# On cancelling $\mathfrak m$-primary ideal of regular local ring $(R,\mathfrak m)$

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.

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

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.
• where did you use $J$ is $\mathfrak m$-primary ? Commented Sep 30, 2018 at 17:31
• 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 $\mathfrak m^2 \ne J^2$ ? Commented Oct 4, 2018 at 13:27