Let's consider more general situation. Let $R$ be a ring and $I\subset R$ be an ideal of it. Fix a homomorphism $$ f: R\to R/I $$ where $f$ is not necessarily the quotient map. Suppose $f$ happens to be an isomorphism of rings, then can we conclude $I=(0)$?
The question is inspired by this post where the answer is negative. But examples there more or less involve some infinite direct products. I doubt that this kind of examples may not exist if we work with some ring like the polynomial algebra $R=\mathbb C[x_1,\dots,x_n]$.
Question: Maybe with some reasonable conditions (like noetherian), we can actually show $I=(0)$.