Example of ring R and $\phi : R \twoheadrightarrow S$ homomorphism such that $\phi(\text{rad}(R)) \neq \text{rad}(S)$

Question

I am looking for an example of a ring R and a surjective homomorphism $\phi : R \twoheadrightarrow S$ such that $\phi(\text{rad}(R)) \neq \text{rad}(S)$, where $\text{rad}(R)$ is the Jacobson radical (intersection of all maximal ideals).

In general, $\phi(\text{rad}(R)) \subset \text{rad}(S)$. In fact, it can be seen that for semi-local rings $\phi(\text{rad}(R)) = \text{rad}(S)$. (If $A$ is a semilocal ring and $f:A\rightarrow B$ is a surjective homomorphism, then $f(\operatorname{rad}A)=\operatorname{rad}B$).

So, any example will have infinitely many maximal ideals. The only (simple) example of a ring with infinitely many maximal ideals that comes to mind is a polynomial ring. For example, the maximal ideals of $\mathbb{C}[X]$ are $\{(x-a) \ | \ a \in \mathbb{C}\}$. Again, the only homomorphism that comes to mind is just the evaluation map $\phi(f(x)) = f(0)$. However, in this case $\phi(\text{rad}(\mathbb{C}[X])) = \phi(0) = 0 = \text{rad}(\mathbb{C})$.

Does anyone have an example?

Answer 1

You are right to choose polynomial rings, but there are lots of other surjective homomorphisms to choose from. For example, the projection $$\pi:\mathbb{C}[x]\to\mathbb{C}[x]/(x^2)$$ has the desired property.

Answer 2

Take the quotient map $k[X]\to k[X]/(X^2)$ for any field $k$. This is a surjective ring homomorphism, but we have $\text{rad}(k[X])=0$ and $\text{rad}(k[X]/(X^2))=(X)\neq 0$.

Answer 3

The simplest example is the unique ring homomorphism $$ \mathbb{Z}\to\mathbb{Z}/p^2\mathbb{Z} $$ where $p$ is a prime. The codomain is local.