# 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.

• You are trying to show that $K$ is a field, not $S$. So finding inverse of $s \in S$ will not help. – Anurag A Aug 15 '19 at 17:33
• @AnuragA fixed. Just typing mistake – Mateus Rocha Aug 15 '19 at 17:42

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.

• Why $sK$ does not contain $1$? – Mateus Rocha Aug 15 '19 at 17:47
• If it did we would have sk=1 for some k in K. But then ks=1 also so k is clearly an inverse to s, but we started with an s without an inverse. – Jonathan Hole Aug 15 '19 at 17:54
• Sorry for another question, but how can I finish the question? I didn't got it – Mateus Rocha Aug 15 '19 at 18:36
• what do you mean? – Jonathan Hole Aug 15 '19 at 19:13

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

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. $$OE\Delta$$.

• Ok, I understood almost everything, but the final part is confuse. Why "$\pi$ is not injective" contradicts my hypothesis? We are dealing with homomorphisms $\phi:R\rightarrow S$, where $S$ is some fixed ring, but $\pi$ maps to $K/uK$ – Mateus Rocha Aug 15 '19 at 23:19
• @MateusRocha: first, $\pi$ is not injective because it maps two distinct elements of $K$, $u$ and $0$, to the zero element of $K/(u)$. As for your second question, you assumed every homomorphism $\phi:K \to S$ is injective; my proof is by contradiction, so when I show $\pi$ is not injective (in the case $K$ is not a field), I go against your assumption. Is that Clear? Cheers! – Robert Lewis Aug 16 '19 at 0:08
• @MateusRocha: I didn't know $S$ was fixed and I'm not sure the result holds if it is. I could be wrong though . . . – Robert Lewis Aug 16 '19 at 0:10
• Maybe it wasn't fixed and I am misinterpretating it – Mateus Rocha Aug 16 '19 at 0:26
• @MateusRocha: maybe. Something to learn about, anyway! – Robert Lewis Aug 16 '19 at 0:30