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 ?

  • $\begingroup$ 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). $\endgroup$ – KReiser Jun 10 at 21:53

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