The Stacks project

Lemma 46.3.7. Let $A$ be a ring. Let $F$ be a module-valued functor on $\textit{Alg}_ A$. Assume that for $B \to B'$ flat the map $F(B) \otimes _ B B' \to F(B')$ is an isomorphism. Let $B$ be a graded $A$-algebra. Then

  1. $F(B) = \bigoplus _{k \in \mathbf{Z}} F(B)^{(k)}$, and

  2. the map $B \to B_0 \to B$ induces map $F(B) \to F(B)$ whose image is contained in $F(B)^{(0)}$.

Proof. Let $x \in F(B)$. The map $p : B \to B[t, t^{-1}]$ is free hence we know that

\[ F(B[t, t^{-1}]) = \bigoplus \nolimits _{k \in \mathbf{Z}} F(p)(F(B)) \cdot t^ k = \bigoplus \nolimits _{k \in \mathbf{Z}} F(B) \cdot t^ k \]

as indicated we drop the $F(p)$ in the rest of the proof. Write $F(a)(x) = \sum t^ k x_ k$ for some $x_ k \in F(B)$. Denote $\epsilon : B[t, t^{-1}] \to B$ the $B$-algebra map $t \mapsto 1$. Note that the compositions $\epsilon \circ p, \epsilon \circ a : B \to B[t, t^{-1}] \to B$ are the identity. Hence we see that

\[ x = F(\epsilon )(F(a)(x)) = F(\epsilon )(\sum t^ k x_ k) = \sum x_ k. \]

On the other hand, we claim that $x_ k \in F(B)^{(k)}$. Namely, consider the commutative diagram

\[ \xymatrix{ B \ar[r]_ a \ar[d]_{a'} & B[t, t^{-1}] \ar[d]^ f \\ B[s, s^{-1}] \ar[r]^-g & B[t, s, t^{-1}, s^{-1}] } \]

where $a'(b) = s^{\deg (b)}b$, $f(b) = b$, $f(t) = st$ and $g(b) = t^{\deg (b)}b$ and $g(s) = s$. Then

\[ F(g)(F(a'))(x) = F(g)(\sum s^ k x_ k) = \sum s^ k F(a)(x_ k) \]

and going the other way we see

\[ F(f)(F(a))(x) = F(f)(\sum t^ k x_ k) = \sum (st)^ k x_ k. \]

Since $B \to B[s, t, s^{-1}, t^{-1}]$ is free we see that $F(B[t, s, t^{-1}, s^{-1}]) = \bigoplus _{k, l \in \mathbf{Z}} F(B) \cdot t^ ks^ l$ and comparing coefficients in the expressions above we find $F(a)(x_ k) = t^ k x_ k$ as desired.

Finally, the image of $F(B_0) \to F(B)$ is contained in $F(B)^{(0)}$ because $B_0 \to B \xrightarrow {a} B[t, t^{-1}]$ is equal to $B_0 \to B \xrightarrow {p} B[t, t^{-1}]$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06UZ. Beware of the difference between the letter 'O' and the digit '0'.