# Proof

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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@Brenner

### References

#### Adapted from CC BY-SA 3.0 Sources:

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