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 |
5
votes
Accepted
First Galois cohomology of Weil restriction of $\mathbb{G}_m$
There is a general argument that is slightly more elementary than what you wrote. By standard properties of Weil restrictions, we have $R(L) = \prod_{\sigma} \mathbb{G}_m(L)$, where the product is tak …
12
votes
Accepted
Brauer groups and field extensions
No: the conic $C:X^2+Y^2+1=0$ splits over the field $L=\mathbb{Q}(x)[y]/(x^2+y^2+1)$, since $(X,Y)=(x,y)$ is an $L$-point of $C$. However $L$ has no subfields algebraic over $\mathbb{Q}$ other than $\ …
7
votes
Accepted
Is co-restriction in Galois cohomology in fact the norm map via Kummer isomorphism?
The corestriction map on cohomology is indeed the norm in degree zero (see Tate's notes on Galois cohomology for example). By a dimension shifting argument, it then easily follows that the corestricti …