# Localisation of two rings which is an integral extension, then integral extension still holds?

Question seems simple, but I just can't find the solution. Let A/B be an integral ring extension and let P be a prime ideal of B. By going-up theorem, there is Q, a prime ideal of A, lying over P. Then the ring of fractions of A localised on Q is still integral on that of B localised on P?

Thank Karl and Matt, the two exemples from you are brilliants. But in the special case for the field of fractions, I think this proposition is true. I don't know if my proof is exact or not.

Let $A/B$ be an integral ring extension, $K=FracA$ the field of fractions for $A$, $k=FracB$ the field of fractions for $B$. Then $K$ is algebraic on $k$.

I have a proof: Let $S=A- \lbrace 0\rbrace, s=B- \lbrace 0\rbrace$ then $S^{-1}A=K, s^{-1}B=k,\mbox{then } k \subseteq{s^{-1}A} \subseteq K.$ $\forall \mbox{ } a/S_0\in K, a\in A, \mbox{ }S_0\in S,$ there is $S_0/s_0X-a/s_0 \in k(S_0/s_0, a/s_0)[X]$ where $s_0 \in s \subseteq B$ so that $a/S_0$ is its root. So $a/S_0$ is algebraic on the field $k(S_0/s_0,a/s_0)$ (In fact, it just lies in this field). Besides, $a/s_0$ and $S_0/s_0$ are algebraic on $k$, since $s^{-1}A$ is integral on $k$. So $k(S_0/s_0,a/s_0)/k$ is finite, then $a/S_0$ is algebraic on $k$.

• This isn't true if I recall correctly. Certainly if $W = B \setminus P$ then $B_P = W^{-1} B \subseteq W^{-1} A$ is integral. But of course $W^{-1} A \subseteq A_P$ is generally not integral. I'd try messing around with the following example to make it explicit. Let $A = k[t]$ and $B = k[t^2]$. Choose $P = \langle t^2 - 1 \rangle$ and $Q = \langle t - 1 \rangle$ say. Then we have $$A_Q = k[t]_{\langle t - 1 \rangle } \supseteq k[t^2]_{\langle t^2 - 1 \rangle} = B_P.$$ – Karl Schwede Dec 19 '13 at 4:15
• Thanks for your answer. Could you examine the proof I give above? – user83008 Dec 20 '13 at 0:12
• It's definitely true for domains localized at the origin, but that's just because only one prime of $A$ lies above $\langle 0_B \rangle$. – Karl Schwede Dec 20 '13 at 2:36
• This question appears to be off-topic. It is answered in the first comment. As for the edit, Math Overflow is not for asking people to check your proofs, especially when we already know the proof is wrong. – David White Dec 20 '13 at 14:51
• Sorry, I didn't know that here we can't ask for the verification for the proof. By the way, this proof I think is right, and it proves a proposition is true in a special case while it is wrong generally according to other experts' opinions. Anyhow, thanks for your reminding about the principles in mathoverflow. – user83008 Dec 20 '13 at 16:12

Let me explain my comment in a bit more detail.

Proposition: Suppose $B \subseteq A$ is an integral extension of domains and $P$ is a prime of $B$ with $Q \subseteq A$ a prime of $A$ lying over $B$. Then $B_P \subseteq A_Q$ is integral if and only if $Q$ is the unique prime of $A$ lying over $B$.

Proof: Suppose first that $Q$ is the unique prime lying over $P$ and let $W = B \setminus P$. Then $B_P = W^{-1} B \subseteq W^{-1} A$ is an integral extension (via an easy computation -- clear denominators as appropriate, find the integral relation, and then put the denominators back).

Claim: $W^{-1} A \cong A_Q$.

Proof of claim: Obviously $W \subset A \setminus Q$ and equality won't help, but we can still have the claim. Indeed, suppose that $Q_1$ is a prime of $A$ corresponding to a prime of $W^{-1} A$. We will show that $Q_1$ is contained in $Q$. Let $P_1 = B \cap Q_1$. If $P_1 \subseteq P$, then by the going up theorem, there exists $Q'$ over $P$ with $Q_1 \subseteq Q'$. But $Q' = Q$ by the uniqueness hypothesis which proves that $Q_1$ is contained in $Q$ as desired. This shows that $W^{-1} A$ is a local ring with maximal ideal $W^{-1} P$. But then by the universal property of localization, $W^{-1} A \cong A_P$ which proves the claim.

The claim obviously shows that $B_P \subseteq A_Q$ is integral. The above direction doesn't need the domain hypothesis.

For the converse direction, suppose that $Q_1, Q_2$ are distinct primes of $A$ lying over $P$. If $W = B \setminus P$ and $B_P = W^{-1} B \subseteq A_{Q_1}$ is integral, then certainly $W^{-1} A \subseteq A_{Q_1}$ is also integral (we are just enlarging our ring of coefficients). Note that if $A$ is not a domain, it doesn't necessarily follow that $W^{-1} A \subseteq A_{Q_1}$ (this is where the domain hypothesis is used). But now obviously we have $(W^{-1} Q_2) A_{Q_1} = Q_2 A_{Q_1} = A_{Q_1}$ and so there is no prime of $A_{Q_1}$ lying over $W^{-1} Q_2$, a contradiction. This proves the proposition.

Remark: In the non-domain case, the proposition can fail. Consider $k \subseteq k \oplus k$ via the diagonal inclusion. There are two primes of the overring lying over $\langle 0_k\rangle$, and localizing $k \oplus k$ at either just yields $k$ again. However, as noted in the proof above, generalizations of this phenomonon are essentially the only way it can fail for non-domains.