If $i:R \rightarrow S$ is an injective, non-surjective ring homomorphism, then $R \simeq S$? > Let $R,S$ be commutative rings with unity, and $i:R \rightarrow S$ be a (unital) ring homomorphism. Suppose $i$ injective, but not surjective. Assume $R,S$ are finitely-generated $k$-algebras for some field $k$. Can there exist a ring isomorphism $f: R\rightarrow S$? (Without the condition that $R,S$ must be finitely-generated $k$-algebras, if $R=S=\oplus_{i \in \mathbb{N}} \mathbb{R}$ and $i$ shifts the coordinates rightwards by one place, I believe we can just take $f=\text{id}_R$. I'm hoping that the finitely-generated $k$-algebra condition is sufficient to preclude the existence of such an isomorphism $f$. If anyone knows a better sufficient/necessary condition, I'd appreciate it.)

How about $R=S=k[X]$, $k$ a field, and $\iota(f(X))=f(X^2)$?