Is $\mathbb{Z}[x]\over \langle x+3\rangle$ field?

Question

Is $\mathbb{Z}[x]\over \langle x+3\rangle $ field?

Can I say $x+3$ is irreducible over $\mathbb{Z}$, so it is field.

Answer 1

But clearly $\Bbb Z[x]/(x+3)\cong \Bbb Z$ and apparently is not a field.

The problem is that irreducible elements do not necessarily generate maximal ideals. This is the case in principal ideal domains, but this domain is not a PID.

Of course, $x+3$ is prime and irreducible, so $(x+3)$ is certainly a prime ideal (and that corroborates the fact that the quotient by it is an integral domain: $\Bbb Z$) but it is not a maximal ideal. There is, for example, the ideal $(x,3)$ which sits above $(x+3)$, and $(x,3)$ is a maximal ideal.

Answer 2

$\mathbb{Z}$ is not a field. If $F$ is a field, then $F[X] / \langle p(x) \rangle $ is a field when $p(x)$ is irreducible over $F$. But $\mathbb{Z}$ is "just" a ring.

Answer 3

Clearly $3\not\in(x+3)$. Now if $1\in \langle x+3\rangle+ \langle3\rangle$, then $1=f(x)(x+3)+g(x)3$. So, $1=f(0)3+g(0)3$, hence $3\mid 1$, a contradiction. Therefore $\langle x+3\rangle+\langle3\rangle\not=Z[x]$, and hence $\langle x+3\rangle$ is not maximal. Then $Z[x]/\langle x+3\rangle$ is not a field.