Let $A$ be a Noetherian ring, $f: A\rightarrow B$ a surjective ring map, then should the induced map on spectra $f^*: Spec(B)\rightarrow Spec(A)$ be an open map? In Atiyah and Macdonald, Chapter 1, Exercise 21, $f^*$ is already a closed map and is a homeomorphism from $Spec(B)$ onto the closed subset $V(\ker f)$ of $Spec(A)$.

Since $f: A\rightarrow B$ trivially satisfies the "going-down property" defined in Chapter 5, Exercise 10 of Atiyah and Macdonald, and if the conclusion of Chapter 7, Exercise 24 is true, then $f^*$ must be an open map, and by Chapter 1, Exercise 21, we only need to show the image of $f^*$, i.e. $V(\ker f)$, is open in $Spec(A)$. Is it true?

So the problem is reduced to

If $A$ is a Noetherian ring, $f: A\rightarrow B$ a surjective ring map, then should the closed subset $V(\ker f)$ of $Spec(A)$ is also open in $Spec(A)$?

Since any closed subset of $Spec(A)$ is of the form $V(\ker f)$ for some surjective ring map $f: A\rightarrow B$, we have also reduced the problem to

In the space $Spec(A)$ (having the Zariski topology), where $A$ is a Noetherian ring, does the collection of open sets and of closed sets coincide?

  • 2
    $\begingroup$ Dear Lao-tzu, As Martin points out in his answer below, you mean surjection, not epimorphism. E.g. $\mathbb Z \to \mathbb Q$ is an epimorphism in the cat. of rings, and the induces map on Specs is neither open nor closed. Regards, $\endgroup$
    – Matt E
    Feb 13, 2014 at 3:06

2 Answers 2


You ask if every closed subset of a noetherian affine scheme is open, which is - of course - absolutely wrong. Always look at examples first! What about $\mathbb{A}^1$?

PS: Actually you mean surjective ring homomorphisms. Ring epimorphisms are quite more general.

  • $\begingroup$ Thanks, I have modified this bug! $\endgroup$
    – Lao-tzu
    Feb 13, 2014 at 3:10
  • $\begingroup$ If the answer of my question is negative, then this mean the conclusion of Chapter 7, Exercise 24 is wrong, even in the particular case when the map is surjection, or equivalently, the $A$-algebra $B$ is generated by a single element $1\in B$. $\endgroup$
    – Lao-tzu
    Feb 13, 2014 at 3:16
  • $\begingroup$ Yes this is kind of weird. $\endgroup$ Feb 13, 2014 at 4:30
  • $\begingroup$ @ Martin Brandenburg Now I know that the "going-down property" should be replaced by what I call the "strong going-down property", which is just that in the sense of the hint. And the conclusion in Chapter 5, Exercise 10 keeps true when we replace b') by this strong form, in fact c') there is exactly equivalent to this strong form. $\endgroup$
    – Lao-tzu
    Feb 13, 2014 at 5:52

For your highlighted questions: a topological space has a nonempty proper subset which is both open and closed iff the space is disconnected. Thus in a (Noetherian) ring with no idempotents $\ne 0, 1$ (e.g. a local ring or a domain), no proper nonempty closed set is open.

The reason for the confusion (both in your post and in the comments under Martin's answer): a surjective ring map, although of finite type, does not satisfy going down - what it does satisfy (trivially) is going up. Geometrically, a surjective ring map corresponds to a closed immersion, which is in particular a closed map.


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