Lemma 90.29.4. With notation and assumptions as in Situation 90.29.1. Assume $\mathcal{F}$ is a deformation category. Then there is a canonical $l$-vector space isomorphism
of infinitesimal automorphism spaces.
Lemma 90.29.4. With notation and assumptions as in Situation 90.29.1. Assume $\mathcal{F}$ is a deformation category. Then there is a canonical $l$-vector space isomorphism
of infinitesimal automorphism spaces.
Proof. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ and denote $x_{l, 0}$ the corresponding object of $\mathcal{F}_{l/k}$ over $l$. Recall that $\text{Inf}(\mathcal{F}) = \text{Inf}_{x_0}(\mathcal{F})$ and $\text{Inf}(\mathcal{F}_{l/k}) = \text{Inf}_{x_{l, 0}}(\mathcal{F}_{l/k})$, see Remark 90.19.4. Recall that the vector space structure on $\text{Inf}_{x_0}(\mathcal{F})$ comes from identifying it with the tangent space of the functor $\mathit{Aut}(x_0)$ which is defined on the category $\mathcal{C}_{k, k}$ of Artinian local $k$-algebras with residue field $k$. Similarly, $\text{Inf}_{x_{l, 0}}(\mathcal{F}_{l/k})$ is the tangent space of $\mathit{Aut}(x_{l, 0})$ which is defined on the category $\mathcal{C}_{l, l}$ of Artinian local $l$-algebras with residue field $l$. Unwinding the definitions we see that $\mathit{Aut}(x_{l, 0})$ is the restriction of $\mathit{Aut}(x_0)_{l/k}$ (which lives on $\mathcal{C}_{k, l}$) to $\mathcal{C}_{l, l}$. Since there is no difference between the tangent space of $\mathit{Aut}(x_0)_{l/k}$ seen as a functor on $\mathcal{C}_{k, l}$ or $\mathcal{C}_{l, l}$, the lemma follows from Lemma 90.29.3 and the fact that $\mathit{Aut}(x_0)$ satisfies (RS) by Lemma 90.19.6 (whence we have (S2) by Lemma 90.16.6). $\square$
Your email address will not be published.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)