kernel maximal ideal If I have a homomorphism $\phi : R \rightarrow S$ between integral domains, how can I show that if the kernel is non-zero then it is a maximal ideal in R? 

The kernel of a ring homorphism $f\colon R\to S$ is maximal if and only if $f(R)$ is a simple ring (has no proper nontrivial two-sided ideals). In the case of a commutative ring with identity $1\neq 0$, a ring is simple if and only if it is a field. 
So a homomorphism $\phi\colon R\to S$ between integral domains (which necessarily have $1\neq 0$) which sends $1$ to $1$ has kernel equal to a maximal ideal if and only if the image is a field. This is not true in general (as Tobias's example shows), but may be the case. E.g., $f\colon\mathbb{Q}[x]\to\mathbb{Q}[x]$ given by $f(p(x)) = p(0)$ has image equal to $\mathbb{Q}$, so the kernel, $(x)$, is maximal.
Since the image will always be an integral domain (being a subring of an integral domain that contains $1$), the kernel is always a prime ideal; if the morphism is not injective, then the kernel is always a proper nontrivial prime ideal.
In some cases, this may imply maximality. For example, if the ring $R$ has Krull dimension $1$ (this includes PIDs that are not fields), or if $R$ is an Artin ring (has the descending chain condition), then the conclusion you want will follow.
