Any prime ideal that is a contraction of an ideal of S is a contraction of a prime ideal of S

Question

I was asked to prove the following: Let $f: R \to S$ be a morphism of rings. Then, any prime ideal $P \subseteq R$ that is a contraction of an ideal of S is a contraction of a prime ideal of S. I've got the following hint:

Consider $\mathcal{F} = \{J \lhd S $ | $ f^{-1}(J) = P\}$ a family of ideals. Then prove that every totally ordered chain of $\mathcal{F}$ has an upper bound and use Zorn's lemma to find a maximal element and prove that this maximal element is prime.

I am a bit stuck proving every totally ordered chain has an upper bound. If $J, K \lhd S$, $f^{-1}(J) = f^{-1}(K) = P$. Then $J + K \lhd S$, but I don't succeed to prove that $f^{-1}(J + K) = P.$ I had the following: if $x \in f^{-1}(J+K)$, then $f(x) = j+k$. But since f is not surjective, this does not imply that x $\in f^{-1}(J) + f^{-1}(K) \subseteq P$.

Moreover, when this is proved, I am also struggling to prove that the maximal element, say Q, is prime.

Answer 1

This answer assumes the ring is commutative.

Hint $(1)$: you did not attempt to use the chain condition, you just considered arbitrary $J,K$. Try to use the standard tricks for finding an upper bound of a chain of ideals.

Let $C$ be a chain of elements of $\mathcal{F}$. Their union will again be an ideal and the preimage of a union is the union of preimages so you can see the union of $C$ will be an upper bound of $C$ in $\mathcal{F}$.

Hint $(2)$: in applying Zorn, you will find a maximal element. Consider the standard tricks used to e.g. show every proper ideal is contained in a maximal/prime ideal. Suppose your maximal element is not prime. We will make an extension of it and then we need to show our extension of the maximal element is in $\mathcal{F}$, contradicting maximality, and to do so we need to show its preimage is $P$. So you play the contradiction game again and suppose there are some elements of the preimage not in $P$ and you try to contradict $P$'s primality as well.

By Zorn, there is a maximal $Q$ of $\mathcal{F}$ (the set is nonempty by assumption). $f^{-1}(Q)=P$ of course. Now we must check primality. Say $ab\in Q$ and $a,b\notin Q$. $Q':=Q+(a)$ extends $Q$ strictly and $f^{-1}(Q')\supseteq P$. By maximality of $Q$, it is not equal to $P$; so there is $x\notin P$ with $f(x)\in Q'$. Necessarily $f(x)\notin Q$. $f(x)=q+sa$ then for some $q\in Q,s\in S$, such that $sa\notin Q$. Similarly there is $y$, $y\notin P$, with $f(y)=q'+tb$ for some $t\in S$ such that $tb\notin Q$ and for some $q'\in Q$.

We conclude that $f(xy)=(q+sa)(q'+tb)=qq'+(sa)q'+(tb)q+(st)(ab)\in Q$ so it follows $xy\in P$. However, neither $x$ nor $y$ are in $P$ so there is a contradiction with primality of $P$.

Answer 2

The empty chain has an upper bound (since the family is non-empty by assumption). If $\{J_i\}$ is a non-empty chain of ideals of $S$ with $f^{-1}(J_i) = P$, then clearly $\bigcup_i J_i$ is also an ideal (using the "chain" property), and we have $f^{-1}(\bigcup_i J_i) = \bigcup_i f^{-1}(J_i) = \bigcup_i P = P$.

By Zorn's Lemma, there is a maximal element $J$ in the family of ideals with $f^{-1}(J) = P$. Let us prove that $J$ is prime. Clearly, $1 \notin J$. Assume $a,b \in S \setminus J$. Then $f^{-1}(J + \langle a \rangle) \neq P$ and $f^{-1}(J + \langle b \rangle) \neq P$. Hence, there are $u,v \in R \setminus P$ with $f(u) \in J + \langle a \rangle$ and $f(v) \in J + \langle b \rangle$. Then $f(uv) = f(u) f(v) \in J + \langle ab \rangle$, so $uv \in f^{-1}(J + \langle ab \rangle)$ and $uv \notin P$. Thus, $f^{-1}(J + \langle ab \rangle) \neq P$ and we see $ab \notin J$.

Alternative, more abstract proof: Choose an ideal $J$ with $P = f^{-1}(J)$. Then $\overline{f} : R/P \to S/J$ is injective. It therefore suffices to prove: If $R$ is an integral domain and $f : R \to S$ is an injective homomorphism, then there is a prime ideal of $S$ that contracts to $0$. Well, $S \otimes_R Q(R)$ is non-trivial, hence has a prime ideal. Its contraction to $S$ does the job.