Proving irreducibility via Eisenstein's criterion

Question

So I wrote down a question incorrectly, and it's certainly highlighted an issue for me. I was supposed to show $X^4+1$ is irreducible in $\mathbb{Q}[X]$. I wrote $P(X)=X^4+4$ instead.

Now I had seen before the trick of using "Set $f(X)=P(X+1)$, and show this is irreducible". So, I did so, obtained $X^4+4X^3+6X^2+4X+5$, which taking $p=2$ is apparently irreducible; $2|4, 2|6$ and $2^2\not|\ 5$

The issue is, I noticed $X^4+4$ IS reducible! It factors to $(x^2-2 x+2) (x^2+2 x+2)$.

I've seen other notes which mention using this idea, and I checked the criterion to be sure I wasn't being silly.. I really don't know what's going on! What apparent restrictions are there to using this method?

Answer 1

Eisenstein's Criterion says that if $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ is a polynomial with integer coefficients, and

(i) $p$ divides $a_i$ for $0\le i\le n-1$, and

(ii) $p$ does not divide $a_n$, and

(iii) $p^2$ does not divide $a_0$,

then $P(x)$ is irreducible over the rationals.

In your case, the prime $2$ does not divide $a_0$, which is $5$. So Condition (i) fails.

Answer 2

Hint $ $ Such errors are avoided if one remembers the motivation for Eisenstein's criterion, that the polynomial reduces $\,{\rm mod}\ p\,$ to a prime power $\,x^n\,$ (which must factor uniquely). Your example does not reduce to such a prime power because the constant coefficient does not vanish $\,{\rm mod}\ p.$