$.

Claim: For all $m$, $J^m \cap (g) = J^m(g)$. We will prove this in a moment.

To see that this is sufficient, write

$$I^m \cap (g) = (J+(g))^m \cap (g) = (J^m + J^{m-1}(g) + \cdots + (g)^m) \cap (g). $$

Intersection distributes over sum, and all terms but the first are already contained in $(g)$, so this is

$$ = J^m \cap (g) + (J^{m-1}(g) + \cdots + (g)^m).$$

By our claim, $J^m \cap (g) = J^m \cdot (g)$, so we can factor (g) back out:

$$ = (J^m + J^{m-1} + J^{m-2}g + \cdots + g^{m-1})\cdot (g)$$

Now $J^m \subset J^{m-1}$, so this is exactly $I^{m-1} \cdot (g)$, as desired.

Proof of claim. First, note that $$J^m \cap (g) = J^m(g) \text{ iff } \mathrm{Tor}^1_R(R/J^m,R/g) = 0.$$

This is explained here. There's probably more than one way to prove that this Tor vanishes in this situation, but here is one proof: we have an isomorphism

$$k[[x_0, \ldots, x_{n-1}]] \to \widehat{R},$$

where $\widehat{R}$ is the completion of $R_p$, that sends $x_0 \mapsto g$, and $x_i \mapsto f_i$ for $i=1, \ldots, n-2$, and $x_{n-1} \mapsto t$, for some choice of uniformizer $t$ for $C$ at our point $p$. (Here I am using that $C$ is smooth. I'm not sure this is strictly necessary.)

Now since $R \to \widehat{R}$ is faithfully flat, $\mathrm{Tor}^1_R(R/J^m,R/g) = 0$ if and only if $\mathrm{Tor}^1_{\widehat{R}}(\widehat{R}/J^m\widehat{R},\widehat{R}/g\widehat{R}) = 0$, if and only if $J^m \widehat{R} \cap g\widehat{R} = J^mg \widehat{R}$.

But now this last equality is obvious, since $J\widehat{R}$ is the monomial ideal $(x_2, \ldots, x_{n-2})$ and $g\widehat{R} = (x_0)$. Both $J^m \widehat{R} \cap g\widehat{R}$ and $J^mg \widehat{R}$ are generated by $\{x_0 y : y$ is a monomial of degree $m$ in $x_1, \ldots, x_{n-2}\}$.