$\phi: F[x] \rightarrow R$ is a ring homomorphism to an integral domain. Why must ker$(\phi)$ be a maximal ideal or (0)?

Question

Here is my working so far:

The kernel of any ring homomorphism is an ideal.

If $\phi$ is injective then ker$(\phi)$ is (0).

I don't know what to do if $\phi$ isn't injective. I know that the maximal ideals of $F[x]$ are generated by monic irreducible polynomials. I was thinking something along the lines of assuming that the kernel is generated by two polynomials, and arguing by contradiction.

Answer 1

1. Make use of a fundamental theorems of ring homomorphisms: $\mathrm{Im\,}\phi \cong F[x]/\ker \phi$.

2. Note that the image is an integral domain (being a subring of one).

3. Think about for which type of ideals $R/I$ can be integral domain.

Answer 2

Hint

Let $I=\ker(\varphi)$ then by the first isomorphism theorem we have $$F[x]/I\cong \operatorname{Im(\varphi)}$$

If we prove that $\operatorname{Im(\varphi)}$ is a field what we can conclude?

Answer 3

You are on the right track. Suppose that the kernel of $\phi$, the ideal I, is generated by reducible polynomial $f=gh$. Then the image of $f$ in R is zero. But because $\phi$ is homomorphism and neither $g$ nor $h$ are inside the kernel, you also get that $\phi(g)\phi(h)=0$. This means that you have two nonzero divisors of zero, $\phi(g)$ and $\phi(h)$ in the integral domain R. This gives a contradiction you were looking for.

Answer 4

Note that for any commutative ring $R$ with unity, $R/I$ is an integral domain $\iff$ $I$ is a prime ideal. The proof is relatively straightforward. Also, use the fact that any ideal $I \subseteq F[x]$ is prime $\iff$ $I = \{0\}$ or $I = \langle f(x) \rangle$ where $f(x)$ is irreducible.