Let $A, B$ be unitary rings and let $f$ be a surjective homomorphism of rings. Then $f(Jac(A)) \subseteq Jac(B)$

Question

Let $A, B$ be unitary rings, and let $J(A)$, $J(B)$ denote the Jacobson's radical of $A$ and $B$. Let $f$ be a surjective homomorphism of rings. Then:

$$f(J(A)) \subseteq J(B)$$

Attempt: It suffices to prove that the preimage of every maximal ideal of $B$ is a maximal ideal in $A$. Let $M \subseteq B$ be an ideal of $B$, and let $f^{⁻1}(M)$ denote its preimage. Suppose $J$ is an ideal of $A$ such that $f^{⁻1}(M) \subseteq J$. Since $f$ is a surjection, this implies $M \subseteq f(J)$. Since $M$ is maximal, we have $f(J)=M$ or $f(J)=B$. If $f(J)=M$, then $J \subseteq f^{⁻1}(M)$ and then $J=f^{-1}(M)$. If $f(J)=B$, then I intended to prove that $1_{A}$ is in $J$, since any surjective homomorphism maps $1_{A}$ to $1_{B}$.The problem is that might be that exist other elements of $A$ mapped to $1_{B}$, and then we can't conclude that $1_{A} \in J$.

My questions:

1.Is the initial problem true?

2.Is the preimage of a maximal ideal, through an epimorphism, also a maximal ideal?

3.Are there any ring homomorphisms $f:A\rightarrow B$ that maps $1_A$ and another element to $1_B$?

Answer 1

1. Yes, it is true, for the reason below:
2. Yes, this is true as well. By the correspondence theorem, the ideals of $ B $ correspond bijectively to the ideals of $ A $ containing $ \ker f $ in an inclusion respecting manner, so a maximal ideal of $ B $ pulls back to a maximal ideal of $ A $. Note that this is not true if we don't require the homomorphism to be surjective: if $ R $ is any commutative ring with a non-maximal prime ideal $ \mathfrak p $, then the preimage of the maximal ideal $ \mathfrak p R_{\mathfrak p} $ under the natural map $ R \to R_{\mathfrak p} $ is $ \mathfrak p $, which is not maximal.
3. Sure, for example, the quotient map $ \mathbf Z \to \mathbf Z/2\mathbf Z $ is one such homomorphism.

Answer 2

For an alternative approach, for a unital ring $R$, you can use the characterization that $Jac(R)$ is the set $\{x\in R:1+RxR\subset R^\times\}$.

Let $x\in Jac(A)$. To see $f(x)\in Jac(B)$, you need $1+bf(x)b'$ is a unit in $B$ for any $b,b'\in B$. Since $f$ is surjective, $b=f(a)$ and $b'=f(a')$ for some $a,a'\in A$. Then $$ 1+bf(x)b'=1+f(a)f(x)f(a')=f(1+axa'). $$

Since $x\in Jac(A)$, $1+axa'$ is a unit in $A$, hence the image under $f$ is a unit in $B$. So $f(x)\in Jac(B)$.