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I have a question regarding the exercise 3.29 in Atiyah-MacDonald. In the previous exercises it was introduced the constructible topology. Basically the space $\DeclareMathOperator{\Spec}{Spec}\Spec(A)$ for a ring $A$ is endowed with a topology such that the closed set are images of $f^\ast(\Spec(B))$ where $f:A\to B$ is a ring homomorphism. It was shown that this approach provides indeed a topology and is finer than the Zarisky topology. Now the question: Let $f:A\to B$ a ring homomorphism is $f^\ast:\Spec(B)\to \Spec(A)$ continuous if $\Spec(A)$ and $\Spec(B)$ are endowed with the constructible topology?

My attempt: Let $g:A\to C$ a ring homomorphism then \begin{equation*} f^{\ast-1}(g^\ast(\Spec(C)))=f^{\ast-1}(g^\ast(\Spec(C))\cap f^\ast(\Spec(B)))= f^{\ast-1}(h^\ast(\Spec(C\otimes_A B))), \end{equation*} where $h:A\to C\otimes_A B$ is the ring homomorphism $h(x)=1_C\otimes f(x)=g(x)\otimes 1_B$. If we now use the mapping $\alpha:B\to C\otimes_A B, b\mapsto 1_C\otimes b$ then one has $h=\alpha\circ f$ hence $h^\ast=f^\ast\circ \alpha^\ast$, but from this follows only $\alpha^\ast(\Spec(C\otimes_A B))\subset f^{\ast-1}(g^\ast(\Spec(C)))$. Is there a way to show equality or is the continuity trivial and I don't see it? Or is the map $f^\ast$ not continuous in general with respect to the constructible topology?

Sniper
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1 Answers1

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Have you done exercise 3.28 yet? If not you should do it first, if so then what you've shown is that the $X_f$ and $V(f)$ sets give a sub-basis for the constructable topology, so to show that $f^\ast$ is continuous it suffices to show that sets of the form $f^{\ast -1}(X_f)$ and $f^{\ast -1}(V(g))$ are closed. That fact is $(i)$ and $(ii)$ of exercise 1.21.

Jim
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