Valuations and separable extensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:33:24Z http://mathoverflow.net/feeds/question/56276 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56276/valuations-and-separable-extensions Valuations and separable extensions Laurent Moret-Bailly 2011-02-22T12:50:23Z 2011-02-22T17:22:22Z <p>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?</p> <p>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).</p> <p>The answer is yes for a discrete valuation, because then (if $F=k$) the completion of $R$ is isomorphic to $k[[t]]$.</p> http://mathoverflow.net/questions/56276/valuations-and-separable-extensions/56288#56288 Answer by vytas for Valuations and separable extensions vytas 2011-02-22T14:54:06Z 2011-02-22T17:22:22Z <p><strike>Assume that $F=k$ and let $p=char k$.</p> <p>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 $\alpha=\bar{\alpha}\in k$.</strike></p> http://mathoverflow.net/questions/56276/valuations-and-separable-extensions/56292#56292 Answer by Hagen for Valuations and separable extensions Hagen 2011-02-22T15:45:29Z 2011-02-22T15:45:29Z <p>What about this: we have to prove that $K$ and every purely inseparable extension $l/k$ are linearly disjoint over $k$.</p> <p>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 $R$.</p> <p>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.</p>