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Li Zhan
Li Zhan
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 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 $k$ 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}}$$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 $p$, andthen show that $ codim(Z,X) \leq codim(Y',Y)$. I

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

 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, $p \in Spec(B), Y'= \bar{{p}}$ and $Z$ be an irreducible component of $f^{-1}(Y')$, whose generic point $q$ maps to $p$, and 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".

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".

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Li Zhan
Li Zhan
  • 95
  • 6

Comparation of dimensions of rings

 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, $p \in Spec(B), Y'= \bar{{p}}$ and $Z$ be an irreducible component of $f^{-1}(Y')$, whose generic point $q$ maps to $p$, and 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".