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The following is a statement in Artin's Algebra (2nd edition p. 489):

Corollary 16.6.5
(a) Every finite extension $K/F$ is contained in a Galois extension.
(b) If $K/F$ is a Galois extension, and if $L$ is an intermediate field, then $K$ is also a Galois extension of $L$, and the Galois group $G(K/L)$ is a subgroup of the Galois group $G(K/F)$.
Proof. Theorem 16.6.4 allows us to replace the phrase "Galois extension" by "splitting field." Then the Corollary follows from Lemmas $\color{blue}{16.3.1}$ and $\color{red}{16.6.2}$.

And the following are the lemmas and theorems quoted in the proof:

Lemma 16.3.1
(a) If $F\subset L\subset K$ are fields, and if $K$ is a splitting field of a polynomial $f$ over $F$, then $K$ is also a splitting field of the same polynomial over $L$.

Lemma $\color{red}{16.6.2}$
(a) The Galois group $G$ of a finite field extension $K/F$ is a finite group whose order divides the degree $[K:F]$ of the extension.
(b) Let $H$ be a finite group of automorphisms of a field $K$. Then $K$ is a Galois extension of its fixed field $K^H$, and $H$ is the Galois group of $K/K^H$.

Theorem 16.6.4
Let $K/F$ be a finite extension and let $G$ be its Galois group. The following kare equivalent:
(a) $K/F$ is a Galois extension, i.e., $|G|=[K:F]$,
(b) The fixed field $K^G$ is equal to $F$,
(c) $K$ is a splitting field over $F$.

Here is my question:

Can anyone explain where does the author use Lemma 16.6.2 in the proof of Corollary 16.6.5.?

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  • $\begingroup$ To the person who downvoted this question: would you please let me know how should I improve my question? $\endgroup$ – Jack Apr 28 '14 at 23:36
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    $\begingroup$ I think he uses Lemma 16.6.2 in part (b) of the corollary to get that $Gal(K/L)$ is a subgroup of $Gal(K/H)$, by noting that $L$ is the fixed field of $Gal(K/L)$. $\endgroup$ – Siddharth Venkatesh Apr 29 '14 at 0:30
  • $\begingroup$ @SiddharthVenkatesh, I agree. It seems that we don't need that lemma at all since $Gal(K/L)$ is a group by definition and it is obviously a subset of $Gal(K/H)$. $\endgroup$ – Jack Apr 29 '14 at 23:53

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