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Following exercises in Atiyah-Macdonald, I am led to prove the following:

Let $\phi:A \rightarrow B$ be a ring homomorphism, and let $X = \operatorname{Spec}{A}$ and $Y = \operatorname{Spec}(B)$ be the prime spectra of $A, B$, endowed with the Zariski topology. We denote by $\phi^*$ the induced map between spectra $\phi^*:Y \rightarrow X$.
Show that if $\phi$ is surjective, then $\phi^*$ is a homeomorphism onto its image.

I have so far shown that $\phi^*$ is a continuous map. I have also shown that when $\phi$ is surjective, $\phi^*$ is injective. Thus $\phi^*$ is a continuous bijection onto its image. It remains for me to show that $\phi^*$ is an open (or equivalently a closed) map. Any tips on how to go about this ?

hm2020
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Mr Lolo
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    Just a quick comment if you are interested towards algebraic geometry. Such a morphism of rings not only induces the map you mentioned at the level of topological spaces, it also induces a morphism at the level of affine schemes $ (\phi^{},\phi'):(SpecB,\mathcal{O}{SpecB})\rightarrow (SpecA,\mathcal{O}{SpecA}) $ which is a closed immersion. That means that $\phi^$ is a homeomorphism onto a closed subset, which is the content of your question, and also the morphism of sheafs is surjective. – Jason V Jan 10 '23 at 18:47

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Question: "I have so far shown that ϕ∗ is a continuous map. I have also shown that when ϕ is surjective, ϕ∗ is injective. Thus ϕ∗ is a continuous bijection onto its image. It remains for me to show that ϕ∗ is an open (or equivalently a closed) map. Any tips on how to go about this ?"

Answer: @Mr Lolo - there is for any commutative unital ring $A$ and any ideal $I \subseteq A$ a 1-1 correspondence between the prime ideals $\mathfrak{p} \subseteq A$ containing $I$ and the prime ideal $\mathfrak{q} \subseteq A/I$. This is used to prove that the morphism $Spec(A/I)\rightarrow Spec(A)$ is a closed immersion. Moreover (Hartshorne, Corr.II.5.10) one proves that any closed subscheme $Z \subseteq Spec(A)$ arise in this way.

Note: If $\phi$ is surjective and the map $\phi^*:Spec(A/I) \rightarrow Spec(A)$ is open it follows $Spec(A)$ is disconnected: There is a non trivial direct sum decomposition $A \cong A_1\oplus A_2$ and you may write

$$Spec(A) \cong Spec(A_1) \cup Spec(A_2)$$

as a disjoint union of non trivial open and closed affine subschemes.

hm2020
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  • Thank you for your answer. I am new to most of this, so it will take some digesting of Hartshorne to completely grasp your answer. Could you briefly explain how this applies to $\phi^*:\operatorname{Spec}(B) \rightarrow \operatorname{Spec}(A)$ when $B$ is not a quotient of $A$ ? – Mr Lolo Jan 05 '23 at 11:30
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    @Mr Lolo - when $\phi$ is surjective it follows $B \cong A/I$: There is a short exact sequence $0 \rightarrow ker(\phi):=I \rightarrow A \rightarrow B \rightarrow 0$. – hm2020 Jan 05 '23 at 11:32
  • Oh yes of course ! Great thank you. – Mr Lolo Jan 05 '23 at 11:34