Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You should copy carefully what is in your book. – Mlazhinka Shung Gronzalez LeWy 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
up vote 2 down vote accepted

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.

share|cite|improve this answer

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 $S$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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