# Notes (part 2): S. Bosch, U Güntzer and R. Remmert, Non-Archimedean Analysis


• $$A^{(\infty)} := \bigoplus_{i=1}^\infty A$$
• the set of zero sequences, $$c(A) := \{(c_i): \lim_{i \to \infty} |c_i| = 0\}$$
• the set of bounded sequences, $$b(A) := \{(b_i)_i: \sup_i |b_i| < \infty\}$$

It is clear that $$A^{(\infty)} \subseteq c(A) \subseteq b(A).$$ If $$A$$ is a ring/algebra over a field/module over a ring, each of these sets inherits the structure from $$A.$$ We would always consider each of these sets equipped with the component-wise supremum norm:
$|(b_i)_i| := \sup_i |b_i|$
If $$A$$ is a normed ring (respectively, field), this makes these sets normed $$A$$-modules (respectively, vector spaces). If $$A$$ is complete, then so are $$b(A)$$ and $$c(A),$$ but not $$A^{(\infty)}.$$

## Examples/Facts:

1. Consider a valued field $$K$$ equipped with a nontrivial valuation.
1. $$c(K)$$ is a $$K$$-vector space of countable type, i.e. it has a dense linear subspace of at most countable dimension (e.g. $$K^{(\infty)}$$). It follows that any weakly $$K$$-cartesian family in $$c(K)$$ is at most countable. (For each element in such a family, replace it by an element of $$K^{(\infty)}$$ sufficiently close to it, then the resulting system would be a weakly cartesian subset of $$K^{(\infty)}.$$ Since weak cartesianity implies linear independence, and since $$K^{(\infty)}$$ has countable dimension, it follows that the original family must be countable.)
2. $$b(K)$$ has an uncountable orthonormal family. (The natural map
$(b(K))^\sim \to \prod_{i=1}^\infty K^\sim$
is surjective and the latter space has uncountable dimension over $$K^\sim.$$ Now pullback an uncountable orthonormal system from the image.)
3. It follows that there is no $$K$$-linear map $$b(K) \to c(K)$$ which is a homeomorphism.
4. On the other hand given any $$a \in K$$ with $$0 < |a| < 1,$$ the mapping
$(b_i)_i \mapsto (a^ib_i)_i$
defines a continuous injective $$K$$-linear map from $$b(K)$$ to $$c(K).$$
2. If a ring is complete, it does not necessarily mean that its field of fractions is complete.
1. E.g. take $$A := k[[x, y]]$$ with the usual (discrete) valuation corresponding to its maximal ideal. The valuation is also induced by the order, i.e.
$|f| := e^{-\ord(f)}$
Pick a power series $$f(t) \in k[[t]]$$ which is not a rational function in $$t,$$ i.e. which is not in the image of the natural map from $$k(t) \to k[[t]]$$ (e.g. one can take the power series for $$e^t,$$ at least if the characteristic of $$k$$ is zero). Then $$f(y^2/x)$$ is in the completion of the field of fractions of $$A,$$ but it can not be represented as $$g/h$$ for $$g, h \in A.$$ [To see this consider the edge of $$f(y^2/x)h$$ corresponding to the inner normal $$(2, 1).$$]
2. Let $$A := k \langle x \rangle$$ be the ring of strictly convergent power series over a field $$k$$ with a complete nontrivial valuation. Then $$A$$ is complete with respect to the Gauss norm. The field of fractions of $$A$$ is the field $$L := k \langle x \rangle [x^{-1}]$$ of strictly convergent meromorphic series. Take $$c \in k$$ such that $$0 < |c| < 1.$$ Then the sequence $$\sum_{j=1}^i c^i x^{-i},$$ $$i = 1, 2, \ldots,$$ is Cauchy but without limit in $$L.$$
3. The algebraic closure of a complete field does not have to be complete.
1. Take the field $$k((t))$$ of Laurent series, which is complete with respect to the valuation induced by $$\ord.$$ Its algebraic closure is the field of “meromorphic” Puiseux series, which is not complete, since the Cauchy sequence
$f_n := \sum_{j=1}^n t^{1/n + n}$
does not converge.
