# Do we have $A_{(\mathfrak{p})} \simeq (A_{(f)})_{\mathfrak{q}}$?

Let $$A = \oplus_{d \ge 0} A_d$$ be a graded ring. Let $$\mathfrak{p}$$ a homogeneous prime of $$A$$. Let $$f$$ be a homogeneous of $$A$$ of degree $$\require{cancel}\cancel{d}$$ $$1$$ such that $$f \not\in \mathfrak{p}$$. Let $$A_{(\mathfrak{p})} = \Bigl\{ \frac{a}{b} : b \not\in \mathfrak{p}, a, b \text{ homogeneous of same degree} \Bigr\} \text{ and } A_{(f)} = \Bigl\{ \frac{a}{f^r} : a \in A_{dr} \Bigr\}.$$ Let $$\mathfrak{q} = \mathfrak{p} A_f \cap A_{(f)}$$. Do we have $$A_{(\mathfrak{p})} \simeq (A_{(f)})_{\mathfrak{q}}$$? We have a well-defined map $$\phi : A_{(\mathfrak{p})} \longrightarrow (A_{(f)})_{\mathfrak{q}}, \frac{a}{b} \longmapsto \frac{a/f^r}{b/f^r}$$ where $$r = \deg a = \deg b$$. Is this map an isomorphism ?

• Yes, these objects are isomorphic, but there are some light issues with your work: you're implicitly assuming $\deg f=1$ here without actually writing it down. You can prove this in general (i.e. not assuming $\deg f=1$) if you want, and then you'll need to alter your map a bit. Constructing an inverse should be very doable - if you're stumped, see Stacks 00JR for what the inverse should be (though with some verification steps omitted). – KReiser Jun 10 at 21:53