# Prime spectrum and the restriction

In Atiyah's commutative algebra, exercise3.21(ii) is the following: Let $f:A \to B$ be a ring homomorphism. Let $X=Spec(A)$ and $Y=Spec(B)$, and let $f^* : Y \to X$ be the mapping associated with $f$. Identifying $Spec(S^{-1}A)$ with its canonical image $S^{-1}X=\{ p | p \cap S = \emptyset \}$ in $X$, and $Spec(S^{-1}B)(=Spec(f(S)^{-1}B))$ with its canonical image $S^{-1}Y=\{q|q \cap f(S) = \emptyset\}$ in $Y$, show that $(S^{-1}f)^* : Spec(S^{-1}B) \to Spec(S^{-1}A)$ is the restriction of $f^*$ to $S^{-1}Y$, and that $S^{-1}Y=f^{*-1}(S^{-1}X)$.

I solved the above problem, but I have a question. In the case of $S=A-p$, then we can reduce $f^*$ to $S^{-1}Y \to S^{-1}X$. But in another problem(Ex5.10(c')), textbook says that $f^*$ reduces to $Spec(B_q) \to Spec(A_p)$ where $p=f^{-1}(q)$. Then the codomains are same(up to isomorphism), but the domains seems different. $S^{-1}Y$ is the set of primes that are disjoint from $f(S)=f(A-p)$, and $Spec(B_q)$ is the set of primes contained in $q$. But I think that the latter is correct since $q' \subset q$ implies $f^{-1}(q') \subset f^{-1}(q)=p$. What is the problem?

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What is $S$ in the first paragraph? – Rasmus Apr 21 '11 at 9:53
@Rasmus $A$ is a ring, $S$ is a multiplicatively closed subset of $A$. If we let $\phi : A \to S^{-1}A$ the canonical homomorphism, $\phi^* : Spec(S^{-1}A) \to Spec(A)$ is a homeomorphism of $Spec(S^{-1}A)$ onto its image in $X=Spec(A)$. Let this image be denoted by $S^{-1}X$. – Gobi Apr 21 '11 at 11:35

There is no problem: the authors study the morphism of topological spaces $f^\ast : Spec(B_q) \to Spec(A_p)$ which is quite legitimate. It can indeed happen that $S^{-1} (Y)$ is bigger than $Spec(B_q)$ and this just means that the map $f^\ast$ is the restriction of the map $S^{-1}Y=f^{*-1}(S^{-1}X) \to S^{-1}X$.
Here is an example to show that indeed $S^{-1}Y$ can be strictly larger than $Spec(B_q)$.
Take $A=k$ ( a field ) , $B=k[T]$ ( the polynomial ring) and $f:k \to k[T]$ the inclusion.
Take $q=(T)$. Then $p=(0)$, $S=k^\ast\subset A=k$ , $f(S)=k^\ast \subset B=k[T]$ and $S^{-1}Y=Y=Spec (k[T])$ is indeed enormously larger than $Spec(B_q)=Spec(k[T]_{(T)})$ , which consists of only the two points $(0)$ and $q=(T)$.
Thank you. The author meant by $f^\ast : Spec(B_q) \to Spec(A_p)$ that the more restricted map than $S^{-1}Y \to S^{-1}X$. I got it. I was trying to prove that these two restrictions are same, but ended up with no gain. – Gobi Apr 21 '11 at 11:46