Graded Rings and Affine $\IG_m$-Schemes

The $\Proj$-construction $\Proj(A)$ associated to an $\IN$-graded ring $A$ is mentioned in most modern (scheme-y) introductions to algebraic geometry. What is often missing is a geometric interpretation of graded rings, obscuring what is happening in the $\Proj$-construction. The answer is that an $\IN$-graded ring is equivalently an affine scheme together with an action of the affine line $\mathbb{A}^1$, viewed as a multiplicative monoid in the category of schemes. This equivalence extends to an equivalence of the slightly larger categories of $\IZ$-graded rings vs. the (opposite) category of affine schemes with an action of the multiplicative group schemes $\IG_m$ ($\IG_m$-schemes for short). If one wants to understand this equivalence between graded rings and schemes with an action, working with $\IZ$-graded rings and affine $\IG_m$-schemes is a bit easier, because groups are easier to work with than monoids.

The trick, which is useful in general, is to observe that the forgetful functor from affine $\IG_m$-schemes to affine schemes has a left adjoint which maps an affine scheme to the free affine $\IG_m$-scheme $\IG_m \times X$ on $X$. If $X$ is an affine $\IG_m$-scheme then the unit of the adjunction $\IG_m \times X \to X$ is given by the action morphism of $X$. We can reduce many questions about the affine scheme with $\IG_m$-action $X$ to the free action of $\IG_m \times X$. Since the coordinate ring of $\IG_m \times X$ is given by the ring of Laurent polynomials $\sheaf{O}(X)[T^{\pm1}]$ this case is relatively easy to handle.

The Functorial View on Affine Schemes

A $\IZ$-functor is a functor $X \from \Ring \to \Set$. If $R$ is a ring, then we call an element $x \in X(R)$ an $R$-valued point of $X$ and write $x \in_R X$. If $F \from Y \to X$ is a morphism (natural transformation) of $\IZ$-functors and $y \in_R Y$, then $F(y) \in_R X$.

The most important $\IZ$-functor is the affine line $\IA^1$ given by $\IA^1(R) := R$. If $X$ is any $\IZ$-functor, then a function on $X$ is a morphism $f \from X \to \IA^1$. We write $\sheaf{O}(X)$ for the set (ignoring some size issues) of functions on $X$. In fact, $\sheaf{O}(X)$ carries the structure of a ring, with the (ring-wise and) point-wise addition and multiplication of functions. If $F \from Y \to X$ is a morphism of $\IZ$-functors, then there is an induced ring homomorphism $F^* \from \sheaf{O}(X) \to \sheaf{O}(Y)$, $f \mapsto F^*(f) := f \circ F$.

If $X$ is a $\IZ$-functor and $x \in_R X$, then there is an induced ring homomorphism $\widetilde{x} \from \sheaf{O}(X) \to R$, $f \mapsto f(x)$ called the character associated to $x$. The $\IZ$-functor $X$ is called an affine scheme if for every ring $R$ the assignment $X(R) \to \Hom(\sheaf{O}(X), R)$, $x \mapsto \widetilde{x}$ is a bijection.

If $A$ is any ring, then there is an affine scheme $\Spec(A)$ called the spectrum of $A$ where $\Spec(A)(R) = \Hom(A, R)$, i.e. the $R$-valued points of $\Spec(A)$ are the $R$-valued characters on $A$.

The affine line $\IA^1$ is an affine scheme. If $X, Y$ are $\IZ$-functors, then $X \times Y$ is the $\IZ$-functor such that $(X \times Y)(R) = X(R) \times Y(R)$. If $X$ and $Y$ are affine schemes, then so is $X \times Y$.

The Multiplicative Group Scheme

We let $\IG_m$ be the $\IZ$-functor given by $\IG_m(R) = R^\units$ for every ring $R$. Every function $f \in \sheaf{O}(\IG_m) := \Hom(\IG_m, \IA^1)$ has the form $f(t) = \sum_{i \in \Z} t^i a_i$ for unique constants $a_i \in \Z$, almost all of which vanish.

More generally, if $X$ is an affine scheme, then every function $f \in \sheaf{O}(\IG_m \times X)$ has the form $f(t, x) = \sum_{i \in \IZ} t^i f_i(x)$ for unique functions $f_i \in \sheaf{O}(X)$, almost all of which vanish.

An other way to phrase this is to say that the morphism $\sheaf{O}(X)[T^{\pm1}] \to \sheaf{O}(\IG_m \times X)$ mapping functions $f \in \sheaf{O}(X)$ to $\pi_2^*(f)$ and mapping $T$ to the function $\IG_m \times X \xrightarrow{\pi_1} \IG_m \to \IA^1$ is an isomorphism.

