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What is the meaning of "fix" in field theory?

Example: I found a definition of field automorphism,

A field automorphism fixes the smallest field containing $1$, which is $\Bbb Q$, the rational numbers, in the case of field characteristic zero.

The set of automorphisms of $F$ which fix a smaller field $F'$ forms a group, by composition, called the Galois group, written $\operatorname{Gal}(F/F')$. For example, take $F'=\Bbb Q$, the rational numbers, and ...

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It means that it acts as identity if restricted to that subfield. In other words it sends each elements of $\mathbb{Q}$ to himself. Hence it "fixes" $\mathbb{Q}$ –  b00n heT Jul 10 at 10:41
    
Look up "fixed point group action" on your favorite search engine. –  blue Jul 10 at 11:56
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It’s not a specially mathematical term. One speaks of the “fixed stars” (as opposed to the planets): they don’t move. Or one can say “I am fixed in my opinion” to indicate that the opinion hasn’t changed. –  Lubin Jul 10 at 13:41
    
Or "fixed ladder", for that matter. The ordinary dictionary definition of "fixed" is "fastened securely". –  Pseudonym Jul 11 at 0:34

3 Answers 3

up vote 4 down vote accepted

Let's get concrete for a moment. Consider the field automorphism:

$$f : \mathbb{C} \rightarrow \mathbb{C}\,\,\hbox{where}\,\,f(a + ib) = a - ib$$

This is commonly known as complex conjugation. (It's obvious that it's an invertible and that the domain and codomain are the same, and you can verify for yourself that it is a homomorphism if you want.)

If $z \in \mathbb{R}$, then $f(z) = z$. $\mathbb{R}$ is a subfield of $\mathbb{C}$, so we say that this subfield is "fixed" by the automorphism.

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I disagree with the other two answers. They tell you what the author meant, however this is not necessarily the meaning of "fixes" in mathematics.

The author is imprecise. If an automorphism $\sigma$ fixes a subfield $S$ it could mean two things.

  • It fixes the elements of the subfield, $\sigma(s)=s$ for all $s\in S$. That is, it fixes the subfield pointwise. So $\sigma$ induces the trivial automorphism of $S$.

  • It fixes the subfield, but moves the elements around within the subfield, $\sigma(S)=S$. That is, it fixes the subfield setwise. So $\sigma$ induces an automorphism of $S$, but not necessarily the trivial automorphism.

For example, in group theory an inner automorphism will fix all normal subgroups setwise but not necessarily pointwise. For an explicit example, think of the semidirect product of $H=\mathbb{Z}_2\times\mathbb{Z}_2$ with $K=\mathbb{Z}_2$ where the action of the non-trivial element of $\mathbb{Z}_2$ on $H$ swaps the two copies of $\mathbb{Z}_2$. This fixes $H$ setwise but not pointwise, although it fixes the "third" copy of $\mathbb{Z}_2$ in $H$ pointwise. (If you don't know about semidirect products, then forget about normal subgroups and just consider $\mathbb{Z}_2\times\mathbb{Z}_2$ with the same automorphism.)

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I don't think I've ever seen "fixed field" mean anything other than the first. –  Hurkyl Jul 10 at 23:38
    
Yeah, wouldn't the second case normally be stated as "$S$ is invariant under $\sigma$"? Do you know of a reference in which "fixed" is used in the second sense? –  Nate Eldredge Jul 11 at 0:00
    
Yes, the usual term for that is "invariant", not "fixed". –  Pseudonym Jul 11 at 0:43
    
@NateEldredge Doing a Google Scholar search for "fixed setwise" automorphism brings up lots of results. My point wasn't that "fixes" doesn't have a precise meaning when talking about fields (I used the word "necessarily" as a get out clause!), but that it doesn't have a precise meaning in mathematics. I have flicked though my mini desk library, and for example Artin's Galois theory and Fraleigh's A first course in abstract algebra both define what they mean by a fixed field. –  user1729 Jul 11 at 8:44
    
(The implication being that because they define "fixed" to have this specific meaning, rather than assuming it, they understand the word "fixed" is ambiguous. The purpose of my answer to point out this ambiguity.) –  user1729 Jul 11 at 8:56

Basically, if $K$ is a field, $f: K \to K$ is a field automorphism then $f$ fixes some subfield $F$ of $K$ iff

$$ f|_{F} = id_F $$

i.e.

$$ \forall x \in F, f(x) = x $$

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