# Some sufficient condition for a Noetherian local ring to be a DVR

Let $$R$$ be a Noetherian local ring with maximal ideal $$\mathfrak{m}$$ and $$k = R/\mathfrak{m}$$ be its residue field.

I would like to show that $$\dim_k{\mathfrak{m}/\mathfrak{m}^2}=1$$ implies that $$R$$ is a discrete valuation ring.

The proof I am currently going through tries to show that $$\bigcap_{n>0} \mathfrak{m}^n = (0)$$ in order to define the discrete valuation. But I do not quite understand why this intersection is exactly $$(0)$$. I suppose one has to use the fact that $$\mathfrak{m}/\mathfrak{m}^2$$ has dimension $$1$$, but I do not know how exactly this can be used.

It is a consequence of Krull's intersection theorem: $$\;\displaystyle\bigcap_n \mathfrak m^n$$ is the set of $$x\in R$$ for which there exists $$m\in\mathfrak m$$ such that $$(1-m)x=0$$. As $$R$$ is local and $$\mathfrak m$$ is its maximal ideal, $$1-x$$ is a unit in $$R$$, so $$x=0$$.
In this case there is an easier direct argument without using Krull's intersection theorem. First, note that by Nakayama's lemma, if $$p\in\mathfrak{m}\setminus\mathfrak{m}^2$$, then $$p$$ generates $$\mathfrak{m}$$ (since $$\mathfrak{m}/\mathfrak{m}^2$$ is $$1$$-dimensional and hence generated by any nonzero element). It follows that $$\mathfrak{m}^n$$ is generated by $$p^n$$, and $$\bigcap_{n>0} \mathfrak{m}^n$$ is the set of elements which are divisible by $$p^n$$ for all $$n$$.
Now suppose $$x\in \bigcap_{n>0} \mathfrak{m}^n$$ is nonzero and let $$x_n$$ be such that $$p^nx_n=x$$ for each $$n$$. I claim that for each $$n$$, $$x_n\not\in(x_{n-1},\dots,x_0)$$, so that the ideals $$(x_0)\subset(x_0,x_1)\subset(x_0,x_1,x_2)\subset\dots$$ are an infinite ascending chain and contradict the assumption that $$R$$ is Noetherian. To prove this, suppose $$x_n\in (x_{n-1},\dots,x_0)$$ and write $$x_n=\sum_{i=0}^{n-1} a_ix_i$$ for $$a_i\in R$$. Multiplying by $$p^n$$, we get $$x=\sum a_ip^{n-i}x$$, so $$(1-\sum a_ip^{n-i})x=0$$. But $$\sum a_ip^{n-i}$$ is divisible by $$p$$ (since $$i$$ stops at $$n-1$$) and hence in $$\mathfrak{m}$$ so $$1-\sum a_ip^{n-i}$$ is a unit. This contradicts the assumption that $$x$$ is nonzero.
(If you additionally assume $$R$$ is a domain, which seems to be intended in your context (otherwise the conclusion that $$R$$ is a DVR is not correct!), there is an even quicker way to formulate this proof. Namely, in that case if $$x$$ is infinitely divisible by $$p$$ there is a unique way to divide it by $$p$$, which means $$x/p$$ is also infinitely divisible by $$p$$. That is, for any $$x\in\bigcap_{n>0} \mathfrak{m}^n$$, $$x/p\in \bigcap_{n>0} \mathfrak{m}^n$$ as well, which means $$p\cdot \bigcap_{n>0} \mathfrak{m}^n=\bigcap_{n>0} \mathfrak{m}^n$$. By Nakayama, this implies $$\bigcap_{n>0} \mathfrak{m}^n=0$$.)