How to know which elements are in the Quotient Ring $\mathbb{Z}[x]/\langle x, 2\rangle$?

I'm asked to prove that the quotient ring $\mathbb{Z}[x]/\langle x, 2\rangle$ is isomorphic to $\mathbb{Z}_2$.

The first thing that came to my mind was to apply the Theorem of Homomorphism of Rings, but to do so I need to construct a homomorphism $\phi: \mathbb{Z} \to \mathbb{Z}_2$, such that $\mathrm{Ker}\phi = \langle x, 2\rangle$ ($\langle x, 2\rangle$ is an ideal of $\mathbb{Z}[x]$). But

What elements are in $\mathbb{Z}[x]/\langle x, 2\rangle$ in the first place?

I know that $\langle x, 2\rangle$ of all the polynomials with even constant term, but I can't figure out how the elements of $\mathbb{Z}[x]/\langle x, 2\rangle$ look like.

Any help would be appreciated.

Solution 1:

To find your morphism, let $f: \mathbb{Z}[X] \to \mathbb{Z}$ with $f(P) = P(0)$. It is easy to check that $f$ is a surjective ring morphism. Moreover, let $\pi$ be the canonical projection $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$.

Thus $\pi\circ f $ is a surjective ring morphism. What is its kernel ? The first isomorphism theorem then concludes the proof.

Solution 2:

Well, the first homomorphism theorem states that $\mathrm{Im} f \cong A/\ker f$ for some morphism $f:A \to B$. In particular, this implies that $(A/J)/(I/J) \cong A/I$. For your isomorphism, note that $k[x]/(x,2) \cong \frac{\mathbb Z[x]/(2)}{(x,2)/(2)}=\mathbb Z_2[x]/(x) =\mathbb Z_2$ where the last isomorphism is induced by the relation $x \mapsto 0$ in the quotient ring.