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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

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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".

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  • $\begingroup$ 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 $\endgroup$
    – Dr. Scotti
    Dec 7, 2021 at 23:56

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