Lemma 93.7.3. In Example 93.7.1 let $P$ be a graded $k$-algebra. Then
are finite dimensional if $P$ is finitely generated over $k$.
Lemma 93.7.3. In Example 93.7.1 let $P$ be a graded $k$-algebra. Then
are finite dimensional if $P$ is finitely generated over $k$.
Proof. We first deal with the infinitesimal automorphisms. Let $Q = P \otimes _ k k[\epsilon ]$. Then an element of $\text{Inf}(\mathcal{D}\! \mathit{ef}_ P)$ is given by an automorphism $\gamma = \text{id} + \epsilon \delta : Q \to Q$ as above where now $\delta : P \to P$. The fact that $\gamma $ is graded implies that $\delta $ is homogeneous of degree $0$. The fact that $\gamma $ is $k$-linear implies that $\delta $ is $k$-linear. The fact that $\gamma $ is multiplicative implies that $\delta $ is a $k$-derivation. Conversely, given a $k$-derivation $\delta : P \to P$ homogeneous of degree $0$, we obtain an automorphism $\gamma = \text{id} + \epsilon \delta $ as above. Thus we see that
as predicted in the lemma. Clearly, if $P$ is generated in degrees $P_ i$, $0 \leq i \leq N$, then $\delta $ is determined by the linear maps $\delta _ i : P_ i \to P_ i$ for $0 \leq i \leq N$ and we see that
as desired.
To finish the proof of the lemma we show that there is a finite dimensional deformation space. To do this we choose a presentation
of graded $k$-algebras where $\deg (X_ i) = d_ i$ and $F_ j$ is homogeneous of degree $e_ j$. Let $Q$ be any graded $k[\epsilon ]$-algebra finite free in each degree which comes with an isomorphsm $\alpha : Q/\epsilon Q \to P$ so that $(Q, \alpha )$ defines an element of $T\mathcal{D}\! \mathit{ef}_ P$. Choose a homogeneous element $q_ i \in Q$ of degree $d_ i$ mapping to the image of $X_ i$ in $P$. Then we obtain
and since $P = Q/\epsilon Q$ this map is surjective by Nakayama's lemma. A small diagram chase shows we can choose homogeneous elements $F_{\epsilon , j} \in k[\epsilon ][X_1, \ldots , X_ n]$ of degree $e_ j$ mapping to zero in $Q$ and mapping to $F_ j$ in $k[X_1, \ldots , X_ n]$. Then
is a presentation of $Q$ by flatness of $Q$ over $k[\epsilon ]$. Write
There is some ambiguity in the vector $(G_1, \ldots , G_ m)$. First, using different choices of $F_{\epsilon , j}$ we can modify $G_ j$ by an arbitrary element of degree $e_ j$ in the kernel of $k[X_1, \ldots , X_ n] \to P$. Hence, instead of $(G_1, \ldots , G_ m)$, we remember the element
where $g_ j$ is the image of $G_ j$ in $P_{e_ j}$. Moreover, if we change our choice of $q_ i$ into $q_ i + \epsilon p_ i$ with $p_ i$ of degree $d_ i$ then a computation (omitted) shows that $g_ j$ changes into
We conclude that the isomorphism class of $Q$ is determined by the image of the vector $(G_1, \ldots , G_ m)$ in the $k$-vector space
In this way we see that we obtain an injection
Since $W$ visibly has finite dimension, we conclude that the lemma is true. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)