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.

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

  • $\begingroup$ Why $sK$ does not contain $1$? $\endgroup$ – Mateus Rocha Aug 15 '19 at 17:47
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    $\begingroup$ 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. $\endgroup$ – Jonathan Hole Aug 15 '19 at 17:54
  • $\begingroup$ Sorry for another question, but how can I finish the question? I didn't got it $\endgroup$ – Mateus Rocha Aug 15 '19 at 18:36
  • $\begingroup$ what do you mean? $\endgroup$ – 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

$(u) = uK = Ku \ne K, \tag 3$


$1_K \notin (u); \tag 4$


$0 \ne u = u1_K \in uK = (u) \Longrightarrow (u) \ne \{0\}, \tag 5$

and we conclude that $(u)$ is a non-trivial, proper ideal of $K$.

Now consider the cannonical projection homomorphism

$\pi:K \to K/(u), \; \pi(k) = k + (u); \tag 6$


$\pi(0) = \pi(u) = (u) \in K/(u), \tag 7$

$\pi$ is not injective, contrary to hypothesis; thus every $u \in K$ is invertible, and $K$ is a field. $OE\Delta$.

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    $\begingroup$ 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 $K/uK$ $\endgroup$ – Mateus Rocha Aug 15 '19 at 23:19
  • $\begingroup$ @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! $\endgroup$ – Robert Lewis Aug 16 '19 at 0:08
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    $\begingroup$ @MateusRocha: I didn't know $S$ was fixed and I'm not sure the result holds if it is. I could be wrong though . . . $\endgroup$ – Robert Lewis Aug 16 '19 at 0:10
  • $\begingroup$ Maybe it wasn't fixed and I am misinterpretating it $\endgroup$ – Mateus Rocha Aug 16 '19 at 0:26
  • $\begingroup$ @MateusRocha: maybe. Something to learn about, anyway! $\endgroup$ – Robert Lewis Aug 16 '19 at 0:30

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