Question: Let $S$ be a graded ring and $ f \in S_+$. Does the ring $ S_{(f)}$ which consists of degree $ 0$ elements of $ S_f$ represent a nice functor?

Motivation: Let $ X = {\rm Spec} A$. Assume that $D(f) \subseteq D(g)$. It is messy to think about the restriction map $ A_g \to A_f $ using formulas. Things are much more transparent in the language of representable functors. Indeed, $A_g$ represents the functor

$$ R \mapsto \{ \text{maps $ A \to R$ which send $ g $ to a unit} \} $$

Since $ D(f) \subseteq D(g) $ we have a natural transformaion

$$ \{ \text{maps $ A \to R$ which send $ f $ to a unit} \} \subseteq \{ \text{maps $ A \to R$ which send $ g $ to a unit} \} $$

The yoneda lemma gives us a ring homomorphism $ A_g \to A_f $ which is the restriction map in the structure sheaf of the affine scheme. I just want to emphasize that I am not saying the yoneda lemma construction is better, it is just easier for me to keep in my head than a bunch of formulas.

Now let $S$ be a graded ring. We construct $ {\rm Proj} S $ by gluing together the affine schemes $ {\rm Spec} S_{(f)}$ where $ f \in S_{+}$ is homogeneous. In order to glue we need ring isomorphisms $$ S_{(fg)} \cong (S_{f})_{\frac{g^{\deg f}}{f^{\deg g}}} $$ If $ S_{(f)}$ represented a nice functor, we could try and mimic the construction of the restriction maps in the structure sheaf of an affine scheme. As it stands, the standard construction of this isomorphism is messy and very hard to remember (for me).


As for your motivation: First of all, $S_f$ satisfies the same universal property in the category of graded commutative rings (in fact, there is a theory of localization for objects or algebras in arbitrary cocomplete tensor categories). From this one deduces a natural isomorphism of graded rings $S_{fg} \cong (S_f)_{g}$. Instead of inverting $g$ here, we can also invert $g^n/f^m$ for any positive integers $n,m$, but we choose $n=\deg(f)$, $m=\deg(f)$ so that the element has degree $0$. Every isomorphism of graded rings induces an isomorphism of the rings in degree $0$, hence $S_{(fg)} \cong (S_{(f)})_{g^n/f^m}$. As you see, this is not messy at all.

  • $\begingroup$ @MArtin Brandenburg: Please, I wish know a reference for "localization for objects or algebras in arbitrary cocomplete tensor categories". $\endgroup$ Sep 15 '13 at 20:54
  • $\begingroup$ @Buschi: For example section 2.2. in Florian Marty's thesis. Generalizations will probably appear in my thesis. $\endgroup$ Sep 15 '13 at 21:32
  • $\begingroup$ @Daniel: Glad that I could help. $\endgroup$ Sep 15 '13 at 21:33
  • $\begingroup$ @Brandemburg: Thank you very much, a very interesting thesis. $\endgroup$ Sep 16 '13 at 11:45

It seems that Martin has provided the answer which Daniel sought. But the question in the title doesn't appear to be answered yet: Is there a universal property for graded localization? More precisely, for the degree-0 part of the localization?

I believe there is one. Let $S$ be an $\mathbb{N}$-graded ring. Let $M \subseteq S$ be a multiplicatively closed set consisting only of homogenous elements. Assume that $M$ contains an element of degree $1$. Let $R$ be an arbitrary ring. Then $$ \mathrm{Hom}_{\mathrm{Ring}}(S[M^{-1}]_0, R) \cong \mathrm{Hom}_{\mathrm{grRing}}(S, R[X])_M / R^ \times. $$ An element of the right hand side is an equivalence class (modulo rescaling) of a graded ring homomorphism $S \to R[X]$ which maps the degree-$n$ elements of $M$ to polynomials whose coefficient of $X^n$ is a unit in $R$.

A ring homomorphism $\varphi : S[M^{-1}]_0 \to R$ gives rise to such an equivalence class $[\psi]$ by picking an element $f \in M \cap S_1$ and defining $\psi(x) := \varphi(x/f^n) X^n$ for homogeneous elements $x$ of degree $n$.

Conversely, such an equivalence class $[\psi]$ gives rise to a ring homomorphism $\varphi$ by setting $\varphi(x/s) := \psi(x)|_{X=1} \cdot \psi(s)|_{X=1}^{-1}$.

I sincerely hope that there is some better way to express this universal property.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.