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Abstract Algebra ring homomorphism

$f:K\rightarrow R$ is a homomorphism from field to ring. To prove is that it is injective or zero.

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    $\begingroup$ Hint: the kernel of $f$ is an ideal of $K.$ $\endgroup$ – Andrew Nov 8 '12 at 16:01
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    $\begingroup$ I am having a hard time believing this is not a duplicate, but I can't find one :/ $\endgroup$ – rschwieb Nov 8 '12 at 16:24
  • $\begingroup$ @rschwieb - I agree. I'm having that deja vu feeling, but cannot locate where. $\endgroup$ – Chris Leary Nov 8 '12 at 17:26

$f(x)=f(y) \iff f(x)-f(y)=0 \overset{f\text{ hom.}}{\iff} f(x-y)=0$.

If there are $x\ne y$ with this property (ie., $f$ is not injective), then multply it by $f\left(\displaystyle\frac1{x-y}\right)$ and conclude that $f(s)=0$ for all $s\in K$, using the fact that $f$ is homomorphism, i.e. behaves friendly with the ring operations and that $s=1\cdot s$.

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A ring homomorphism $f: R \to R'$ is injective if and only if its kernel is $\{0\}$.

Also, the kernel of a ring homomorphism is an ideal.

Now let $f: K \to R$ be a ring homomorphism from a field into a ring. The only ideals in a field $K$ are $\{0\}$ and $K$ itself. Hence $f$ can either have kernel $\{0\}$ or $K$ itself. In the former case, $f$ is injective, in the latter case it is the zero map.

Hope this helps.

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    $\begingroup$ thanks, but actually the fact that zero set and K are the only ideals also has to be proven $\endgroup$ – nakajuice Nov 9 '12 at 12:27
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    $\begingroup$ @starovoitovs It sure does : ) $\endgroup$ – Rudy the Reindeer Nov 9 '12 at 12:35

Andrew's hint in the comments to your question is a good way to look at it, especially since it generalizes to simple rings (rings with exactly two ideals, both of them trivial).

So, you can prove the statement: If $f:R\rightarrow S$ is a homomorphism of rings, and $R$ is a simple ring, then $f$ is injective.

This of course covers fields as a special case. Moreover, it is best looked at with Andrew's hint: "Show $\ker(f)$ is an ideal of $R$."

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