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Let $L:K$ be a field extension, let $\mathrm{Gal}(L:K)$ be the Galois group of $L$ over $K$ (by which I mean the group of all automorphisms of $L$ that fix every element of $K$), and let $\mathcal{E}$ be the set of all field extensions of $K$ contained in $L$. Now define the function $\varGamma$ on $\mathcal{E}$ as $$ \varGamma(E)=\{\alpha \in \mathrm{Aut}(L) : \alpha(z) = z,\; \forall z \in E\} $$

My textbook 1, in the middle of a proof, begins a paragraph with "[s]ince $\varGamma$ is one-to-one, ...", and just runs with it.

I don't recall seeing a proof of this assertion anywhere in the text up to this point, and the assertion is not obvious to me.

Given the context, it is also not clear to me whether the author is claiming that $\varGamma$ (as defined above) is one-to-one (a) unconditionally, or (b) given the conditions of the theorem he is in the middle of proving (namely, that "$L$ is a separable normal extension of a field $K$, with finite degree $n$").

In either case, I'd like to see a proof of the injectivity of $\varGamma$.


1 John M. Howie, Fields and Galois theory, p. 121.

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    $\begingroup$ $\Gamma$ isn't a map into the Galois group, but into the set of subgroups of the Galois group. $\endgroup$ – mercio Mar 23 '18 at 14:36
  • $\begingroup$ @mercio: right you are; I'll fix my post. $\endgroup$ – kjo Mar 23 '18 at 14:40
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This is the fundamental theorem of the Galois theory. The extension has to be a Galois extension.

https://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory

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  • $\begingroup$ OK, I see. D'oh. Thanks! $\endgroup$ – kjo Mar 23 '18 at 14:25

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