2. The algebraic closure of $$\qq_p$$ is not complete. Both this and the preceding example are special cases of the following:
Lemma [BGR 3.4.3/1]. Let $$K$$ be a field with a complete nontrivial valuation and $$K_a$$ be its algebraic closure. Assume $$[K_a:K] = \infty.$$ Then $$K_a$$ is not complete.
Proof. If the separable closure $$K_{sep}$$ of $$K$$ in $$K_a$$ had finite degree over $$K,$$ then it would be complete, and therefore, since $$K_{sep}$$ is dense in $$K$$ [BGR 3.4.1/6], it would follow that $$K_a = K_{sep}$$ is also of finite degree over $$K,$$ which is a contradiction. So $$[K_{sep}: K] = \infty.$$ Choose a sequence $$x_1 = 1, x_2, \ldots$$ of elements in $$K_{sep}$$ which are linearly independent over $$K$$. Choose $$c_2, c_3, \ldots \in K \setminus \{0\}$$ such that
• $$\lim_i |c_ix_i| = 0$$
• $$|c_{i+1}x_{i+1}| < \min\{|c_ix_i|, r(\sum_{j=2}^i c_jx_j)\}$$
(where $$r(\alpha)$$, for $$\alpha$$ separable over $$K$$ of degree > 1, is the minimum of $$|\alpha – \beta|$$ over all roots $$\beta \neq \alpha$$ of the minimal polynomial of $$\alpha$$ over $$K.$$) Note that $$r(\sum_{j=2}^i c_jx_j)$$ is well defined for $$j \geq 2,$$ since $$\sum_{j=2}^i c_jx_j \in K_{sep} \setminus K,$$ since all $$c_j$$ are nonzero and $$x_j$$ constitute a basis of $$K_{sep}$$ over $$K.$$ Then if $$\sum_{j=2}^\infty c_jx_j$$ converges to an element $$x \in K_a,$$ then by assumption,
$|x – \sum_{j=2}^i c_jx_j| = |\sum_{j=i+1}^\infty c_jx_j| \leq |c_{i+1}x_{i+1}| < r(\sum_{j=2}^{i}c_ix_i)$
so that Krasner’s lemma implies that $$K(x) \supseteq K(\sum_{j=2}^i c_jx_j).$$ It follows that $$K(x) \supseteq K(x_2, x_3, \ldots),$$ so that $$[K(x):K] = \infty,$$ which is a contradiction, since $$x$$ is algebraic over $$K.$$ This completes the proof. In fact, if $$x$$ is the element in the completion of $$K_a$$ which represents $$\sum_{j=2}^\infty c_jx_j,$$ then it follows (from the arguments in the proof of Krasner’s lemma [BGR 3.4.2/2]) that $$K(x)$$ indeed contains $$K(x_2, x_3, \ldots),$$ and in particular, $$x$$ is transcendental over $$K.$$
4. A (countable) basis of $$\tilde V$$ over $$\tilde K$$ may not lift to a Schauder basis of a vector space $$V$$ over a (valued field) $$K.$$ Let $$K$$ be a field with a nontrivial non-discrete valuation. Choose a Schauder basis $$\{x_1, x_2, \ldots,\}$$ of $$V := c(K)$$ such that $$|x_i| = 1$$ for all $$i.$$ Choose a sequence $$\lambda_2, \lambda_3, \ldots$$ in $$K$$ such that $$0 < |\lambda_i| < 1$$ and $$\lim_i \lambda_2 \lambda_3 \cdots \lambda_i \neq 0.$$ Set
$y_i := x_i – \lambda_{i+1}x_{i+1}$
Then $$\tilde y_i = \tilde x_i$$ for all $$i$$, i.e. $$\tilde y_1, \tilde y_2, \ldots,$$ constitute a $$\tilde K$$-basis of $$\tilde V.$$ However, $$x_1$$ can not be written as a convergent sum $$x_1 = \sum_{i=1}^\infty a_i y_i$$ because then
$x_1 = \sum_{i=1}^\infty a_i(x_i – \lambda_{i+1}x_{i+1}) = a_1x_1 + \sum_{i=2}^\infty (a_i – a_{i-1}\lambda_i)x_i$
which means $$a_1 = 1$$ and $$a_i = a_{i-1}\lambda_i = \lambda_2 \lambda_3 \cdots \lambda_i$$ for $$i \geq 2.$$ It follows that $$\lim_i |a_iy_i| = \lim_i|a_i| \neq 0.$$
5. A complete non perfect field. The field $$k((t))$$ of Laurent series is complete with respect to the valuation induced by order in $$t.$$ However, $$k((t))$$ does not contain a $$p$$-th root of $$t$$ if the characteristic $$p$$ of $$k$$ is positive.
