Proof

We only provide a sketch of a proof:

From the lemma about the one-to-one correspondence of ideals in a factor ring and a commutative ring it follows that the prime ideals in the factor ring \(R/{I}\) correspond over the ideal \({J}\mapsto q^{-1}({J})={J}+{I}\) to the prime ideals of \(R\) containing the ideal \({I}\).
Therefore, the spectrum function \(q^{ * }\) is bijective and has the image being the set \(V(I)\) of all ideals of \(R\) containing \(I\).
It is also closed, since for any ideal \({K}\subseteq R/{I}\) and a prime ideal \({J}\subseteq R/{I}\) we have \({K}\subseteq {J}\) if and only if \[{K}+{I}=q^{-1}({K})\subseteq {J}+{I}.\] Therefore the image of \(V({K})\) equals the image of \(V({K}+{I})\) and thus, it is closed.

Note that for any prime ideal \({J}\) and an element \(f\in R\) we have \(f\not \in {J}\) if and only if \({J}\) is disjoint to the multiplicative system \(\left\{f^{n}{|}\,n\in \mathbb {N} \right\}\).
From that note, it follows together with \((2)\) that the map is injective and its image equals \(D(f)\).
The same argument applied on \(g\in R\) respectively \({\frac {g}{1}}\in R_{f}\) reveals that the image of \(D(g)\subseteq \operatorname {Spek} \left(R_{f}\right)\) equals \(D(fg)\) and is therefore open.
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Brenner, Prof. Dr. rer. nat., Holger: Various courses at the University of Osnabrück