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, $\phi(R)=R$.
Best Answer
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.