I want to show that

$f$ has the going-down property $\Leftrightarrow$ For any prime ideal $\mathfrak{q}$ of $B$, if $\mathfrak{p}=\mathfrak{q}^c$, then $f^{*}:\textrm{Spec}(B_{\mathfrak{q}}) \rightarrow \textrm{Spec}(A_{\mathfrak{p}})$ is surjective.

I have proved ($\Leftarrow$), but there's something wrong in ($\Rightarrow$).

pf of ($\Rightarrow$): First, I understood $ \textrm{Spec}(A_{\mathfrak{p}}) = \{\mathfrak{p}' \in \textrm{Spec}(A) | \mathfrak{p}' \subset \mathfrak{p} \}. $ Let $\mathfrak{p}' \subset \mathfrak{p}$. Then $f(\mathfrak{p}') \subset f(\mathfrak{p})$ are prime ideals in $f(A)$. From $f(\mathfrak{p})=f(f^{-1}(\mathfrak{q}))=\mathfrak{q} \cap f(A)$, since f has the going-down property, there exists $\mathfrak{q}' \subset \mathfrak{q}$ such that $\mathfrak{q}' \cap f(A) = f(\mathfrak{p}')$. Now $f^{*}(\mathfrak{q}')=f^{-1}(\mathfrak{q}')=f^{-1}(\mathfrak{q}' \cap f(A))=f^{-1}(f(\mathfrak{p}'))$

If $\mathfrak{p}' \supset \ker f$, then $f^{-1}(f(\mathfrak{p}'))=\mathfrak{p}'$, so the proof is done. But isn't it possible that $\mathfrak{p}'$ does not contain $\ker f$? But $f^{*}$ to be surjective, $\textrm{Spec}(A_{\mathfrak{p}})$ must consists of contracted ideal. Is the problem wrong?

  1. Is going-down property actually defined for arbitrary ring homomorphism? I have the impression that we only talk about it for inclusion of rings.

  2. In what you wrote, it's wrong to say that $f(p')$ is prime - this is only true if $p'$ contains $ker f$.

  3. A counterexample to your statement would be to consider $k[x] \to k \to k[x]$, where the first map is evaluation map at $0$, and the second map is inclusion. Then the preimage of $(0)$ is $(x)$. $(x)$ contains $(0)$ but is not the pre-image of any prime ideal.

  4. The statement is true if $f$ is an inclusion.

  • $\begingroup$ Thanks for your answer. In Atiyah's introduction to commutative algebra, $f: A \to B$ is said to have the going-down property if the conclusion of going-down theorem holds for $B$ and its subring $f(A)$. $\endgroup$ – Gobi Apr 18 '11 at 7:30
  • $\begingroup$ I found a link, which contains similar question to mine. link. But there's no answer to it. $\endgroup$ – Gobi Apr 19 '11 at 2:29
  • $\begingroup$ is there anything in my answer that is not good enough? I said that if $f$ is inclusion, your problem is right, otherwise it's wrong. $\endgroup$ – Soarer Apr 25 '11 at 4:12
  • $\begingroup$ Your answer is good, but there seems to be definition of going-down property about any ring homomorphism $f$. I answered what I guess. Anyway I accepted your answer. $\endgroup$ – Gobi Apr 26 '11 at 1:39
  • 1
    $\begingroup$ This answer is wrong. The going-down property is defined for arbitrary ring homomorphisms and the counterexample in 3 is not one - it does not satisfy either hypothesis. $\endgroup$ – RghtHndSd Apr 22 '16 at 12:26

I concluded like this.

"$f$ has the going-down property" means that for any prime ideals $p \supset p'$ in $A$ and $q$ in B s.t $f^{-1}(q)=p$, there exists $q'$ in B s.t $q \supset q'$ and $f^{-1}(q')=p'$. Then it is obviously equivalent to that for any prime ideal $q$ of $B$, if $p=f^{-1}(q)$, then $f^∗:Spec(B_q)→Spec(A_p)$ is surjective.


We actually don't need the argument with the kernel, since the contraction of any prime in $f(A)$ is a prime in $A$. (Then just use Gobi's argument). With a bit more detail:

By going-down property, we want to show that whenever we have $p'\subseteq p$ in Spec $f(A)$ and a $q\in $Spec $B$ such that $q\cap f(A)=p$, we can find some $q'\subseteq q$ such that $p'=q'\cap f(A)$. But $a'=f^{-1}(p')$ is some prime ideal in $A$, and is clearly contained in $a=f^{-1}(p)$, another prime in $A$. So from the surjectivity condition, since $a$ is the contraction of $q$, there is some prime in Spec$(A_p)$, that is, some $q'\subseteq q$ such that $f^*(q')=a'$. It is clear then that $q'\cap f(A)=p'$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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