Affine $\IG_m$-schemes

Let $X$ be an affine scheme and let $\rho \from \IG_m \times X \to X$ be a morphism of schemes. For $(t, x) \in_R \IG_m \times X$ we write $t \cdot x := \rho(t, x) \in_R X$. We say that $\rho$ defines an action of $\IG_m$ on $X$ if we have $1 \cdot x = x$ for all $x \in_R X$ and $s \cdot (t \cdot x) = (s \cdot t) \cdot x$ for all $s, t \in_R \IG_m$, $x \in_R X$. (This can also be described as the equality of two morphisms $\IG_m \times \IG_m \times X \rightrightarrows X$.) We call the pair $(X, \rho)$ an affine $\IG_m$-scheme and agree to write simply $X$ instead of $(X, \rho)$.

If $X$ and $Y$ are affine $\IG_m$-schemes, a morphism $F \from X \to Y$ of affine $\IG_m$-schemes is a morphism of affine schemes which satisfies $F(t \cdot y) = t \cdot F(y)$ for all $t \in_R \IG_m$ and $y \in_R Y$.

$\IG_m \times X$ as an affine $\IG_m$-scheme

Let $X$ be any affine scheme. Then $\IG_m \times X$ is an affine $\IG_m$-scheme with the action of $\IG_m$ given by $s \cdot (t, x) := (st, x)$ for $s, t \in_R \IG_m$ and $x \in_R X$.

Suppose that $X$ already carries the structure of an affine $\IG_m$-scheme with action given by $\rho \from \IG_m \times X \to X$. We view $\IG_m \times X$ as an affine $\IG_m$-scheme as described above. Then the action morphism $\rho \from \IG_m \times X \to X$ is a morphism of affine $\IG_m$-schemes. Indeed, we have $\rho(s \cdot (t, x)) = \rho(st, x) = (st) x = s(tx) = s \cdot \rho(t, x)$.

Homogeneous Functions

Let $X$ be an affine $\IG_m$-scheme. We call a function $f \in \sheaf{O}(X)$ homogeneous of degree $d \in \IZ$ if we have $f(t \cdot x) = t^d \cdot f(x)$ for all $(t, x) \in_R \IG_m \times X$. We write $\sheaf{O}(X)_d \subset \sheaf{O}(X)$ for the set of all homogeneous functions of degree $d$ on $X$. The set $\sheaf{O}(X)_d \subset \sheaf{O}(X)$ is obviously an additive subgroup. Furthermore, if $f \in \sheaf{O}(X)_d$ and $g \in \sheaf{O}(X)_e$, then we have $(fg)(tx) = f(tx) g(tx) = t^df(x) t^eg(x) = t^{d + e}(fg)(x)$, i.e. $f \cdot g \in \sheaf{O}(X)_{d + e}$.

Let $F \from Y \to X$ be a morphism of affine $\IG_m$-schemes and let $f \in \sheaf{O}(X)_d$ be a homogeneous function. Then for $t \in_R \IG_m$ and $y \in_R Y$ we have $F^*(f)(ty) = f(F(ty)) = f(t F(y)) = t^df(F(y)) = t^d F^*(f)(y)$, i.e. $F^*(f) \in \sheaf{O}(Y)_d$.

Lemma 1.

1. Let $X$ be an affine scheme and view $\IG_m \times X$ as an affine $\IG_m$-scheme as above. Let $f \in \sheaf{O}(\IG_m \times X)$ be a function on $\IG_m \times X$ and let $f_i \in \sheaf{O}(X)$ be such that $f(t, x) = \sum_{i \in \IZ} t^i f_i(x)$ for all $(t, x) \in_R \IG_m \times X$ (recall). Then $f$ is homogeneous of degree $d$ if and only if $f_i = 0$ for $i \neq d$ in which case we have $f(t, x) = t^d f_d(x)$ for $t \in_R \IG_m$, $x \in_R X$. (In other words, the degree $d$ homogenous functions on $\IG_m \times X$ are the degree $d$ homogeneous Laurant polynomials of $\sheaf{O}(\IG_m \times X) \cong \sheaf{O}(X)[T^{\pm 1}]$.)

2. Let $X$ be an affine $\IG_m$-scheme. We view $\IG_m \times X$ as an affine $\IG_m$-scheme and $\rho \from \IG_m \times X \to X $ as a morphism of affine $\IG_m$-schemes as above. Then a function $f \in \sheaf{O}(X)$ is homogeneous of degree $d$ if and only if the function $\rho^*(f) \in \sheaf{O}(\IG_m \times X)$ is homogeneous of degree $d$ in which case we have $\rho^*(f)(t, x) = t^d f(x)$ for $t \in_R \IG_m$, $x \in_R X$.

