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Assume that $(A,m)$ is a Noetherian normal local domain, $K = Quot(A) \subset E, F$ Galois extensions of $K$. If $B=\overline{A}^{E}$, $C=\overline{A}^F$, and $D=\overline{A}^{EF}$ and we choose primes $m_B, m_C$, and $m_D$ (in the corresponding rings) over $m$, then is it true that the separable part of residue field of $m_D$ is generated by the separable parts of the residue fields of $m_B$ and $m_C$ over the residue field of $m$?

The phrase 'separable part' is intended to describe the maximal separable extension of the residue field of $m$ contained in the residue field of $q$ where $q$ is one of $m_B, m_C, $ or $m_D$.

Thanks.

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    $\begingroup$ If $A$ is not complete or, more generally, henselian the integral closures are not necessarily local anymore. But the formulation of your questions seem to imply that you would like $B$, $C$, and $D$ to be local? $\endgroup$ – Torsten Wedhorn Mar 15 '11 at 10:28
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    $\begingroup$ Thanks. I've edited it to be more clear that I am not assuming that $A$ has any special properties. $\endgroup$ – PJT Mar 15 '11 at 17:31
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Counterexample based on Hagen's remark: take $A=\mathbb{R}[[t]]$, $B=\mathbb{R}[[\sqrt{t}]]$, $C=\mathbb{R}[[\sqrt{-t}]]$. Both have residue field $\mathbb{R}$, but $D$ contains a square root of $-1$ (namely $\sqrt{-t}/\sqrt{t}$), so its residue field must be $\mathbb{C}$.

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No answer, just a simple starting point: forget about the rings and look at two finite extensions $l_1/k$ and $l_2/k$. Then the separable closure of $k$ in the compositum $l_1.l_2$ indeed is the compositum of the separable closures $l_1^s$ and $l_2^s$ of $k$ in $l_1$ and $l_2$: by definition the extension $l_1/l_1^s$ is purely inseparable, hence the extension $(l_1.l_2^s)/(l_1^s.l_2^s)$ is purely inseparable too. The same holds for $(l_2.l_1^s)/(l_1^s.l_2^s)$ hence for $(l_1.l_2)/(l_1^s.l_2^s)$.

So one has to search for counterexamples in the case $k(m_D )\neq k(m_B ).k(m_C )$. For example one could start with a case in which the ring compositum $B_{m_B}\cdot C_{m_C}$ is not normal.

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