6. There are valued fields $$K$$ admitting algebraic extensions $$L \neq K$$ such that $$K$$ is dense in $$L$$ with respect to the spectral norm. In particular, $$L$$ is not weakly cartesian over $$K.$$ In addition, there are such examples with $$L$$ purely inseparable over $$K$$, in which case the spectral norm on $$L$$ is a valuation.
1. E.g. one can take $$K$$ to be the separable closure of a complete non perfect field $$k$$ and $$L$$ to be the algebraic closure of $$k.$$ $$K$$ is dense in $$L$$ since there are separable elements arbitrarily close to every element in $$L.$$
2. For a slightly more explicit example, start with fields $$k$$ of characteristic $$p > 0$$ with a nontrivial valuation such that the field $$k^{p^{-1}}$$ of $$p$$-th roots over $$k$$ is of infinite degree over $$k.$$ E.g. one can take for $$k$$ the field of fractions (or, if completeness is required, the completion of the field of fractions) of the formal power series ring in infinitely many variables over the finite field of $$p$$ elements. Let $$L$$ be the field of fractions of the ring $$k^{p^{-1}}\langle t \rangle$$ of strictly convergent power series over $$k^{p^{-1}}$$ equipped with the Gauss norm. Let $$A$$ be the set of all series $$\sum_i a_i t^i \in k^{p^{-1}}\langle t \rangle$$ such that the coefficients $$a_i$$ generate a finite extension of $$k.$$ The field $$K$$ of fractions of $$A$$ is dense in $$L$$ (since e.g. it contains all polynomials over $$k^{p^{-1}}$$). Moreover, since $$L \subseteq K^{p^{-1}},$$ it follows that $$L$$ is algebraic over $$K$$ and the Gauss norm on $$L$$ is identical to the spectral norm of $$L$$ with respect to the Gauss norm on $$K.$$ To see that $$L \neq K,$$ choose a sequence $$h_i \in k^{p^{-1}}$$ such that $$\lim_i |h_i| = 0$$ and $$[k(h_1, h_2, \ldots, ): k] = \infty.$$ Then $$\sum_i h_it^i \in L \setminus K.$$
3. For an even more explicit example with discrete valuation and finite extension, start with a field $$k$$ of characteristic $$p > 0$$ such that there is $$f(t) \in k[[t]]$$ which is not algebraic over $$k(t)$$ (e.g. one can take $$k$$ and $$f$$ as in Schmidt’s example in the preceding post). Let $$K := k(x,y^p) \subset L := k(x,y).$$ As in that example, equip $$L$$ with the norm $$|\ |_f$$ induced by the valuation on $$k((t))$$ via (the pullback of) the homomorphism $$k[x,y] \to k[[t]]$$ given by $$x \mapsto t,$$ $$y \mapsto f(t).$$ Since $$g^p \in K$$ for each $$g \in L,$$ it follows that the norm on $$L$$ is identical to the spectral norm of $$L$$ with respect to the norm on $$K.$$ Writing $$f(t) := \sum_i a_it^i,$$ one gets that $$y – \sum_{j=1}^i a_ix^i \to 0$$ as $$i \to \infty,$$ so that $$K$$ is dense in $$L.$$
7. The above phenomenon is impossible if $$L/K$$ is separable or $$K$$ is perfect. Since the the algebraic closure of a perfect field $$K$$ is weakly $$K$$-cartesian [BGR 3.5.1/4].
