If every non-zero ring homomorphism $\phi: K\rightarrow S$ is injective, $K$ is field If every ring homomorphism $\phi: K\rightarrow S$ is injective, $K$ is field.
PS: $K$ is a commutative ring with unity, and my definition of homomorphism includes $\phi(1_{K})=1_{S}$.
My idea is to fix an arbitrary $k\in K$ and construct a homomorphism $\phi$ such that, by $\phi,$ I can find the inverse $k^{-1}$... But I don't got it.
 A: Let $I$ be a nonzero ideal in $K$, consider the quotient map $K \rightarrow K/I$. 
A: If $K$ isn't a field then there is an $s\in K\backslash\{0\}$ without an inverse. Then the non-zero ideal $sK$ does not contain $1$, and the quotient map $\phi: K \rightarrow K/sK$ is injective.
A: Our colleague Mindlack certainly has the right idea, +1; here we flesh out some details:
If $K$ is a commutative unital ring which is not a field, then it is possessed of a non-zero, non-invertible element; that is,
$\exists 0 \ne u \in K, \; \forall v \in K, uv \ne 1_K, \tag 1$
or, equivalently,
$\exists 0 \ne u \in K, \; \not \exists v \in K, uv = 1_K; \tag 2$
it follows that the principal ideal
$(u) = uK = Ku \ne K, \tag 3$
since
$1_K \notin (u); \tag 4$
furthermore
$0 \ne u = u1_K \in uK = (u) \Longrightarrow (u) \ne \{0\}, \tag 5$
and we conclude that $(u)$ is a non-trivial, proper ideal of $K$.
Now consider the cannonical projection homomorphism
$\pi:K \to K/(u), \; \pi(k) = k + (u); \tag 6$
since
$\pi(0) = \pi(u) = (u) \in K/(u), \tag 7$
$\pi$ is not injective, contrary to hypothesis; thus every $u \in K$ is invertible, and $K$ is a field.  $OE\Delta$.
