Find the order of the quotient ring $\mathbb Z[x]/J$

Question

Let $J=\{f(x)\in \mathbb Z[x]:6\mid f(0)\}$. Show that $J$ is an ideal of $\mathbb Z[x]$, but not a prime ideal of $\mathbb Z[x]$. Also find the order of the quotient ring $\mathbb Z[x]/J$.

I know that any element of the quotient ring $\mathbb Z[x]/J$ is of the form $g(x)+J$, where $g(x)\in \mathbb Z[x].$ But from here how can we find the order of the ring $\mathbb Z[x]/J$ ?

Answer 1

Note that $J$ is the ideal $(6,X)$: if $f(X)$ has constant term divisible by $6$ then $f(X)=Xg(X)+6k\in (6,X)$, and conversely. Then the quotient is

$$\frac{\Bbb Z[X]}{(6,X)}\simeq \Bbb Z_6$$ The isomorphism is induced by the map that sends $f(X)=\sum a_i X^i$ to $\overline {f(0)}\in \Bbb Z_6$. Since $\Bbb Z_6$ is not a domain, the ideal is not prime.

Answer 2

Define the ring homomorphism $\varphi\colon \mathbb Z[x]\rightarrow \mathbb Z/6\mathbb Z$ by $f\mapsto f(0) + 6\mathbb Z$. Show that $\operatorname{Ker}\varphi = J$ and deduce $\mathbb Z[x]/J \cong \mathbb Z/6\mathbb Z$.