# Proof of second form of the going-up theorem

Let $$A ⊆ B$$ be two rings and suppose that $$B$$ is integral over $$A$$. Show that if $$p_1 ⊂ p_2 ⊂ \dots ⊂ p_n$$ is a chain of prime ideals of $$A$$, and $$q_1 ⊂ q_2 ⊂ \dots ⊂ q_m$$ (with $$m < n$$) is a chain of prime ideals of $$B$$ such that $$q_i$$ “lies over” $$p_i$$ (i.e. $$q_i∩A = p_i$$ for $$1 ≤ i ≤ m$$), then the second chain can be extended to $$q_1 ⊂ q_2 ⊂ \dots ⊂ q_n$$ so that this remains true, i.e. $$q_i ∩ A = p_i$$ for $$1 ≤ i ≤ n$$.

We can just prove it for $$n=m+1$$. From the first form of the going-up theorem, we obtain that exists a prime $$p_n$$ such that $$p_n\cap A=q_n$$: the aim is to prove that $$q_m\subset q_n$$. Notice that since the $$A\hookrightarrow B$$ is injective, we have $$\operatorname{dim} B/q_m=\operatorname{dim} A/p_m$$; plus, since $$B$$ is integral over $$A$$, also $$B/q_m$$ is integral over $$A/p_m$$, so $$\operatorname{ht} q_n/q_m\le\operatorname{ht}p_n/p_m$$. Putting together these observations we get $$\operatorname{dim} B/q_m=\operatorname{dim} A/p_m\ge \operatorname{ht}p_n/p_m\ge \operatorname{ht}q_n/q_m.$$ Now I don't know how to formalize well my intuition (if it is correct): I'm thinking that if $$q_n$$ didn't contain $$q_m$$, then $$q_n/q_m=B/q_m$$, and $$\operatorname{ht}B/q_m=\operatorname{dim} B/q_m+1$$, contradicting the inequalities above. Is it ok or extending the height to an ideal that is not proper doesn't make sense? Thanks in advance

Question: "Is it ok or extending the height to an ideal that is not proper doesn't make sense?"

Answer: Since $$Q_{m+1}⊆B/q_m$$ is a prime ideal in $$B/q_m$$ it follows the lift $$q_{m+1}⊆B$$ of $$Q_{m+1}$$ is a prime ideal containing $$q_m$$, with the property that $$q_{m+1}∩A=p_{m+1}$$.

Note: There is a 1-1 correspondence between prime ideals $$Q \subseteq B/q_m$$ and prime ideals $$q \subseteq B$$ with $$q_m \subseteq q$$. (Atiyah-Macdonald Proposition 1.1).

Comment: "I don't understand, it seems to me that you are using the fact that we want to prove. When you say that Qm+1 is a prime ideal in B/qm, how to you know that Qm+1 is a proper ideal? (I'm assuming that Qm+1 is qm+1/qm). If qm+1 doesn't contain qm then Qm+1 is B/qm. I suppose that I'm wrong somewhere but I can't find where – Dorian 8 hours ago"

This is the "going up theorem": If $$A \subseteq B$$ is integral and if $$p \subseteq A$$ is a prime ideal, we get an integral extension $$A_p \subseteq B_p$$. Hence any maximal ideal $$n \subseteq B_p$$ (here we must believe in Zorns lemma) will have $$n \cap A_p$$ to be maximal, hence $$n \cap A_p=pA_p$$. Let $$\beta: B \rightarrow B_p$$. It follows $$q:=\beta^{-1}(n)\subseteq B$$ is maximal with $$q\cap A=p$$. See Theorem 5.10 in Atiyah-Macdonald.

Note: Zorns lemma guarantees that there is a maximal ideal $$n\neq (1) \subseteq B_p$$. If you assume $$A,B$$ to be finitely generated over a Dedekind domain you do not need Zorns lemma. Then you can use the Hilbert Basis theorem.

For rings that are not finitely generated over a Dedekind domain you can use "Zorn's lemma" to prove results that are counterintuitive. Hence some people view the lemma as "controversial".

• I don't understand, it seems to me that you are using the fact that we want to prove. When you say that $Q_{m+1}$ is a prime ideal in $B/q_m$, how to you know that $Q_{m+1}$ is a proper ideal? (I'm assuming that $Q_{m+1}$ is $q_{m+1}/q_m$). If $q_{m+1}$ doesn't contain $q_{m}$ then $Q_{m+1}$ is $B/q_m$. I suppose that I'm wrong somewhere but I can't find where Dec 7, 2021 at 23:56