Proof. If we have $f(t, x) = t^d f_d(x)$ with $f_d \in \sheaf{O}(X)$, then we have $f(s \cdot (t, x)) = f(st, x) = (st)^d f_d(x) = s^d(t^d f_d(x)) = s^d f(t, x)$ which shows that $f$ is homogeneous of degree $d$. Conversely, suppose that $f(t, x) = \sum_{i \in \IZ} t^i f_i(x)$ and $f$ is homogeneous of degree $d$. Then for all $(t, x) \in_R \IG_m \times X$ we have $$ \begin{split} \sum_{i \in \IZ} t^i f_i(x) &= f(t, x) \\ &= f(t \cdot (1, x)) \\ &= t^d f(1, x) \end{split} $$ Comparing coefficients we see that $f_i = 0$ for $i \neq d$ while $f_d(x) = f(1, x)$.

Now we prove the second statement. If $f \in \sheaf{O}(X)$ is homogeneous of degree $d$, then the pull-back $\rho^*(f)$ is homogeneous of degree $d$ by the above. So assume that $\rho^*(f)$ is homogeneous of degree $d$. By the first statement there is a function $g \in \sheaf{O}(X)$ such that $\rho^*(f)(t, x) = t^d g(x)$ for $(t, x) \in_R \IG_m \times X$. In other words, we have $f(t \cdot x) = t^d g(x)$. Plugging in $t = 1$ we see that $f(x) = g(x)$ for all $x \in_R X$, i.e. $f = g$. Now the equation $f(t \cdot x) = t^d f(x)$ shows exactly that $f$ is homogeneous of degree $d$. ∎

Pregraded and Graded Rings

Let $A$ be a ring. Suppose that for each $d \in \IZ$ there is an additive subgroup $A_d \subset A$ and that for $d, e \in \IZ$ we have $A_d \cdot A_e \subset A_{d + e}$. Then we call $A$ together with the family of subgroups $\{A_d\}_{d \in \IZ}$ a pregraded ring. A morphism $\varphi \from A \to B$ of pregraded rings is a ring homomorphism such that $\varphi(A_d) \subset B_d$ for all $d \in \IZ$.

We have seen in the paragraph above that the coordinate ring of an affine $\IG_m$-scheme is canonically pregraded and that if $F \from Y \to X$ is a morphism of affine $\IG_m$-schemes, then $F^*(f) \from \sheaf{O}(X) \to \sheaf{O}(Y)$ is a morphism of pregraded rings.

We call a pregraded ring $A$ a graded ring if $A$ is the internal direct sum $A = \bigoplus_{i \in \IZ} A_i$ of its homogeneous components $A_i$.

Proposition 2. Let $X$ be an affine $\IG_m$-scheme. Then the ring $\sheaf{O}(X)$ with the induced pregrading defined in the section above is a graded ring.

Proof. Let $f \in \sheaf{O}(X)$ be a function on $X$. Using that $\sheaf{O}(\IG_m \times X) \cong \sheaf{O}(X)[T^{\pm 1}]$, we find functions $f_i \in \sheaf{O}(X)$ such that $f(t \cdot x) = \rho^*(f)(t, x) = \sum_{i \in \IZ} t^i f_i(x)$ for $(t, x) \in_R \IG_m \times X$. Pluggin in $t = 1$, we find $f(x) = \sum_{i \in \IZ} f_i(x)$, i.e. $f = \sum_{i \in IZ} f_i$. Our goal is to show that the $f_i$ are homogeneous of degree $i$. For $s, t \in_R \IG_m$ and $x \in_R X$ we have $$ \begin{split} \sum_{i \in \IZ} s^i t^i f_i(x) &= \sum_{i \in IZ} (st)^i f_i(x) \\ &= f((st)x) \\ &= f(s(tx)) \\ &= \sum_{i \in IZ} s^i f_i(tx). \end{split} $$ Using the unique representation of functions on $\IG_m \times (\IG_m \times X)$ and comparing coefficients we find that for each $i \in \IZ$ we have $t^i f_i(x) = f_i(tx)$ which is what we wanted to show.

