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 polynomials
Search options not deleted
user 4008
Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.
15
votes
Accepted
Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent?
This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"
- 22.6k
7
votes
solutions to equation mod a prime
If you rewrite the equation as $b^2=-a^2/(a^2+1)$ as Will did you can continue as follows: The equation can be rewritten as
$$\left(\frac{b}{a}\right)^2 = -(a^{-2}+1)$$
(excepting $a=b=0$). Putting $x …
- 22.6k
3
votes
Accepted
The irreducibility of an algebraic variety
Then in the quotient ring, the images of the variables are invertible so we may replace the polynomial ring with the ring of Laurent polynomials whose spectrum is an algebraic torus. …
- 22.6k