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Is the map $f^*:\mathrm{Spec}(S) \to \mathrm{Spec}(R)$ injective or surjective where $f: R \to S$ is a ring homomorphism.

I know that if $P$ is a prime ideal in $S$, then $f^{-1}(P)$ is a prime ideal in $R$.

I think it is injective but cannot show why.

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    Inst it well known that its injective iff f is injective ? As maps of sheaves, I mean, but maybe you mean as maps of the underlying space ? – 2017-01-28
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    The continuous map $f^*: Spec B \to Spec A$ is surjective iff every prime ideal of $A$ is contracted, and if every prime of $B$ is an extended ideal, then $f^*$ is injective (the converse doesn't hold — see $k[t^2,t^3] \subset k[t]$). Moreover, if $B$ is flat over $A$, then $f^*$ is surjective iff $a^{ec}=a$ for any ideal $a$ of $A$, iff $B$ is faithfully flat over $A$ [Problems 16 and 20, section 3 in Atiyah MacDonald]. – 2017-01-28

2 Answers 2

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Take $$k\to k[x]$$ then $\rm{spec} \ k$ has one element, so it cannot be injective.

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This from a book Algebraic geometry and arithmetic curves, Qing Liu, p.28.

Lemma: Let $\varphi:A\to B$ be a ring homomorphism and $spec(\varphi):spec(B)\to spec(A)$ then if $\varphi$ is surjective then $spec(\varphi)$ induces a isomomorphism from $spec(B) $ onto the closed subset $V(ker \varphi)$ of $spec(A).$

EX: Let $A$ be a ring then the quotient homomorphism $\varphi:A\to A/I$ induces a homomorphism $spec(\varphi):spec(A/I)\to spec(A)$ and $ spec(A/I)\cong V(I)\subseteq spec(A) $ where $V(I):=\{p\in spec(A); I\subseteq p \}$

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    if $\varphi$ is bijective then $ker\varphi=0$ then $spec(B) \cong V(0)=spec(A)$ – 2017-01-28