# Why is a strict $p$-ring whose residue ring is a field necessarily local?

Let $A$ be a strict $p$-ring. Recall that this means $A$ is $p$-adically separated and complete, $p:A\rightarrow A$ is injective, and $A/pA$ is a perfect $\mathbf{F}_p$-algebra. If $A/pA$ is a field, then $A$ is known to be a discrete valuation ring. The only way I have seen this proved is to use the actual construction of the $p$-Witt ring $W(A/pA)$, and I'm wondering if there is a way to prove this using only the definition of a strict $p$-ring.

In general, a ring $B$ is a discrete valuation ring if and only if it is local, max-adically separated, and its maximal ideal is principal and non-nilpotent (this is basically Proposition 2 in Serre's Local Fields, although in place of the separated hypothesis, he has a Noetherian hypothesis). If $A$ is a strict $p$-ring with $A/pA$ a field, then $pA$ is maximal, and non-nilpotent by definition. Also, $A$ is $p$-adically separated by definition. But I can't think of a simple way to prove that $A$ is local, i.e., that $pA$ is the unique maximal ideal of $A$. I suppose one could work out the universal formulas which define the ring operations on such a ring and use them to characterize the non-units as precisely the elements of $pA$, but I was hoping there might be a way to avoid this.

The reason I expect such a thing might be possible is that Serre claims that "a complete discrete valuation ring, absolutely unramified, with perfect residue field $k$, is nothing other than a strict $p$-ring with residue ring $k$." He doesn't give any argument for why a strict $p$-ring whose residue ring is a field is local. This leads me to believe I'm missing a potentially obvious argument.

I apologize if this question doesn't seemed well-defined. The ring operations on a strict $p$-ring can be derived from the definitions (starting by proving the existence of a multiplicative system of representatives, then getting series expansions, etc.), so, proving locality from this, strictly speaking, is just using the definitions, but I would say there's considerable work involved (or at least considerable messiness). Basically I'm trying to avoid writing down the polynomials that give the ring structure.

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## 1 Answer

If $x\not\in pA$, then there is some $y$ such that $xy\equiv1$(mod $p$). Then $x$ is invertible mod $p^n$ for any $n$ since $(xy)^{p^n}\equiv 1$(mod $p^n$). Thus $x$ is invertible.

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Great! Thank you! –  Keenan Kidwell Jul 7 '12 at 15:13