I'm reading Galois Theory part of "Abstract Algebra" by Dummit and Foote. I have a question of the proof of Corollary 23 (p.594), which says the existence of Galois closure.

Corollary 23. Let $E/F$ be any finite separable extension. Then $E$ is contained in an extension $K$ which is Galois over $F$ and is minimal in the sense that in a fixed algebraic closure of $K$ any other Galois extension of $F$ containing E contains K.

proof: There exists a Galois extension of $F$ containing $E$, for example the composite of the splitting fields of the minimal polynomials for a basis for $E$ over $F$. Then the intersection of all the Galois extensions of $F$ containing $E$ is the field $K$.

The previous theorem says that the intersection of two Galois extensions is Galois and I think this proof used that. But are there finitely many such Galois extensions? Is it obvious?

And why do we need the condition of separability here?

It's very hard for me to read Chapter 14 Galois theory of this book, unlike the previous chapters. Espeically Chapter 14.4 and Chapter 14.7.

I have read "A first course in Abstract algebra" by J.B.Fraleigh. I feel that the approaches of these two books are quite different. Fraleigh's book contains Isomorphism extension theorem and counts the number of embeddings. But Dummit and Foote's book does not contain such contents. Do you know some references that combine these two viewpoints?


1 Answer 1


You should know that an extension is Galois iff. it is normal and separable. If $E/F$ is not separable, then $L/F$ is not separable for any $L/E$, so no Galois closure could exist. The heart of this theorem is actually the statement that there is an extension of $E$ s.t. $L/F$ is Galois (and in fact also $L/E$ is Galois).

Ok, so what if $E/F$ is separable (and finite)? Then we can take the compositum $L$ they mention. By definition of normal, $L$ is normal over $F$. Because $E/F$ is separable, all these minimal polynomials are separable and we find $L$ is separable over $F$ as well; thus, $L/F$ is Galois. So at least one such extension exists.

But in fact $L$ is already the Galois closure! If $L'\le K$ contains $E$ and is Galois over $F$, it is also Galois over $E$ and necessarily must contain all the roots of those minimal polynomials of those basis elements for $E/F$, by normality. Thus, $L'$ contains every such splitting field and contains $L$.

  • $\begingroup$ Wonderful answer! Thank you. Did you use that compusitum of separable extensions is again separable in the second paragraph? $\endgroup$
    – MLe
    Jan 24 at 13:45
  • $\begingroup$ Oh, I guess it can be deduced from the definition and construction directly, too. $\endgroup$
    – MLe
    Jan 24 at 14:17
  • 1
    $\begingroup$ @MLe You're welcome; as I mentioned in the post, $L$ is by construction separable over $F$ since it is generated by roots of separable polynomials over $F$. I'm not too familiar with Galois theory in positive characteristic so I don't want to confidently claim composita of separable extensions are always separable in case it's not true. $\endgroup$
    – FShrike
    Jan 24 at 14:52

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