# universal finite differential module of affinoid algebra

Let $k$ be a value field (archimedean), for example $k = \mathbb{Q}_p$, the p-adic field. The free Tate algebra is $$T_n := \left\{ \ \sum a_I X^I, \ a_I \in k, \ a_I \rightarrow 0 \text{ as } |I| \rightarrow \infty \ \right\}$$.

I want to compute the universal finite differential module $\Omega^f_{T_n/k}$ of $T_n$ over $k$. For any $A$-algebra $B$, the "universal finite differential module" $\Omega^f_{B/A}$ is a finitely generated $B$-module with an $A$-derivation $d : B \rightarrow \Omega^f_{B/A}$ such that for any $A$-derivation $d^{'} : B \rightarrow M$ with $M$ being a finite generated $B$-module, there exists a $B$-module homomorphism $\phi : \Omega^f_{B/A} \rightarrow M$ such that $d^{'} = \phi \circ d$. It doesn't always exist.

I would like to show that $\Omega^f_{T_n/k}$ is the free $T_n$ module of rank $n$. Let me explan the case $n=1$. Similar to the case of formal power series ring $k[[ X ]]$, given a $k$-derivation $d_1 : T_1 \rightarrow M$ to a finitely genterated $T_1$-module $M$, we definte $\phi : T_1 * dX \rightarrow M$ by sending $dX$ to $d_1 X$ and extend it $T_1$-linearly. We want to show that $d_2 := d_1 - \phi \circ d = 0$. $d_2$ is still a $k$-derivation and one shows that $d_2 (f) = 0$ if $f \in k[X]$. In the case of $k[[X]]$, we know for any $f \in k[[X]]$, $d_2 (f) \in (X^r)M$ for any $r > 0$. By Krull Intersection Theorem, we know there is a $g \in (X)$ such that $(1-g) N = 0$, here $N := \cap_{ r > 0 } (X^r)M$. Since for $g \in (X)$, $1-g$ is invertible in $k[[X]]$, we know $N = 0$ and hence $d_2 = 0$.

But in the case of $T_1$, for $g \in (X)$, $1-g$ may not be invertible in $T_1$. For example, $g=X$, its inverse is $1+X+X^2+ \cdot \cdot \cdot$ which is not in $T_1$.

So the question is how to prove $d_2 = 0$. In p.64 of the book Rigid analytic geometry and its applications by Jean Fresnel and Marius van der Put, there is a proof. Instead of considering only the ideal (X), they consider any maximal ideal $m$ of $T_1$. They said that any maximal ideal $m$ is generated by polynomials ( It's ok.), hence $d_2 ( T_1)$ is contained in $m^r M$ for any $r > 0$ for which I don't understand why.

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I got an idea. Let $\overline{k}$ be the algebraic closure of $k$ and extend everything to be over $\overline{k}$, i.e extend the $k$-derivations $d_2$ to $D_2: T_n \otimes_{k}\overline{k} \rightarrow M \otimes_{k} \overline{k}$. Since the natural morphism $M \rightarrow M \otimes_{k} \overline{k}$ is injective, in order to show that $d_2 = 0$, we only need to show that $D_2 = 0$. So we can assume that $k$ is algebraically closed at the beginning.

One can show that the residue field $T_n/m$ is a finite extension of $k$ for any maximal $m$. In our case, since $k$ is algebraically closed, we have $T_n/m = k$. It follows that any maximal ideal $m$ is of the form $m = ( X - \lambda )$ for some $\lambda \in k$ ( with $| \lambda | < 1$) . For any $f \in T_1$ and any $r > 0$, we can write $f$ as $f = \sum_{i=0}^{r} \ a_i (X - \lambda)^i + (X - \lambda )^{r+1} g(X)$ with $a_i \in k$ and $g(X) \in T_1$. Using the rules of derivations and the fact that $d_2$ is zero on polynomials, we get $d_2 (f) \in m^{r+1}M$. It follows that $d_2 (f) \in \cap_{r>0} I^rM$, here $I = \cap_{m \in \text{Max}(T_1)} m$ is the Jacobson radical. By Krull Intersection Theorem, we get the result.

I apology for answering my own question. It will be appreciated for other solutions.

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