MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be a field, $A$ be an integral domain, $B \subset A$, and $A, B$ are both finitely generated $k$ algebra. Let $p \subset B$ be a prime ideal. Suppose there exists prime ideals $q \subset A$, such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion. Then, is it true $ dim A_q \leq dim B_p$ ?

For the geometric meaning, it comes from the exercise of Chapter 2, 3.22(a) of Hartshorne, where: Let $ f: Spec(A) \to Spec(B) $ be a dominant morphism, $p \in Spec(B), Y'=$ {$ \bar{ p }$} (the closure of {$p$}) and $Z$ be an irreducible component of $f^{-1}(Y')$, whose generic point $q$ maps to $p$, then show that $ codim(Z,X) \leq codim(Y',Y)$.

I guess, everything translates faithfully to the above algebra fact except "$f$ dominant " is weakend by " $ B \to A$ is injective".

share|cite|improve this question
Try using Noether Normalization on $B$ to get $p$ generated by $dim B_p$ elements, and then perhaps try to apply Krull's Hauptidealsatz to $pA_q$. – Konstantin Ardakov May 18 '11 at 9:43
A map of finitely generated $k$-algebra domains $B \to A$, induces a dominant map $f:\text{Spec} A \to \text{Spec} B$ if and only if $B \to A$ is injective. For the direction you are confused about (injective => dominant) see Exercise 21(v) in Atiyah-Macdonald. – Karl Schwede May 18 '11 at 14:14
That should have read "Exercise 21(v) in the first chapter of Atiyah-Macdonald". – Karl Schwede May 18 '11 at 16:12
Dear Karl Schwede, thank you! The direction you point out used to make me unease when translate it into algebra facts. – Li Zhan May 19 '11 at 3:03
Dear Konstantin Ardakov,could you say more about how to apply Noether Normalization to $B$ to get $p$ generated by $dim(B_p)$ elements? After konwing this, it will be perfect for applying Krull's theorem, because taking into account the minimality of $q$ , one can prove prime ideal $qA_q$ is just minimal over $pA_q$ . – Li Zhan May 19 '11 at 3:42
up vote 5 down vote accepted

Dear Li, first of all I think that when you write "... such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion", you mean "... and $q$ is a minimal ideal...".

The answer to your question is given, I think, by the following more general result
Theorem Let $\phi: B\to A$ be a morphism of noetherian rings and ${\frak q} \subset A$ a prime ideal with inverse image ${\frak p}\subset B$. Then we have the following formula $$dim A_{\frak q}\leq dim B_{\frak p} +dim (A_{\frak q}\otimes_B \kappa(\frak p)) $$

Notice that there is no mention of injectivity for $\phi$ , nor of a field nor of finite generation of $A$ or $B$.
Now, if $\frak q$ happens to be to be -as in your case- the generic point of one of the irreducible components of the fibre at $\frak p$ of the morphism $Spec(\phi): Spec(A) \to Spec(B)$, then the local ring of the fiber at $\frak q$, namely $A_{\frak q}/{\frak p}A_{\frak q} = A_{\frak q}\otimes_B \kappa(\frak p)$, is zero-dimensional (see reminder below) and you get the formula you wished.

Reminder The local ring of the generic point of an irreducible scheme is a ring having only one prime ideal (its nilpotent radical) and thus has dimension zero. If the scheme is also reduced, the local ring of its generic point is a field.

Matsumura, Commutative Algebra, Theorem 19, page 79
Matsumura, Commutative Ring Theory, Theorem 15.1, page 116

An example In view of Li's comment it might be of interest to some users to see an example.
Let $k$ be a field , $B=k[t]$ and $A=k[t,X,Y]/(t-XY)=k[t,x,y]$. Let $\phi:B\to A$ be the inclusion. Then the fibre of ${\frak p} =(t)\in Spec(B)$ is the subscheme $F=V(t) \subset Spec(A)$. Please note that, even though $A$ and $B$ are domains, the fibre has two irreducible components with generic points ${\frak q}=(t,x)$ and ${\frak q}'=(t,y)$. The potentially confusing fact is that to the "physical" point $Q={\frak q}$ (say) are associated two local ring. On the one hand the local ring of $Q$ in the scheme $Spec(A)$, which is $ \mathcal O_{Spec(A), Q}=A_{{\frak q}}$ $=A_{(t,x)}$ . And on the other the local ring of $Q$ in the scheme $F$, which is $\mathcal O_{F,Q}=A_{{\frak q}}/tA_{{\frak q}}=k(y)$, a field as expected.

share|cite|improve this answer
Dear Georges Elencwajg, You are right! Thank you! There are two commnets: 1. I do mean $q $ can be any minimal prime satisfies that condition, but I guess, it is not hard to prove actually this is the only prime by the uniqueness of generic point of integral scheme. 2. I don't quite understand the ring $A_q \otimes_B k(p)$ is the local ring of the generic point of fiber. But for me, I can prove this ring is isomorphism to the ring $A_q / pA_q$, and the by the minimality of $q$, its dim is zero. – Li Zhan May 19 '11 at 9:30
Dear Li, I have added a canonical isomorphism $A_{\frak q}/{\frak p}A_{\frak q} = A_{\frak q}\otimes_B \kappa(\frak p)$, which might illuminate somewhat the description of the local ring of q seen as a point in the fiber, in contrast to it being seen as the local ring of q in the scheme Spec(A). I have also added an example to show that even if you start with two domains, the fibre may be reducible. Since that complication does not arise in your case, the example may not be of great relevance to you, but it might interest other readers. – Georges Elencwajg May 19 '11 at 16:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.