1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why $sK$ does not contain $1$? $\endgroup$ – Mateus Rocha Aug 15 '19 at 17:47
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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. $OE\Delta$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.