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 |
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.
32
votes
Accepted
Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto i...
No, this can't happen. One way to prove this is via Hilbert irreducibility: The polynomial $p(x) - t$ is irreducible over $\mathbb Q[x,t]$, so there are infinitely many specializations $t = c$ with $c …
8
votes
Accepted
Uniqueness of differences of roots of polynomials over finite field
Over $\mathbb F_p$, the polynomial $f(x) = x^p - x - 1$ is irreducible, and if $\alpha$ is a root, then the full set of roots is $R = \{\alpha + c : c \in \mathbb F_p\}$. In particular, the difference …
6
votes
Accepted
Is there a uniform family of polynomials $f_p(x) =x^2 + a(p)x + b(p)$ such that $f_p(x)\in \...
No, this can't happen: For any polynomials $a,b \in \mathbb Z[t]$, there are infinitely many primes $p$ such that $f_p(x) := x^2 + a(p)x + b(p)$ is reducible modulo $p$. …