Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Great! Thank you! –  Keenan Kidwell Jul 7 '12 at 15:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.