# Is every injective ring homomorphism an automorphism?

It seems a very interesting, innocuous and dubious result to me. Is the following true?

Consider the injective ring homomorphism $$\phi : R \rightarrow R$$. Is it true that $$\phi$$ is an isomorphism? I am taking ring homomorphism to mean $$f(ab)=f(a)f(b)$$ and $$f(a+b) = f(a)+f(b)$$ with $$f(1)=1$$.

My attempt: $$R \cong \phi (R) \subset R$$. So, $$\phi(R)=R$$.

No, consider $$R=\Bbb Z[X]$$. Let $$\phi$$ be defined by $$X\mapsto X^2$$, i.e. $$\phi\left(\sum_{i=0}^n a_iX^i \right)=\sum_{i=0}^n a_iX^{2i}$$ Then $$\phi$$ is injective but its image does not contain $$X$$.
It is also not true if we replace 'injective' with 'surjective'. To see this, let $$R=\Bbb Z[X_1,X_2,X_3,\dots]$$ where $$\phi$$ is given by $$X_1\mapsto 0$$ and $$X_i\mapsto X_{i-1}$$ for $$i>1$$.
So rings can be isomorphic to both proper subrings and proper quotients.

• your notation is a bit confusing, at first I thought you mean each polynomial goes to it's square, i.e $\phi(p(x)) = p(x) \cdot p(x)$ while you mean unique $\phi$ that maps $x$ to $x^2$. Maybe you can reword it a bit
Jan 30, 2021 at 0:25
• However, it is true if one replaces 'injective' by 'surjective' and requires the ring to be Noetherian. Jan 30, 2021 at 8:45
• @Marktmeister Good point! Jan 30, 2021 at 9:05

Your reasoning $$R \cong \phi (R) \Rightarrow \phi(R)=R$$ suggests me that you are misinterpreting cardinality with sets. For e.g. $$tan^{}(\cdot) : (-\frac{\pi}{2}, \frac{\pi} {2} ) \rightarrow \mathbb{R}$$ is a bijective function but the sets are not equal.

Important: the following answer is assuming $$|R|<\infty$$, otherwise this isn't true in general.

Consider the ring homomorphism $$\varphi: R\to R$$. Then, if you want to prove it's an isomorphism, you just need to prove it's injective or it's surjective (you don't need to prove both).

Why is this true? Because, since $$\varphi$$ goes from $$R$$ to $$R$$ ($$R$$ goes to himself), the cardinality of the origin set and the destiny set are the same (because they are the same set). So:

• If $$\varphi$$ is injective, every different $$x\in R$$ gives a different $$\varphi (x)$$, in fact $$|R|$$ different images, but the cardinality of the destiny set is $$R$$ so it's also surjective, hence a bijection.

• If $$\varphi$$ is surjective, then every $$y\in R$$ has some preimage $$x$$, so in order for that to be true you need $$|R|$$ different preimages, the cardinality of the origin set, so then $$\varphi$$ is also injective, hence a bijection.

So, when facing an homomorphism that goes from one finite set to himself, you just need to prove one of the two conditions to prove it's a bijection (since the other one will be a direct consequence of the first).

• Are you assuming the ring has finite cardinality? Jan 29, 2021 at 15:58
• Not necessarily true when $R$ is infinite, see leoli's answer. Jan 29, 2021 at 15:59
• @JasonDeVito yes, I forgot to say it, i'll edit. Thanks! Jan 29, 2021 at 15:59