Now we want to show that if we have $f = \sum_{i \in \IZ} f_i$ with $f_i \in \sheaf{O}(X)$, then the $f_i$ are uniquely determined by $f$, or in other words that $f = 0$ implies $f_i = 0$ for all $i \in \IZ$. Using that the $f_i$ are homogeneous and that $f = \sum_{i \in \IZ} f_i$ we get $\rho^*(f)(t, x) = \sum_{i \in \IZ} f_i(tx) = \sum_{i \in \IZ} t^i f_i(x)$ for $(t, x) \in_R \IG_m \times X$. Since $\sheaf{O}(\IG_m \times X) \cong \sheaf{O}(X)[T^{\pm1}]$, it follows that the $f_i$ are uniquely determined by $\rho^*(f)$ and therefore by $f$. ∎

How to Recover the Action from the Grading

Let $X$ be an affine $\IG_m$-scheme. Recall that a point $x \in_R X$ is completely determined by its character $\widetilde{x} \from \sheaf{O}(X) \to R$ given by $\widetilde{x}(f) := f(x)$. Now let $x \in_R X$ and $t \in_R \IG_m$. If we write $f = \sum_{i \in \IZ} f_i$ with $f_i \in \sheaf{O}(X)_i$, then we have $$ \begin{split} \widetilde{tx}(f) &= f(tx) \\ &= \sum_{i \in \IZ} f_i(tx) \\ &= \sum_{i \in \IZ} t^i f_i(x). \end{split} $$ This shows that the action of $\IG_m$ on $X$ is determined completely by the induced grading on $\sheaf{O}(X)$.

The Affine $\IG_m$-Scheme Associated to a Graded Ring

Let $A$ be a graded ring. Recall that $\Spec(A)$ is an affine scheme with $\Spec(A)(R) = \Hom(A, R)$ the space of $R$-valued characters on $A$.Given $\chi \in_R \Spec(A)$ and $t \in_R \IG_m$, we define $t \cdot \chi \in_R \Spec(A)$ by $(t \cdot \chi)(\sum_{i \in \IZ} f_i) = \sum_{i \in \IZ} t^i \chi(f_i)$ where $f = \sum_{i \in \IZ} f_i$ with $f_i \in A_i$. It is clear that $t \cdot \chi \from A \to R$ is compatible with addition. If $f_i \in A_i$ and $g_j \in A_j$ then we have $(t \cdot \chi)(f_i g_j) = t^i \chi(f_i) t^j \chi(g_j) = t^{i + j} \chi(f_i g_j)$. Using linearity it follows that $t \cdot \chi \from A \to R$ is indeed a character. If $s, t \in_R \IG_m$ and $\chi \in_R \Spec(A)$ then we have $$ \begin{split} (s(t\chi))(f) &= \sum_{i \in IZ} s^i(t\chi)(f_i) \\ &= \sum_{i \in \IZ} s^i t^i \chi(f_i) \\ &= \sum_{i \in \IZ} (st)^i \chi(f_i) \\ &= ((st)\chi)(f). \end{split} $$ This shows that the morphism $\IG_m \times \Spec(A) \to \Spec(A)$ given by $(t, \chi) \mapsto t \cdot \chi$ equips $\Spec(A)$ with the structure of an affine $\IG_m$-scheme.

Lemma 3. Let $A$ be a graded ring. Denote by $\rho \from \IG_m \times \Spec(A) \to \Spec(A)$ the action $(t, \chi) \mapsto t \cdot \chi$ described above. Recall that an element $f \in A$ induces a function $\widetilde{f} \from \Spec(A) \to \IA^1$ given by $\widetilde{f}(\chi) := \chi(f)$ for $\chi \in_R \Spec(A)$ and that this induces an isomorphism of rings $\varphi \from A \to \sheaf{O}(\Spec(A))$, $f \mapsto \widetilde{f}$.

Denote by $\pi_2 \from \IG_m \times \Spec(A) \to \Spec(A)$ the second projection. By our remarks earlier, there is an isomorphism of rings $\psi \from A[T^{\pm1}] \to \sheaf{O}(\IG_m \times \Spec(A))$ such that $\psi(\sum_{i \in \IZ} f_i T^i) = \sum_{i \in \IZ} \widetilde{T}^i \pi_2^*(\widetilde{f}_i)$ where $\widetilde{T} \in \sheaf{O}(\IG_m \times \Spec(A))$ is given by $\widetilde{T}(t, \chi) := t \in_R \IA^1$ for $(t, \chi) \in_R \IG_m \times \Spec(A)$.

