Why is this not a field?

Question

I am wondering why $\mathbb{Z}[x]/\langle x-1\rangle$ is not a field.

I understand why this wouldn't be a field in one respect:

We know $\mathbb{Z}[x]/\langle x-1\rangle$$\simeq \mathbb{Z}$ using the evaluation ring homomorphism at $1$ and the first isomorphism theorem. So since $\mathbb{Z}[x]/\langle x-1\rangle$ is isomorphic to a non-field, we get that $\mathbb{Z}[x]/\langle x-1\rangle$ is not a field.

But isn't $x-1$ irreducible over $\mathbb{Z}[x]$? Would this not then imply that $\mathbb{Z}[x]/\langle x-1\rangle$ is a field?

Answer 1

The theorem is that, ${\bf{F}}[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible, but the condition here is that ${\bf{F}}$ is a field.

Here ${\bf{Z}}$ is not a field, so the theorem is not correctly applied.

Answer 2

There is a related theorem you may be thinking of: if $\mathfrak{m}$ is a maximal idea in a commutative ring $R$ (with $1$), then $R/\mathfrak{m}$ is a field.

Proof: If $0\neq \overline{a}\in R/\mathfrak{m}$, then $\langle a,\mathfrak{m}\rangle$ is an ideal containing $\mathfrak{m}$. Since $\mathfrak{m}$ is maximal, this implies $\langle a,\mathfrak{m}\rangle = R$, so there is a $b\in R$ with $1 = ab+m$ for some $m\in \mathfrak{m}$. Then $\overline{a}\overline{b} = 1\in R/\mathfrak{m}$.$\square$

But note that in your case, $\langle x-1\rangle\subseteq \mathbb{Z}[x]$ is not maximal. For example. $\langle x-1, 2\rangle$ is a proper ideal which strictly contains $\langle x-1\rangle$.