Ring homomorphism inquiry.

$R$ is a field. Let $\phi: R \to S$ be a homomorphism of rings. Show that $\phi$ is injective. Also provide an example of a non-injective homomorphism if $S$ is a field but $R$ is not a field.

Attempt: So here is what we know already. Since $R$ is a field, it is commutative, unitarian, and has units. $\forall x,y \in R$ $\phi(x+y) = \phi(x) + \phi(y)$ and $\phi(xy) = \phi(x) \cdot \phi(y)$. Now, I know that the kernel of $\phi$ is defined as $ker(\phi) = \{\alpha \in R | \phi(\alpha) = 0' \}$ where $0' \in S$. Thanks for the help.

• "and has units" - Every ring has units, such as the multiplicative identity $1$. However, what distinguishes a field is that all nonzero elements are units. – Karl Kronenfeld Apr 15 '13 at 5:04
• The claim is false for $S=0$. – Martin Brandenburg Apr 15 '13 at 13:36

Hint (for the second part): You are given a field $S$ and are to find a ring $R$ and a non-injective homomorphism from $R$ into $S$. A good example of such an $R$ isn't going to be pulled out of thin air, but it's going depend on $S$ somehow. Do you know of a way to construct new rings from old ones?