Let $k_s|K|k$ be a tower of field extensions and $K|k$ be finite and separable ($k_s$ is a separable closure of $k$). There exists $\alpha\in K$ such that $K=k[\alpha]$. Then the splitting field $L$ of the minimal polynomial of $\alpha$ contains $K$ and $L|k$ is a finite Galois extension.

Now I think that the following should be true:

If $\sigma_1,\sigma_2\in\mathrm{Gal}(k_s|k)$ and $\sigma_1|_K=\sigma_2|_K$ then $\sigma_1|_L=\sigma_2|_L$.

  • $\begingroup$ peter a g has given a good counterexample, but it might useful to notice that your question amounts to the following: If an automorphism fixes one root of a polynomial (in this case, the root $\alpha$ of the minimal polynomial) then must it fix all the roots? In this form, it's pretty easy to guess that the answer is no, and that there might be a counterexample with just two other roots, which are interchanged by an automorphism fixing $\alpha$. That's what peter a r built explicitly. $\endgroup$ – Andreas Blass Aug 16 '15 at 1:27

No - that's false in general: if $G ={\rm Gal} (L/k)$, and $H$ the subgroup that fixes $K$, it is not always true that $H=1$, which is equivalent to what you are saying (because, for $\sigma_1$ and $\sigma_2 \in G$, $\sigma_1|_K = \sigma_2|_K$ is equivalent to $\sigma_1 = \sigma_2 h$, for some $h\in H$).

A concrete example: $k=\mathbb Q$, $K = k ( \alpha)$, where $\alpha^3 =2$. Then there are three roots, $\alpha_1 = \alpha, \alpha_2, \alpha_3$ (of the polynomial $f(x) = x^3 -2$, and $L$ the splitting field of $f$ over $k$). Identify $G$ with the symmetric group on the symbols 1,2, 3: the subgroup $H$ (of order 2) generated by the 2-cycle $(23)$ is the subgroup of $G$ that fixes $K$.

As the Galois group ${\rm Gal} (k_s/k)$ is the inverse limit of the Galois groups of the finite (Galois) extensions, your guess is not correct!

  • $\begingroup$ Thx. Is this a subbasis of the abs. Galois group: $\{\sigma\in\mathrm{Gal}(k_s|k) \;|\; \sigma|_L=\sigma'\}$ for $\sigma'\in\mathrm{Gal}(L|k)$, $L|k$ finite Galois extension. $\endgroup$ – guest Aug 16 '15 at 2:30
  • $\begingroup$ I was trying to understand why the set $\{\sigma\in G\;|\;\sigma\circ\phi = \phi\}$ is open for $\phi\in\mathrm{Hom}_k(L,k_s)$, $L|k$ finite separable extension. $\endgroup$ – guest Aug 16 '15 at 2:37
  • $\begingroup$ I think you are asking about lemma 2.2 here: math.ucdavis.edu/~osserman/classes/250C/notes/infinite.pdf $\endgroup$ – peter a g Aug 16 '15 at 2:47
  • $\begingroup$ So for $\phi=i_L$ the group would simplify to $\{\sigma\in G\;|\;\sigma|_L=1_L\}$. It would be the preimage of the group $\{\sigma\in\mathrm{Gal}(K|k)\;|\; \sigma|_L=1_L\}$ where $K$ is the splitting field of $L$, right? $\endgroup$ – guest Aug 16 '15 at 3:55
  • $\begingroup$ To let you know, I got it now. Thanks a lot, mate. I tried to force it to be an element of the subbasis for whatever reason. I think the name Galois closure made me think it was true. Stupid me. $\endgroup$ – guest Aug 16 '15 at 4:34

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