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, $\phi(R)=R$.


3 Answers 3


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.

  • $\begingroup$ 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 $\endgroup$
    – RiaD
    Jan 30, 2021 at 0:25
  • 3
    $\begingroup$ However, it is true if one replaces 'injective' by 'surjective' and requires the ring to be Noetherian. $\endgroup$ Jan 30, 2021 at 8:45
  • $\begingroup$ @Marktmeister Good point! $\endgroup$
    – leoli1
    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).

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

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