TITLE: Ring epimorphism $f:R\rightarrow S$, $R$ has finitely many maximal ideals, then $f(J(R))=J(S)$.
QUESTION [5 upvotes]: Suppose $R$ and $S$ are commutative rings with unit, and $f:R\rightarrow S$ is an epimorphism. Prove that: $$f(J(R))\subseteq  J(S).$$ If $R$ has finitely many maximal ideals, then prove that:  $$f(J(R))=J(S).$$

I know how to show that $f(J(R))\subseteq  J(S)$ but for the inverse I don't know what should I do, so please help me.
Recall that $J(R)$  is Jacobson radical of $R$.

REPLY [4 votes]: Let $\mathfrak{m}_i$, i=1,...,n be the set of maximal ideals in $R$. As $f$ is an epimorphism, every maximal ideal in $S$ is of the form $f(\mathfrak{m_i})$ for a unique $i$. Also note that for an epimorphism every $f(\mathfrak{m})$ is either a maximal ideal or all of $S$:
Let $f(a)$ be in $S$, but not in $f(\mathfrak{m})$ (in particular $a \notin \mathfrak{m}$). Then it is $f(\mathfrak{m}) + f(a) = f(\mathfrak{m}+a) = S$
Note that two non-equal maximal ideals are always relatively prime. We get
$f(\bigcap_{i=1}^{n} \mathfrak{m}_i) = f(\prod_{i=1}^n \mathfrak{m}_i)= \prod_{i=1}^n f( \mathfrak{m}_i) = \bigcap_{i=1}^{n} f(\mathfrak{m}_i) $
(generally (in the infinite case) we cannot change from intersections to products, so we only get an inclusion)