Prove that field $\mathbb{Z}[x]/(x,3)$ is isomorphic to field $\mathbb{Z}/3\mathbb{Z}$.

Question

I don't realize what to do. Could you give me some hints or point to solution?

I tried to construct surjective homomorphism with $Ker(\phi)=(x, 3)$ by intuition to use First Isomorphism Theorem for Rings, but couldn't find such.

Answer 1

Hint: You know such a homomorphism should send $x$ to $0$ (because $x$ is in the kernel), and $1$ to $1$ (because $1$ is not in the kernel, and there is no other way to make $\phi(1)^2=\phi(1)$). There aren't many choices left after that.

Answer 2

Well, you can use the third isomorphism theorem as follows: $$ \mathbb{Z}[x]/(x,3) \cong (\mathbb{Z}[x]/(x))/((x,3)/(x)) \cong \mathbb{Z}/(3) \cong \mathbb{Z}_3.$$ Here $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ and $(x,3)/(x) \cong (3)$.