If every non-zero ring homomorphism $\phi: K\rightarrow S$ is injective, $K$ is field

Question

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.

Answer 1

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.

Answer 2

Let $I$ be a nonzero ideal in $K$, consider the quotient map $K \rightarrow K/I$.

Answer 3

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. $OE\Delta$.