Why is the map $ D(f) \rightarrow \text{Spec}(R_{f}) $ continuous?

Question

Let $ R $ be a commutative ring and $ D(f) = \lbrace \mathfrak{p} \in \text{Spec}(R) \;|\; f \notin \mathfrak{p} \rbrace. $

I am having difficulty showing that $ \psi_{f}: D(f) \rightarrow \text{Spec}(R_{f}) $ is continuous.

Any help would be appreciated.

Answer 1

$\DeclareMathOperator{\spec}{Spec}$$D(f)$ and $\spec(R_f)$ are (exactly) the same provided you're viewing them both as open subschemes of $\spec(R)$. It is possible to consider $D(f)$ only as an open subset without the additional sheaf structure, in which case $\spec(R_f)$ is a subscheme of $\spec(R)$ whose underlying topological space is $D(f)$.

As sets, $D(f)$ and $\spec(R_f)$ are both the set of prime ideals not containing $f$.

A distinguished open set of $\spec(R_f)$ looks like $$D(g/f^n) = D(g/1) = \{\mathfrak p \in \spec(R_f) : \mathfrak p \not\ni g/1\}$$ for some element $g/f^n \in R_f$.

A distinguished open set of $D(f)$ (in the subspace topology) looks like $$D(g) \cap D(f) = \{\mathfrak p \in D(f) : \mathfrak p \in D(g)\}.$$

All I'm doing is writing down what the (distinguished) open sets are.

Answer 2

Hint:

A basis of open sets in $\operatorname{Spec}R_f$ is the set of $D_{g/1}$. Show that $$\psi_f^{-1}\bigl(D_{g/1}\bigr)=D_f\cap D_g=D(fg).$$