TITLE: Ireducible polynomial over $\mathbb Z_4$
QUESTION [2 upvotes]: How can you prove that $f(x)=X^2+1$ is irreducible over $\mathbb Z_4$, the quotient ring?
We know that $\mathbb Z_4$ admits divisors of $0$, as $2*2=0$, so any elemanary approach using $h\times g=f$ doesn't necessarily lead us to $\text {deg}(h) = \text {deg}(g) = 1$ (as would have normally happened in $\mathbb Z$ or $\mathbb  R$).

REPLY [1 votes]: Suppose we have a factorization $x^2+1=(x^2p + ax+b)(x^2q+cx+d)$ in $(\mathbb{Z}/4)[x]$, with $p,q\in (\mathbb{Z}/4)[X]$.
By unique factorization in $(\mathbb{Z}/2)[x]$, where $x^2+1=(x+1)^2$, this means that $a,b,c,d$ are odd, and $p=2f$, $q=2g$ are even.
Without loss of generality, assume $d\equiv 1\pmod{4}$.  Multiplying out, we can see that $bd \equiv 1\pmod{4}$, so $b\equiv d\equiv 1$.  But then the linear term is $a+c\equiv 0$.  Perhaps switching the order of $f$ and $g$, this leads to the factorization:
$$x^2+1 = (2x^2f + x + 1)(2x^2g-x+1)$$
Multiplying out leads us to the equation $2x^2f(-x+1)+2x^2g(x+1) \equiv 2x^2\pmod 4$, or $f(-x+1) + g(x+1) \equiv (f+g)(x+1)\equiv 1\pmod{2}$, which is impossible by unique factorization.
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