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Let $R$ be a valuation ring containing a field $k$, with residue field $F$ and quotient field $K$. Assume $F/k$ is separable. Is $K/k$ separable?

I have convinced myself that (for a positive answer) it is enough to treat the case of a rank one valuation ("height one" in Bourbaki's terminology), with $F=k$ and $K/k$ finitely generated. (Instaed of the latter condition, we may assume $K$ complete).

The answer is yes for a discrete valuation, because then (if $F=k$) the completion of $R$ is isomorphic to $k[[t]]$.

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up vote 4 down vote accepted

What about this: we have to prove that $K$ and every purely inseparable extension $l/k$ are linearly disjoint over $k$.

Let $x_1,\ldots ,x_r\in l$ be $k$-linearly independent elements and assume $0=a_1x_1+\ldots +a_rx_r$ for some elements $a_i\in K$. We can divide by the coefficient $a_j$ with the least value and thus assume $a_1,\ldots ,a_r\in R$ with at least one coefficient being a unit of $R$.

There is a unique valuation ring $S$ of $K.l$ dominating $R$. Taking residues modulo the maximal ideal of $S$ yields a linear combination $0=\overline{a_1}x_1+\ldots +\overline{a_r}x_r$ in the field $F.l$. The separability of $F/k$ yields $\overline{a_i}=0$ for all $i$, a contradiction.

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Beautifully simple. Thanks! – Laurent Moret-Bailly Feb 23 '11 at 15:15

Assume that $F=k$ and let $p=char k$.

If $K/k$ is inseparable then there exists $\alpha\in K$ such that min. poly of $\alpha$ over $k$ is equal $g(X^{p^n})$ with $g$ irreducible separable. Since $R$ is a valuation ring, either $\alpha$ or $\alpha^{-1}$ is in $R$. We may assume that $\alpha\in R$ by symmetry. Let $\bar{\alpha}$ be the image of $\alpha$ in $k$. Then $\bar{\alpha}^{p^n}$ is a root of $g$ in $k$, hence $g(X)=X-\bar{\alpha}^{p^n}$, which implies $(\alpha-\bar{\alpha})^{p^n}=0$ in $R$. Since $R$ is a domain $\alpha=\bar{\alpha}\in k$.

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There are inseparable extensions containing no inseparable algebraic elements, e.g. the extension $E/k(x,y)$ where $E$ is the fraction field of $k(x,y)[t,u]/(t^p-xu^p-y)$. – Laurent Moret-Bailly Feb 22 '11 at 15:19
Thanks, for pointing it out. – vytas Feb 22 '11 at 17:26
I'm sorry, I don't quite follow. I thought both "$L/K$ is separable" and "$L/K$ is inseparable" implicitly assume that the field extension is algebraic? Otherwise how can we talk about whether the statement "for any $a\in L$, the minimal polynomial of $a$ over $K$ is separable" is true? – Zev Chonoles Feb 22 '11 at 18:33
The notion of 'separable extension' has been generalized to transcendental extensions: An arbitrary extension $F/K$ is called separable if $F$ is linearly disjoint from the maximal purely inseparable extension of $K$, over $K$. – Jizhan Hong Feb 22 '11 at 20:33
@Zev, have a look at chapter II.15 of Zariski-Samuel Commutative Algebra. – vytas Feb 23 '11 at 9:00

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