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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?

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  • $\begingroup$ Nope... not fixed... my question remains..... PLEASE take the time to post well-formed questions. $\endgroup$
    – David G. Stork
    Apr 23, 2020 at 0:10
  • $\begingroup$ Sorry, I meant P(X). $\endgroup$
    – Van-Sama
    Apr 23, 2020 at 0:16
  • $\begingroup$ Ummm.... this is getting tedious: $P(X)$ (in your comment) or $P(x)$??? PLEASE take the time to write a careful question. $\endgroup$
    – David G. Stork
    Apr 23, 2020 at 0:18
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    $\begingroup$ 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 $q(x)=1$. $\endgroup$
    – Arturo Magidin
    Apr 23, 2020 at 0:19

1 Answer 1

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$\!\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.

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