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I'll first state some theorems

Corollary 5.9 [Atiyah-Macdonald pg. 61] Let $A \subset B $ be rings, $B$ integral over $A$. Let $q,q'$ be prime ideals in $B$, such that $q\subset q'$. If $q^c=q'^c=p$, then $q = q'$.

Theorem 5.10 [Atiyah-Macdonald pg.62] Let $A \subset B$ be rings, $B$ integral over $A$, and let $p$ be a prime ideal of $A$. Then there exists a prime ideal $q$ of $B$ such that $q\cap A = q^c = p$.

My question is that, can I somehow combine the above two theorems to conclude the following:

If $A \subset B$ are rings, $B$ integral over $A$. Then for every prime ideal $p \subset A$, there is a unique prime ideal $q \subset B$ such that $q \cap A = p$.

One might proceed as follows: "by Theorem 5.10, there is atleast one $q$ such that $q \cap A = p$. If $q'$ also satisfies $q' \cap A = p$, then..."

But to use Corollary 5.9, I either need to show that $q \subset q'$ or $q' \subset q$. Both of which are not "natural". So is this true or is there a counterexample?

user26857
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General Grievous
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    Theorem 5.10 says that the contraction map $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ is surjective, but Corollary 5.9 does not say that the contraction map is injective, but instead that it reflects the property of an inclusion of prime ideals being an equality. – Geoffrey Trang Jun 14 '20 at 17:52

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It is not true. There are many counterexamples. In fact, there can even be infinitely many primes in $B$ lying over a given prime in $A$, as shown here: Can infinitely many primes lie over a prime?.

Here is an example which I used recently (for another related purpose in an assignment). Let $k$ be a field. Consider the ring homomorphism $k\lbrack t\rbrack\to k\lbrack x,y\rbrack /(xy)$ with $t\mapsto x+y$ (or alternatively, the inclusion $k\lbrack x + y\rbrack\subset k\lbrack x,y\rbrack /(xy)$). This map is integral because $x^2 - x(x+y)=0$ in $k\lbrack x,y\rbrack /(xy)$ and symmetrically for $y$. But in $k\lbrack t\rbrack$, the maximal ideal $(t+1)$ has both $(x,y+1)$ and $(x+1,y)$ lying over it. In case you know some algebraic geometry, here is a visualization of the map at the level of varieties:

enter image description here

The ring $k\lbrack x,y\rbrack /(xy)$ corresponds to the union of the coordinate axes, whereas $k\lbrack t\rbrack$ is the red affine line in the drawing. The squiggly arrows indicates the map of varieties corresponding to the map of algebras I gave. Note how every point (maximal ideal) along the red line has two points (maximal ideals) lying over it -- except for the point at the origin.

o.h.
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