In these notes by Richard Vale, he approaches the notion of a radical of a module axiomatically, by saying the following. A radical is an assignment, to each R-module $M$, of a submodule $\tau(M) \triangleleft M$, such that
- $f(\tau(M)) \subset \tau(N)$ for all $f: M \to N$.
- For all $M$, $\tau(M/\tau(M)) = 0$.
He then goes on to prove some nice properties one can derive from these assumptions.
My question is: in the particular case of the Jacobson radical, I thought $f(J(R)) \subseteq J(S)$ was not true in general, and only becomes true when $f$ is a surjective ring map. So what gives?
If $f:R\to S$ is a morphism of rings, and you view it as a morphism if R-modules, then it is true that it maps the radical of R into the radical of S, but of S viewed as an R-module.
The thing is, the radical of S as an R-module is not the same thing as the radical of S as an S-module.