Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with galois-theory
Search options not deleted
user 9833
Galois theory, named after Évariste Galois, provides a connection between field theory and group theory. Using Galois theory, certain problems in field theory can be reduced to group theory, which is, in some sense, simpler and better understood.
1
vote
When is a bilinear form equivalent to a trace form?
I think the starting point should be the algebra of selfadjoint operators, i.e. the subalgebra $A\subset End_K(V)$ which consists of operators $f$ such that
$$\langle fx,y\rangle=\langle x, fy\rangle …
- 6,891
3
votes
Accepted
Extension field $\mathbb{C}(t,u)$ over $\mathbb{C}(t^n,u^n)$
What the above argument proves is that the extension is either trivial, or not Galois. My first guess was that it isn't Galois, but it was a stupid mistake! In fact, it is trivial.
Denote $z=u+it$. T …
- 6,891