Show that as a polynomial in $\mathbb{Q}[x]$, $P(x)$ is irreducible.

Question

Let $P(x)$ be a polynomial in $\mathbb{Z}[x]$ of degree 5 such that $P(1)=3$ and $P=(x-1)^5\bmod 3$. Show that as a polynomial in $\mathbb{Q}[x]$, $P(x)$ is irreducible.

So here I find that $$P(x)=(x-1)^5+3=x^5-5x^4+10x^3-10x^2+5x+2.$$ How can I show this irreducible? Can I use Eisenstein's Criterion? How would I do that? For instance, if my prime is 5 then it divides 4 of the 6 terms but what about those other 2?

Nope... not fixed... my question remains..... PLEASE take the time to post well-formed questions. — David G. Stork, Apr 23 '20 at 00:10
Ummm.... this is getting tedious: $P(X)$ (in your comment) or $P(x)$??? PLEASE take the time to write a careful question. — David G. Stork, Apr 23 '20 at 00:18
It’s not $P(x)=(x-1)^5+3$; the fact that it is congruent to $(x-1)^5$ modulo $3$ tells you there is a polynomial $q(x)$ such that $P(x) = (x-1)^5 + 3q(x)$; you have no warrant to assert that $q(x)=1$. — Arturo Magidin, Apr 23 '20 at 00:19

Bill Dubuque · Answer 1 · 2020-04-23T00:59:08.930

1

$\!\bmod 3\!:\ p \equiv (x\!-\!1)^5\ $ so Eisenstein applies to $ f(x) := p(x\!+\!1) \equiv x^5\,$ by $\,3^2\nmid f(0)\! =\! p(1)\!=\! 3$

Remark $ $ This is the standard shifted Eisenstein criterion, e.g. see here for motivation.

edited Apr 23 '20 at 00:59

answered Apr 23 '20 at 00:49

Bill Dubuque

272,048

Show that as a polynomial in $\mathbb{Q}[x]$, $P(x)$ is irreducible.

1 Answers1