How does $(\mathfrak {B_2} / \mathfrak {B_1}) \cap (R/ \mathfrak p) = (0)$?

Question

Let $S/R$ be an integral extension. Let $\mathfrak {B_1}, \mathfrak {B_2} \in \mathrm {Spec} (S)$ with $\mathfrak {B_1} \subset \mathfrak {B_2}$ are given. If $\mathfrak {B_1} \cap R = \mathfrak {B_2} \cap R$ then $\mathfrak {B_1} = \mathfrak {B_2}.$

In my book the proof goes like as follows $:$

Let $\mathfrak p := \mathfrak {B_1} \cap R =\mathfrak {B_2} \cap R.$ Then $S/\mathfrak {B_1}$ is integral over $R/\mathfrak p$ and $\mathfrak {B_2}/\mathfrak {B_1}$ is a prime ideal of $S/\mathfrak {B_1}$ which intersects $R/\mathfrak p$ in $(0).$ So we must have $\mathfrak {B_1} = \mathfrak {B_2}.$

But I don't understand why $(\mathfrak {B_2} / \mathfrak {B_1}) \cap (R/ \mathfrak p) = (0)$ and how does it imply that $\mathfrak {B_1} = \mathfrak {B_2}$?

Please help me in this regard.

Thank you very much for your valuable time.

EDIT $:$ I think that I got the second implication if we assume that $(\mathfrak {B_2} / \mathfrak {B_1}) \cap (R/ \mathfrak p) = (0).$

Let us first state a lemma.

Lemma $:$ Let $S/R$ be an integral extension. Let $J$ be an ideal of $S$ with $I:= J \cap R.$ If $J$ contains a non-zero divisor of $S$ then $I \neq (0).$

I know this lemma. Now let us try to prove the second implication by this lemma.

In this case $S/\mathfrak {B_1}$ is integral over $R/\mathfrak {p}$ and $(\mathfrak {B_2} /\mathfrak {B_1}) \cap (R/\mathfrak {p}) = (0).$ So it follows from the above lemma that $\mathfrak {B_2}/\mathfrak {B_1}$ doesn't contain any non-zero divisor of $S.$ But if $\mathfrak {B_2}/ \mathfrak {B_1}$ contains a non-zero element then it would be a non-zero divisor of it by the primality of $\mathfrak {B_1}.$ This shows that $\mathfrak {B_2}/ \mathfrak {B_1} = (0) \implies \mathfrak {B_1} =\mathfrak {B_2}.$

But how do I prove the first implication. I am still thinking about it.

Answer 1

Observe that $$R/\mathfrak p = \{r + \mathfrak {B_1} : r \in R \}.$$ because $R/\mathfrak p$ can be identified by it's image under the inclusion map $$\varphi : R/\mathfrak p \hookrightarrow S/\mathfrak {B_1}$$ defined by $$\varphi (r + \mathfrak p) = r + \mathfrak {B_1},\ r \in R.$$ and also $$\mathfrak {B_2} / \mathfrak {B_1} = \{x + \mathfrak {B_1} : x \in \mathfrak {B_2} \}.$$ Now suppose that $s \in (\mathfrak {B_2}/\mathfrak {B_1}) \cap (R/\mathfrak p).$ Then $\exists$ $r \in R$ and $x \in \mathfrak {B_2}$ such that $s=r+\mathfrak {B_1} = x + \mathfrak {B_1}.$ This shows that $$r-x \in \mathfrak {B_1} \implies r \in x+\mathfrak {B_1} \subseteq \mathfrak {B_2} + \mathfrak {B_1} = \mathfrak {B_2}$$ since $\mathfrak {B_1} \subseteq \mathfrak {B_2}.$ So we have $r \in \mathfrak {B_2} \cap R=\mathfrak p.$ Therefore $s = r+\mathfrak {B_1} = 0 + \mathfrak {B_1},$ since $r \in \mathfrak p \subseteq \mathfrak {B_1}.$ This implies that $$(\mathfrak {B_2}/\mathfrak {B_1}) \cap (R/\mathfrak p) = (0).$$

Answer 2

Lemma: Let $A$ be a domain and let $B$ be an integral $A$-algebra which is a domain. If a prime $\mathfrak{q}$ in $B$ contains no elements of $A$ except for $0$ then it must be $0$.

Proof: Suppose $b$ is not $0$ but contained in $\mathfrak{q}$. $b \in A[b]$, which is a finite extension of $A$ since $b$ is integral. Let $\sum_{i =0}^n a_i b^i =0$ be the minimal polynomial of $b$. $a_0$ cannot be $0$ as then $b \sum_{i =1}^n a_i b^{i-1} = 0$, wheras $B$ is a domain. Note that $\sum_{i =1}^n a_i b^{i-1}$ cannot be $0$ as its degree is less than that of the minimal polynomial. Therefore $b \left( \sum_{i =1}^n a_i b^{i-1} \right) $ is a nonzero element of $A$, but also an element of $\mathfrak{q}$. How absurd!

Note: In your notation, we have a domain $R/\mathfrak{p}$ contained in a domain $S$. There is an ideal $\mathfrak{q} = \mathfrak{B}_2 / \mathfrak{B}_1$. The preimage of $\mathfrak{B}_2$ and $\mathfrak{B}_1$ under the map $A \rightarrow B$ is $\mathfrak{p}$. By the noether isomorphism theorems (I think it's called the $4$th one), this implies that the preimage of $\mathfrak{B}_2 / \mathfrak{B}_1$ and $\mathfrak{B}_1 / \mathfrak{B}_1 = \overline{0}$ in $S / \mathfrak{B}_1$ to an ideal in $A / \mathfrak{p}$ is $\mathfrak{p} / \mathfrak{p} = \overline{0}$ (make sure to know those Noether isomorphism theorems very well!). Now take $A = R/\mathfrak{p}$, $B = S / \mathfrak{B}_1$, $\mathfrak{q} = \mathfrak{B}_2 / \mathfrak{B}_1$, and apply the lemma.

---

Here's another way to show this theorem, but it requires you know localization:

Theorem: Let $B/A$ be an integral extension of rings. Let $\mathfrak{p}$ be a prime, and let $\mathfrak{q}$ and $\mathfrak{q}'$ be primes over $\mathfrak{p}$ in $B$, with $\mathfrak{q} \subset \mathfrak{q}'$. Then $\mathfrak{q} = \mathfrak{q}'$.

The claim is true if and only if it is true of $A/\mathfrak{p} \subset B / \mathfrak{p} B$. So, without loss of generality, we may assume $\mathfrak{p} = 0$. Another reduction step: the claim is true if and only if it is true for $A_{\mathfrak{p}} \subset (A - 0)^{-1} B$. So, without loss of generality, we may assume $A$ is a field. For a last reduction step, the claim is true if and only if it is true for $A \rightarrow B / \mathfrak{q}$ (this map sends $1$ to $1$, so it's injective since $A$ is a field). So we may assume $\mathfrak{q} = 0$. After all this reduction, we just need to show that:

Lemma: Let $k$ be a field, and let $A$ be a domain over $k$ which is integral over $k$. Then $A$ is a field.

Proof: Take a nonzero element $a \in A$. Then $a$ is contained in some finite algebra $B \subset A$ over $k$ as it is integral over $k$. The map $\mu_a : B \rightarrow B$ sending $b$ to $ab$ is injective because $B$ is a domain. Therefore it is surjective (consider the vector-space dimension of $B$). But that means that there is an element $c$ such that $ca = 1$ in $B$ (we used surjectivity). That's just saying there is an inverse for $a$, though.