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 fields
Search options not deleted
user 2757
Fields as algebraic objects. For vector and tensor fields, use eg. [dg.differential-geometry]. For physical fields, use eg. [mp.mathematical-physics] or [quantum-field-theory].
24
votes
Accepted
When f(x)-a and f(x)-b yield the same field extension
The answer is that over a number field $k$, an equivalence class can be finite, and in fact it is usually so for $f$ of moderately large degree. Consider $f(x):=x^7+x$ over $\mathbf{Q}$, for example. …
- 23.8k
11
votes
Accepted
Surjectivity of bilinear forms.
The answer to Question B is no, as I'll show below.
Let $U=V=\mathbf{Q}^3$ and $W=\mathbf{Q}^4$. Define
$$\beta((u_1,u_2,u_3),(v_1,v_2,v_3))=(u_1 v_1,u_2 v_2,u_3 v_3, (u_1+u_2+u_3)(v_1+v_2+v_3)).$$
…
- 23.8k
15
votes
Accepted
Factorization of an irreducible polynomial in the field extension it defines
Let us show that for the partition $2+1+1=4$, there is no such $f$.
If $f$ were inseparable, then over $K$ it would factor as a constant times $(x-\alpha)^5$. If $f$ were separable, its Galois group …
- 23.8k
4
votes
A question on function fields (extending my previous question)
The complete set of solutions consists of
$$(1,0), (0,1),$$
$$(a,x), (x,a), (1/a,-x/a), (-x/a,1/a), (1/x,-a/x), (-a/x,1/x),$$
$$(b,y), (y,b), (1/b,-y/b), (-y/b,1/b), (1/y,-b/y), (-b/y,1/y).$$
Let $C$ …
- 23.8k