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I was recently reading these notes, where it is proved (a theorem of Kaplansky-Schilling) that a field that admits two distinct valuations with respect to which it is henselian is separably closed. A henselian field is the same thing as the data of a field $K$ and a discrete valuation ring $R \subset K$ which is henselian (in the sense of satisfying Hensel's lemma, or such that every finite $R$-algebra splits). Motivated by this, here is a generalization I am curious about:

Suppose $R_1, R_2 \subset K$ are distinct henselian, integrally closed subrings of $K$ such that $K$ is the quotient field of either. Does it follow that $K$ is separably closed? (I am not assuming $R_1, R_2$ are discrete valuation rings.)

I'd like to apply a similar argument as in the notes, where the approximation theorem is applied to the inclusion from $K$ into the product of the completions. So that would be like the inclusion of $R = R_1 \cap R_2$ into the product $\hat{R_1} \times \hat{R_2}$. If this is dense, and if the quotient field of $R$ is $K$, then I believe we're done ($K$ is separably closed) as before. Namely, given a separable irreducible polynomial in $R[X]$, we approximate it by something close in the $R_1$-topology to make it still irreducible over $R_1$, and approximate in the $R_2$-topology it by something over $R_2$ that obviously has a root. The combination of irreducibility (by the first approximation) and Hensel's lemma (by the second approximation) will show that the approximated polynomial must have degree one.

Here's what I think needs to be true for the argument to work:

  1. $R = R_1 \cap R_2$ should be dense in $\hat{R}_1 \times \hat{R_2}$ (or just in $R_1 \times R_2$).
  2. The quotient field of $R$ should be $K$.

I don't know if these are automatically true, or when they are true (or if my reasoning is correct).

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up vote 3 down vote accepted

In general the answer is no for a trivial reason: the localization of a henselian ring is henselian too. Thus if $R_1$ is a henselian local ring of dimension $>1$, then you can chose $R_2=(R_1)_p$ for a non-maximal prime $p$ to get a counter example.

However it is also known that if a field $K$ carries two distinct henselian valuations of rank $1$, i.e. having a value group embedable into the reals or equivalently having a $1$-dimensional valuation ring, then $K$ is separably closed: Schmidt, F. K. Mehrfach perfekte Körper. Math. Ann. 108 (1933), l-25.

So in your conjecture you could assume $R_1$, $R_2$ to be $1$-dimensional, integrally closed, henselian domains with the same fraction field. I do not know whether then the conjecture is true.

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Hm, I had never realized that the localization of a henselian ring is henselian as well. Thanks! – Akhil Mathew Sep 6 '11 at 13:03

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