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

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

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

*Think about for which type of  ideals  $R/I$ can be integral domain. 
A: 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?
A: 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.
A: 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.
