How to show $\mathbb{Z}[x]/\left<2,x\right>$ is isomorphic to $\mathbb{Z}_2$

I'm having quite a bit of trouble figuring out why $\mathbb{Z}[x]/\left<2,x\right>$ is isomorphic to $\mathbb{Z}_2$. So far I have figured out there is an onto map $\zeta: \mathbb{Z}\rightarrow\mathbb{Z}[x]$ given by $\zeta(n) = n$ (as a polynomial with degree 0), and that there's another onto map $\phi: \mathbb{Z}[x]\rightarrow\mathbb{Z}[x]/\left<2,x\right>$ given by $\phi(n) = n + (x) + (2)$, so then $\mathbb{Z}/\ker\phi$ is isomorphic to $\mathbb{Z}[x]/\left<2,x\right>$ by the isomorphism theorem.

I believe $\ker\phi =\{n\in\mathbb{Z}:n = e \} = \{n\in\mathbb{Z}: n + (2) + (x) = 0 + (2) + (x)\} = (2) + (x)$, but I could be wrong about that. In any event, this is as far as I can go.

Does what I've worked out even help? If not, what's a better approach to take?

• The inclusion $\Bbb Z\to\Bbb Z[x]$ is injective, not onto. – blue Apr 23 '14 at 1:42

$\Bbb Z[x]/\langle x,2\rangle\cong (\Bbb Z[x]/\langle x\rangle)/(\langle x,2\rangle/\langle x \rangle)\cong \Bbb Z / \langle 2\rangle$
• @ManolisLyviakis "Z[x]/<x,2> is isomorphic to Z[x]/<x,2>" is a tautology! The first $\cong$ is exactly what you are quoting in your comment, and the second $\cong$ is just expressing that $\Bbb Z[x]/\langle x\rangle\cong \Bbb Z$ and $\langle x,2\rangle/\langle x\rangle\cong \langle 2\rangle$. – rschwieb Jun 13 '14 at 14:35
One can easily argue $\{0,1\}$ is a system of coset representatives.