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?


3 Answers 3


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.

  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, 2011 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, 2011 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. $\endgroup$
    – user325
    Apr 25, 2011 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, 2011 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, 2016 at 12:26

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 $q'\cap f(A)=p'$.


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