If $A$ is a semilocal ring and $f:A\rightarrow B$ is a surjective homomorphism, then $f(\operatorname{rad}A)=\operatorname{rad}B$.
I know that if $A$ is a semilocal ring and if $I_{1},\dots, I_{n}$ are all of its maximal ideals, then $\operatorname{rad}A=I_{1}\cdots I_{n}=I_{1}\cap\cdots\cap I_{n}$. Let $y\in\operatorname{rad}B$, then there is $x\in A$ such that $f(x)=y$. I would like to show that $x\in\operatorname{rad}A$.
Any hint?
Since $B \cong A / \operatorname{ker}(f)$, the maximal ideals of $B$ are the maximal ideals of $A$ that contain $\operatorname{ker}(f)$.
Without loss of generality, suppose that $I_1,\dots,I_m$ contain $\ker(f)$ while $I_{m+1},\dots,I_n$ do not contain $\ker(f)$. Let $y$ be in the radical of $B$. Then $y$ is inside $\cap_{k=1}^m I_k = \prod_{k=1}^m I_k$. Then $y$ is a sum of products $y_1 \cdots y_m$, with $y_k \in f(I_k)$. There exist $x_k \in I_k$ such that $f(x_k) = y_k$. Hence $y_1 \cdots y_m = f(x_1) \cdots f(x_m) = f(x_1 \cdots x_m)$.
Now for $j>m$ we have $I_j + \ker(f) = A$ and so $u_j + \xi_j = 1$ for $u_j \in I_j$ and $\xi_j \in \ker(f)$. Can you continue from here?
