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Proof

(related to Lemma: Fiber of Prime Ideals Under a Spectrum Function)

By assumption, \(\varphi \colon R\longrightarrow S\,\) is a ring homomorphism between two commutative rings and \[\varphi ^{ * }\colon \cases{\operatorname {Spek} \left(S\right)\longrightarrow \operatorname {Spek} \left(R\right),\cr J\longmapsto \varphi ^{ * }(J)}\] is the corresponding spectrum function. For the fiber of a prime ideal \(I\in \operatorname {Spek} \left(R\right)\) under the spectrum function, we want to show the following properties:

\(1\) It equals \(\operatorname {Spec} (S/IS)\).

This follows immediately from the proposition.

\(2\) It equals all prime ideals \(J \in \operatorname {Spec} \left(S\right)\) with \(I S\subseteq J\) and with \(J \cap \varphi (R\setminus I)=\emptyset \).

For a prime ideal \({J}\lhd S\) we have \(\varphi ^{-1}({J})={I}\) if and only if

• \(\varphi ({I})\subseteq {J}\) and
• \(\varphi (R\setminus {I})\subseteq S\setminus {J}\).

The first condition is equivalent to \({I}S\lhd {J}\) und the second to \(\varphi (R\setminus {I})\cap {J}=\emptyset \,\).

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References

Adapted from CC BY-SA 3.0 Sources:

1. Brenner, Prof. Dr. rer. nat., Holger: Various courses at the University of Osnabrück