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.
 A: 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. 
