Given $\varphi : A \to B$ a commutative rings homomorphism, $J$ an ideal in $B$, an element $b \in B \setminus J$ and $U=\{ x \in A : \varphi(x)b \in J \}$, show that:
- If $J$ is a prime ideal in $B$, then $U$ is a prime ideal in $A$.
- If $J$ is a maximal ideal in $B$, then $U$ is a maximal ideal in $A$.
Since each question relies on the fact that $U$ is an ideal, I'll prove that first.
We have that $0_A \in U$, because $\varphi(0_A)b=0_Bb=0_B$ and, since $J$ is an ideal in $B$, $0_B \in J$.
For each $u, v \in U$ and for each $y \in A$, we have:
- $u-v \in U$
$\varphi(u-v)b=(\varphi(u)-\varphi(v))b=\varphi(u)b-\varphi(v)b$
Since $\varphi(u)b, \varphi(v)b \in J$ and $J$ is an ideal, it follows that $\varphi(u)b-\varphi(v)b \in J$, which means $u-v \in U$.
- $uy \in U$
$\varphi(uy)b=\varphi(u)\varphi(y)b=(\varphi(u)b)\varphi(y)$
As $\varphi(u)b \in J$, it follows, for the properties of $J$ as an ideal, that $(\varphi(u)b)\varphi(y) \in J$, that is $uy \in U$.
This proves that $U$ is an ideal in $A$.
1.
Let $c, d \in A$ such that $cd \in U$.
This means that $\varphi(cd)b \in J$, by hypothesis $J$ is a prime ideal, so we have: $\varphi(cd) \in J$ or $b \in J$.
Since $b$ is chosen in $B \setminus J$, it has to be $\varphi(cd) \in J$. This can be rewritten as:
$\varphi(cd)=\varphi(c)\varphi(d)$
As we observed before, it has to be: $\varphi(c) \in J$ or $\varphi(d) \in J$. Let's suppose $\varphi(c) \in J$. This allows us to write $\varphi(c)b \in J$, which means $c \in U$.
It follows that $U$ is a prime ideal in $A$, and of course the choice of $\varphi(c) \in J$ or $\varphi(d) \in J$ makes no difference, since we are working with commutative rings.
Now, I want to ask you whether my proofs up to this point are correct and if you could give some help with the second part, since I don't know how to exploit the fact that $J$ is maximal to prove that $U$ is maximal, too.