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Let $\mathbf{Q}_p \langle X_1,\dots,X_n \rangle$ be the $n$-variable Tate algebra, i.e. the subalgebra of $\mathbf{Q}_p[[X_1,\dots ,X_n]]$ of power series which converge on the closed unit polydisk in $\mathbf{C}_p^n$. Is $\mathbf{Q}_p \langle X_1, X_2 \rangle$ projective as a $\mathbf{Q}_p \langle X_1 \rangle$-module? Note that $\mathbf{Q}_p \langle X_1 \rangle$ is a regular ring of dimension one.

I am well aware that this is the "wrong question", since it's a mild (severe?) sin to forget the topological structures of the objects in question, but the answer is "yes" in a suitable category of Banach $\mathbf{Q}_p \langle X_1 \rangle$-modules, so I wonder...

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  • $\begingroup$ By Bass' theorem, if it is projective it is free, but I'm not sure this helps much. Do you know the answer for rings of power series? $\endgroup$
    – Jérôme Poineau
    Commented Feb 17, 2012 at 9:58
  • $\begingroup$ For ring a power series, it actually follows from Konstantin Ardakov's answer. $\endgroup$
    – Jérôme Poineau
    Commented Feb 18, 2012 at 10:15

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I would guess not. Here is some evidence towards this guess.

Let $S$ be a $p$-adically complete ring of characteristic $0$ and let $M$ be $p$-adically complete $S$-module which is free over $S$. Then I claim that $M$ must be finitely generated over $S$.

To see this, suppose that $X$ is an infinite free generating set for $M$ over $S$: $M = \bigoplus_{x \in X} Sx$. Choose some infinite sequence $x_0, x_1, \cdots$ of elements in $X$. Then the partial sums $\sum_{i=0}^n p^i x_i$ form a Cauchy sequence in the $p$-adic topology of $M$ but do not converge to an element of $M$, a contradiction.

So $\mathbb{Z}_p \langle x,y \rangle$ cannot be free over $\mathbb{Z}_p \langle x \rangle$.

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  • $\begingroup$ Isn't $S := \mathbb{Q}_p$ and $M := \mathbb{C}_p$ (completion of the algebraic closure of $\mathbb{Q}_p$) a counterexample for the assertion that $M$ must be finitely generated ? $\endgroup$
    – Ralph
    Commented Feb 17, 2012 at 1:18
  • $\begingroup$ @Konstantin Ardakov: How do you know that the series does not converge to an element of M? $\endgroup$
    – Jérôme Poineau
    Commented Feb 17, 2012 at 9:43
  • $\begingroup$ Ralph: your $S$ is not $p$-adically complete. "$R$ is $p$-adically complete" means "$R$ is the projective limit of $R/p^nR$" so $\mathbf{Z}_p$ is fine but $\mathbf{Q}_p$ isn't. $\endgroup$
    – Kevin Buzzard
    Commented Feb 17, 2012 at 10:04
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    $\begingroup$ @Jerome Poineau: suppose for a contradiction it converges to some limit $m \in M$. Then $m = \sum_{i=1}^r s_iy_i$ for some finite collection of elements $y_1,\ldots,y_r$ in $X$. Now the sequence converges to $\sum_{i=0}^r p^ix_i$ in $M / p^{r+1}M$ so $\sum_{i=1}^r s_iy_i \equiv \sum_{i=0}^rp^ix_i \mod p^{r+1}M$. Equate coefficients: the left hand side has at most $r$ non-zero entries in $S/ p^{r+1}S$, whereas the right hand side has $r+1$ non-zero entries. $\endgroup$
    – user91132
    Commented Feb 17, 2012 at 11:22
  • $\begingroup$ @AK: I'm sorry, but I still don't get it. How do you equate coefficients? Are you saying that $X$ is free in $M/p^{r+1}M$? $\endgroup$
    – Jérôme Poineau
    Commented Feb 17, 2012 at 12:13
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Let me try to adapt Konstantin Ardakov's answer to the case of Tate algebras over a valued field $k$.

If $k$ is trivially valued, then $k\langle T_1,T_2\rangle = k[T_1,T_2]$ is free over $k\langle T_1\rangle = k[T_1]$.

If $k$ is not trivially valued, there exists an element $\alpha \in k$ with $0 < |\alpha|<1$. If $k\langle T_1,T_2\rangle$ were projective over $k\langle T_1\rangle$, it would be free by Bass' theorem on big projective modules. So there would be a basis $(x_1,\dots)$. By mutiplying by powers of $\alpha$, we may assume that the $x_i$'s have norm at most 1. The series $\sum_{n\ge 1} \alpha^n T_1^n x_n$ is Cauchy, hence convergent in $k\langle T_1,T_2\rangle$. Now we can reduce modulo $T_1^m$ for $m$ big enough to get a contradiction just as in Konstantin Ardakov's answer.

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