Show that $\Bbb Z_5[x]/\langle x^2+1\rangle$ is not an integral domain

Question

Show that $\Bbb Z_5[x]/\langle x^2+1\rangle$ is not an integral domain.

My approach:

Since $x^2+1 = (x-2)(x-3)$ is in $\Bbb Z_5[x]$ it is reducible in $\Bbb Z_5[x]\implies\langle x^2+1\rangle $ ideal won't be maximal ideal $\implies$ $\Bbb Z_5[x] / \langle x^2+1\rangle $ wouldnt be a Field and can i say that it wouldnt be an integral domain also as every field is an integral domain.

Answer 1

As pointed out by Vercassivelaunos, you have actually showed that $\frac{ \mathbb{Z}_5 [x]}{ \langle x^2 + 1 \rangle}$ is not an integral domain by identifying two zero divisors. Since $x^2 + 1 = (x-2)(x-3)$ is $\mathbb{Z}_5 [x]$, we have that

$$ \big( (x-2) + \langle x^2+1 \rangle \big) \big( (x-3) + \langle x^2+1 \rangle \big) = x^2+1 + \langle x^2+1 \rangle = \langle x^2+1 \rangle,$$

the $0$-element in $\frac{ \mathbb{Z}_5 [x]}{ \langle x^2 + 1 \rangle}$.

It is true as you point out that for a field $\mathbb{F}$, the quotient $\frac{\mathbb{F}}{ \langle p(x) \rangle}$ is a field if and only if $ \langle p(x) \rangle$ is a maximal ideal, that is exactly if $p(x)$ is irreducible over $\mathbb{F}$. However, to check if it is an integral domain, you may use the result that

$\frac{\mathbb{F}}{ \langle p(x) \rangle}$ is an integral domain if and only if $\langle p(x) \rangle$ is a prime ideal.

Answer 2

Question: "Show that $Z_5[x]/⟨x^2+1⟩$ is not an integral domain

Answer: Note that $(x+2)(x+3)=x^2+5x+6 \cong x^2+1\cong 0$. The ideals $I:=(x+2),J:=(x+3) \subseteq A:=\mathbb{F}_5[x]$ are coprime and maximal and $IJ=(x^2+1)$, hence

$$A/IJ \cong A/I \oplus A/J \cong \mathbb{F}_5 \oplus \mathbb{F}_5$$

which is a direct sum of fields. The zero divisors are elements on the form $(a,0), (0,a)$ with $a \neq 0: (a,0)(0,a)=(0,0)=0$.