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Let $(R,\mathfrak m)$ be a regular local ring (https://en.wikipedia.org/wiki/Regular_local_ring) . Let $J$ be an $\mathfrak m$-primary ideal such that $J^2=\mathfrak mJ$. Then is it true that $J=\mathfrak m$ ?

I know that regular local rings UFD. If $\mathfrak m$ is principal, then due to Noetehrian ness of $R$, $R$ actually becomes a PID and the claim easily follows. But I am unable to show the claim in general.

Please help.

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  • $\begingroup$ Seems to me you can do this with Nakayama applied to $R/J^2$. $\endgroup$ Commented Sep 30, 2018 at 0:23
  • $\begingroup$ @JohnBrevik: could you please elaborate how ...? $\endgroup$
    – user521337
    Commented Sep 30, 2018 at 1:41
  • $\begingroup$ I retract my previous comment. Sorry about that; shouldn't have answered before I had time to work it out in detail. $\endgroup$ Commented Sep 30, 2018 at 21:50

1 Answer 1

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By induction on the dimension $d$ of $R$. If $d=0$ is trivial, and for $d=1$ your argument of course works, and it is a case of the induction step that follows. We suppose $J\ne\{0\}$, hence not all powers of $\mathfrak m$ contain $J$ (Krull's theorem).

We first claim that $J\not\subset\mathfrak m^2$. The easy argument in a power series ring applies in general through the graded ring $G_{\mathfrak m}(R)=\oplus_r \mathfrak m^r/\mathfrak m^{r+1}$. Let $x_1,\dots,x_d$ be a regular system of parameters of $R$, so that $\mathfrak m=(x_1,\dots,x_d)$ and $k[t_1,\dots,t_d]\to G_{\mathfrak m}(R):t_i\mapsto x_i$ is an isomorphism (regularity, $k=R/\mathfrak m$). Now, let $\mathfrak m^r$ be the biggest power containing $J$ and pick $y\in J\setminus\mathfrak m^{r+1}$. Consider $\bar y\in\mathfrak m^r/\mathfrak m^{r+1}$ and $\bar x_1\in\mathfrak m/\mathfrak m^2$. In $k[t_1,\dots,t_d]$ these are homogeneous polynomials of degrees $r$ and $1$ respectively, hence the product $\bar x_1\bar y$ is a nonzero homogeneous polynomial of degree $r+1$. In other words, $x_1y\notin\mathfrak m^{r+2}$. On the other hand, $x_1y\in\mathfrak mJ=J^2\subset\mathfrak m^{2r}$, so that $2r\le r+1$ and $r=1$.

Once the claim is proven, pick $y\in J\setminus\mathfrak m^2$. Then $y\neq0$ mod $\mathfrak m^2$, and there is a regular system of parameters $z_1,\dots,z_d$ with $z_d=y$. We can go now to $R/(y)$, which is a regular local ring of dimension $d-1$. There we apply induction to the quotient ideals $\overline{\mathfrak m}$, $\bar J$ to get they coincide, that is, $\mathfrak m=J+(y)$, and since $y\in J$ we are done.

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  • $\begingroup$ where did you use $J$ is $\mathfrak m$-primary ? $\endgroup$
    – user521337
    Commented Sep 30, 2018 at 17:31
  • $\begingroup$ I think it is not needed. $\endgroup$
    – Jesus RS
    Commented Sep 30, 2018 at 18:26
  • $\begingroup$ I see... I'll think of it! $\endgroup$
    – Jesus RS
    Commented Sep 30, 2018 at 21:53
  • $\begingroup$ I see you don't mark it as solves. There is anything missing? $\endgroup$
    – Jesus RS
    Commented Oct 3, 2018 at 14:29
  • $\begingroup$ Do you have an example of a Noetherian local ring $(R,\mathfrak m)$ with an ideal $J$ such that $\sqrt J=\mathfrak m, J^2=\mathfrak m J$, but $\mathfrak m^2 \ne J^2$ ? $\endgroup$
    – user521337
    Commented Oct 4, 2018 at 13:27

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