I am having a confusion with terminology. If $L/K$ is not Galois, what is the meaning of "the Galois group of $L$ over $K$"? I have two guesses:

1) It is the field automorphisms of $L$ that fix $K$.

2) It is the Galois group of the Galois closure of $L$ over $K$.


  • 1
    $\begingroup$ There is none. You can speak of the automorphism group that fixes the base field, though. That is to say: it is (1). $\endgroup$ Aug 31, 2014 at 23:53
  • $\begingroup$ If you are seeing this somewhere, it would be useful if you gave an excerpt for context. $\endgroup$
    – KCd
    Sep 1, 2014 at 1:15
  • $\begingroup$ Ian Stewart, in the book Galois Theory (third edition), defines the Galois group as in @user55600's item 1. I have seen it defined as in item 2 in some Sage programs. $\endgroup$
    – j0equ1nn
    Oct 22, 2014 at 5:36

2 Answers 2


Indeed, (2) seems to be standard convention. Wikipedia defines it that way, and I have run across this and become confused before myself.


As an example take the extension $F=\mathbb{Q}(\theta)$ where $\theta=\sqrt[3]{2}$, it is a field extension of degree 3 with basis $\{1, \theta, \theta^2\}$. But it is not the splitting field of $x^3-2$ (see cubic example), it does not contain all the roots of $x^3-2$. Every element of the Galois group associated with this polynomial would map $F$ "outside" of F so can't be considered an automorphism of $F$. If $p$ is a polynomial having a square integer as discriminant then $F$ would be the splitting field of $p$, and the Galois group is the cyclic group of order $3$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .