# If a function is one to one, and its kernel is not zero, why can't it be a ring homomorphism?

My book gives a corollary stating this, but I'm trying to convince myself that it is true.

Corollary 17.6: A ring homomorphism is one-to-one if and only if its kernel is $\{0\}$. -A First Course in Abstract Algebra (2nd ed.) by Anderson and Feil

I figured it out while typing this.

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You should copy carefully what is in your book. –  ABC Nov 5 '13 at 2:26
From what you say, $f(0) \neq 0$, so it cannot be a homomorphism. –  Prahlad Vaidyanathan Nov 5 '13 at 2:29
I second ABC's comment. How do you define a kernel of a function that is not a homomorphism? –  Tunococ Nov 5 '13 at 2:50
@Tunococ The kernel can still be defined as the preimage of zero. It just won't be an ideal anymore. –  Arthur Nov 5 '13 at 2:57
@Arthur I see. It's just strange (to me) to see the term "kernel" instead of "preimage of $0$" in this context. –  Tunococ Nov 5 '13 at 3:10

Say we have a one-to-one ring homomorphism $f: R \rightarrow R'$. Then $f(0) = 0$, and as $f$ is one-to-one the kernel contains only zero.
Say we have a ring $R$ and a ring $S$ and a one-to-one function $f:R \to S$. If $f$ is a ring homomorphism, then $f$ is a bijection between $R$ and $f(R)$, and a bijective homomorphism is an isomorphism. Therefore the kernel of $f$ must be zero.
Conversely, if the kernel of $f: R \to S$ is zero, then $f$ is an isomorphism from $R$ to $f(R)$, and so must be a one-to-one map into the larger ring $S$.