Now finally for the claim: Define $\omega \from A \to A[T^{\pm1}]$ by $\omega(\sum_{i \in \IZ} f_i) = \sum_{i \in \IZ} T^i f_i$ for $f_i \in A_i$. Then $\omega$ is a ring homomorphism and the diagram $$ \begin{CD} A @>\varphi>> \sheaf{O}(\Spec(A))\\ @VV\omega V @VV\rho^*V\\ A[T^{\pm1}] @>\psi>> \sheaf{O}(\IG_m \times \Spec(A)) \end{CD} $$ commutes.

Proof. Since $\varphi$ and $\psi$ are isomorphisms and $\rho^*$ is a homomorphism it will be enough to show the commutativity of the diagram to conclude that $\omega$ is a homomorphism. Given $f = \sum_{i \in \IZ} f_i \in A$ with $f_i \in A_i$ we have on the one hand $$ \begin{split} (\rho^*(\varphi(f))(t, \chi) &= \rho^*(\widetilde{f})(t, \chi) \\ &= \widetilde{f}(t \cdot \chi) \\ &= (t \cdot \chi)(f) \\ &= \sum_{i \in \IZ} t^i \chi(f_i) \end{split} $$ for $(t, \chi) \in \IG_m \times \Spec(A)$. Here the last equation is due to the definition of the action.

On the other hand we have $$ \begin{split} (\psi(\omega(f))(t, \chi) &= (\psi(\sum_{i \in \IZ} f_i T^i))(t, \chi) \\ &= (\sum_{i \in \IZ} \widetilde{T}^i \pi_2^*(\widetilde{f_i}))(t, \chi) \\ &= \sum_{i \in \IZ} t^i \widetilde{f_i}(\chi) \\ &= \sum_{i \in \IZ} t^i \chi(f_i). \end{split} $$ This shows that $\rho^*(\varphi(f)) = \psi(\omega(f))$ which finishes the proof. ∎

Lemma 4. Let $A$ be a graded ring. View $\Spec(A)$ as an affine $\IG_m$-scheme as explained above. Recall that an element $f \in A$ induces a function $\widetilde{f} \from \Spec(A) \to \IA^1$ given by $\widetilde{f}(\chi) := \chi(f)$ for $\chi \in_R \Spec(A)$. We claim that the function $\widetilde{f}$ is homogeneous of degree $d$ if and only if $f \in A_d$.

Proof. We remark that we can view $A[T^{\pm1}]$ as a graded ring where $(A[T^{\pm1}])_d$ consists of all $\sum_{i \in \IZ} f_i T^i$ where $f_i = 0$ for $i \neq d$. Then the isomorphism $\psi \from A[T^{\pm1}] \to \sheaf{O}(\IG_m \times \Spec(A))$ from above is an isomorphism of graded rings by the first item of Lemma 1. By construction of the map $\omega$ in Lemma 3 we have that $f \in A$ lies in $A_d$ if and only if $\omega(f)$ lies in $(A[T^{\pm1}])_d$. This is the case if and only if $\psi(\omega(f)) = \rho^*(\varphi(f)) \in \sheaf{O}(\IG_m \times \Spec(A))$ is homogeneous of degree $d$. By Item 2 of Lemma 1 this is the case if and only if $\varphi(f) = \widetilde{f} \in \sheaf{O}(\Spec(A))$ is homogeneous of degree $d$. ∎

The Equivalence

Let $X$ be an affine $\IG_m$-scheme. Equip $\sheaf{O}(X)$ with the grading where $\sheaf{O}(X)_d$ consists of homogeneous functions of degree $d$. Now we can view $\Spec(\sheaf{O}(X))$ as an affine $\IG_m$-scheme as explained in the section above. The canonical isomorphism of affine schemes $\eta \from X \to \Spec(\sheaf{O}(X))$ which maps a point $x \in_R X$ to its character $\widetilde{x} \in_R \Spec(\sheaf{O}(X))$ is an isomorphism of $\IG_m$-schemes as follows from the computation in this section above.

Conversely suppose that $A$ is a graded ring. Equip $\Spec(A)$ with the structure of an affine $\IG_m$-scheme as in the section above. Then view $\sheaf{O}(\Spec(A))$ as a graded ring with the induced grading. We have a canonical isomorphism of rings $\epsilon \from A \to \sheaf{O}(\Spec(A))$ where an element $f \in A$ is mapped to the function $\widetilde{f} \from \Spec(A) \to \IA^1$ given by $\widetilde{f}(\chi) = \chi(f)$ for $\chi \in_R \Spec(A)$. By Lemma 4 this map is an isomorphism of graded rings.

We can summarize that we have constructed an adjoint equivalence of categories between the category of affine $\IG_m$-schemes and the opposite category of the category of ($\IZ$-)graded rings.