8. Every perfect or complete field $$K$$ is weakly stable, i.e. each finite extension equipped with the spectral norm is weakly $$K$$-cartesian.
9. The preceding observation suggests that the weak stability of a field $$K$$ boils down to whether $$K_\infty := \bigcup_{n \geq 0} K_n$$ is weakly cartesian over $$K$$, where
$K_n := \{x \in \bar K: x^{p^n} \in K\}$
(where $$\bar K$$ is the algebraic closure of $$K$$), since $$K_\infty,$$ being perfect, is weakly stable (i.e. $$\bar K$$ is weakly cartesian over $$K_\infty$$). Whether $$K_\infty$$ is weakly $$K$$-cartesian depends on whether each $$K_n$$ is weakly $$K$$_cartesian, which in turn (via the Frobenius homomorphism $$x \mapsto x^{p^{n-1}}$$) depends on whether $$K_1 = K^{p^{-1}}$$ is weakly $$K$$-cartesian. This is the content of [BGR 3.5.3/1].
10. The field of rational functions over a field in finitely many variables is weakly stable with respect to the valuation induced by degree or order [BGR 3.5.3/4] (this essentially follows from the previous observation together with some computation).
11. A characteristic zero (in particular, separable) complete field $$K$$ and a finite algebraic extension $$L/K$$ which is weakly $$K$$-cartesian, but not $$K$$-cartesian; in other words, weak stability $$\not \Rightarrow$$ stability. Start with the field $$\qq_2$$ of $$2$$-adic numbers. Let $$\alpha_0 := 2, \alpha_1, \alpha_2, \ldots$$ be elements algebraic over $$\qq_2$$ such that $$\alpha_i^2 = \alpha_{i-1}$$ and $$k_i := \qq_2(\alpha_i) = \qq_2(\alpha_0, \ldots, \alpha_i).$$ Let $$K$$ be the completion of the field $$\bigcup_i k_i$$ and $$L := K(\sqrt 3).$$
Lemma [BGR 3.6.1]. $$[L:K] = 2,$$ $$|L| = |K|,$$ and $$L^\sim = K^\sim = \ff_2,$$ the finite field of two elements. In particular, $$L$$ is not $$K$$-cartesian.
The proof is mainly by computation. I tried to solve for $$x^2 = 3$$ over $$K.$$ It seems that one can write $$\sqrt 3$$ as an infinite sum of the form $$\sum_i c_i 2^{\alpha_i}$$ with $$|c_i| = 1$$ and $$0 = \alpha_0 < \alpha_1 < \alpha_2 < \cdots$$ such that $$\lim_i \alpha_i = 1.$$ This suggests that $$|\sqrt 3, K| = 2^{-1}$$ but there is no $$a \in K$$ such that $$|\sqrt 3 – a| = 2^{-1}.$$ In particular, $$K$$ is not strictly closed in $$L.$$
12. The above phenomenon is impossible in the following cases:
1. If the valuation on $$K$$ is discrete, since if $$K$$ is discretely valued and complete, then every finite dimensional normed vector space $$V$$ over $$K$$ is $$K$$-cartesian. (The completeness of $$K$$ implies that $$V$$ is weakly $$K$$-cartesian (so that every $$K$$-subspace of $$V$$ is closed) and then the discreteness of the valuation on $$K$$ implies that every closed $$K$$-subspace of $$V$$ is strictly closed, so that $$V$$ is $$K$$-cartesian.)
2. If $$K^\sim$$ has zero characteristic, since in that case $$K$$ is stable [BGR 3.6.2/13]. (The main reason – which follows from Zariski-Samuel Vol 2, Chapter VI, Sections 11 (remark following the proof of Theorem 19) and 12 (corollary to Theorem 24) – is that in that case $$\sum_i e_if_i = n$$ for all finite extensions of $$K$$.)