Peter Roquette  17. Juli 1996

On convergent power series

We consider the following situation:

K a field equipped with a non-archimedean absolute value | · | which

is assumed to be complete

K[[T]] the ring of formal power series in one variable T over K
K((T)) the field of quotients of K[[T]]

K[[T]]_{r} for r > 0, the subring of those power series which are convergent

in the disc |t| ≤ r ^{1})

K[[T]]_{0} = ⋃_{r>0} K[[T]]_{r} the “ring of convergent power series” over K
K((T))_{0} its field of quotients

Our aim is to present a proof of the following

Theorem 1 K((T))_{0} is algebraically closed in K((T)). Moreover, K((T)) is a regular field extension of K((T))_{0} and, hence, K((T))_{0} is existentially closed in K((T)).

Remark: In an earlier version of this manuscript dated 13 May 1996, the second part of the above theorem, concerning the separability of K((T)) over K((T))_{0}, had been proved only under the additional hypothesis that K is of finite degree of inseparability. The general proof as given here is due to F.V. Kuhlmann.

We denote by ord(z) the initial degree of the power series z ∈ K[[T]]. This defines a discrete, additively written valuation of K((T)) over K for which K((T)) is complete, hence henselian. We shall see below that

K[[T]] ∩ K((T))_{0} = K[[T]]_{0}  (1)

which means that K[[T]]_{0} is the valuation ring in K((T))_{0} belonging to the initial degree valuation. First we are going to show:

Proposition 2 K((T))_{0} is Henselian with respect to the initial degree valuation. Consequently K((T))_{0} is separably algebraically closed within K((T)).

1) Here and in the following, r > 0 will always denote a real number in the value group of K.

On convergent power series  17.7.96  Seite 2

Proof : We consider a polynomial

f(Y) = a_{0} + a_{1}Y + ⋯ + a_{k}Y^{k}  (2)

with coefficients a_{i} ∈ K[[T]]_{0}. We assume that

f(0)(0) = 0 and  f′(0)(0) ≠ 0  (3)

and have to show that there exists y ∈ K[[T]]_{0} such that

f(y) = 0 and y(0) = 0 .  (4)

Observe that f(0) = a_{0} and f′(0) = a_{1} are convergent power series; thus f(0)(0) = a_{0}(0) and f′(0)(0) = a_{1}(0) denote the constant coefficients of those power series.

The usual procedure to construct such zero y of f(Y) is by means of Newton’s iteration method: Starting with y_{0} = 0 as initial value we define successively

y_{n+1} = y_{n} − f′(y_{n})^{−1}f(y_{n})  (n = 0, 1, 2, …)  (5)

It is shown that y = lim_{n→∞} y_{n} exists and f′(y) ≠ 0; then it is seen from the definition (5) that y is a zero of f(Y). Moreover, it is seen that y(0) = 0.

If in the above, the notion of “limit” is understood in the sense of the initial degree valuation ord( ·) then one can only deduce that the limit is contained in the ord-completion of K[[T]]_{0}, but this is K[[T]] and hence this argument does not lead to any information about the limit belonging to K[[T]]_{0}. Therefore, we have to regard the y_{n} as a convergent sequence with respect some other suitable valuation, as follows.

By definition, K[[T]]_{0} is the union of the rings K[[T]]_{r} for r → 0. A power series

z = c_{0} + c_{1}T + c_{2}T^{2} + ⋯ ∈ K[[T]]_{0}  (6)

is convergent in the disc |t| ≤ r if and only if

lim_{ν→∞} |c_{ν}|r^{ν} = 0 .

If this is the case then

∥z∥_{r} = max_{ν} |c_{ν}|r^{ν}  (7)

does exist. If the residue class field of K is infinite then it is well known that ∥z∥_{r} is the norm of uniform convergence in the disc, i.e.,

∥z∥_{r} = max_{|t|≤r} |z(t)| .

In any case, whether the residue field of K is infinite or not, ∥ · ∥_{r} is an absolute value of K[[T]]_{r}, and K[[T]]_{r} is complete with respect to this absolute value. ∥ · ∥_{r} is briefly called the functional norm on the r-disc.

This shows that Proposition 2 is an immediate consequence of:

Proposition 3 If r is sufficiently small then all the y_{n} of Newton’s sequence (5) are contained in K[[T]]_{r} and the sequence y_{n} converges with respect to the functional r-norm ∥ · ∥_{r}. Hence its r-norm limit y = lim_{n→∞} y_{n} is contained in K[[T]]_{r}. Moreover, we have y(0) = 0 .

This proposition is well known from elementary analysis under the name of “Theorem of implicit functions”.^{2}) Nevertheless let us present a proof here.

Before starting with the proof proper let us note:

Lemma 4 Let z ∈ K[[T]]_{r} and suppose ∥z−1∥_{r} < 1. Then z^{−1} ∈ K[[T]]_{r} and ∥z∥_{r} = ∥z^{−1}∥_{r} = 1.

Proof : We put z = 1 − u, then ∥u∥_{r} < 1 and hence the geometric series

(1 − u)^{−1} = ∑_{0≤ν<∞} u^{ν}

converges with respect to ∥ · ∥_{r}; we conclude that z^{−1} = (1 − u)^{−1} ∈ K[[T]]_{r}. The relations ∥z∥_{r} = 1 and ∥z^{−1}∥_{r} = 1 are implied by the non-archimedean property of the valuation ∥ · ∥_{r}. □

Corollary 5 If z ∈ K[[T]]_{0} and z(0) ≠ 0 then z^{−1} ∈ K[[T]]_{0} .

Proof : We put z = z(0)·z^{∗} and see that the invertibility of z is equivalent to the invertibility of z^{∗}. Writing again z instead of z^{∗} we may assume from the start that z(0) = 1:

z = 1 + c_{1}T + c_{2}T^{2} + ⋯ = 1 − u .

z and hence u are convergent in some disc |t| ≤ r_{0}. For r ≤ r_{0} we have

∥u∥_{r} = max_{ν≥1} |c_{ν}|r^{ν} ≤ \frac{r}{r_{0}} · max_{ν≥1} |c_{ν}|r_{0}^{ν} ≤ \frac{r}{r_{0}} · ∥z∥_{r_{0}} .

Hence from Lemma 4 we conclude:

r < \frac{r_{0}}{∥z∥_{r_{0}}} ⇒ ∥u∥_{r} < 1 ⇒ z^{−1} ∈ K[[T]]_{r} .

In other words: if r is sufficiently small then z^{−1} belongs to K[[T]]_{r} and hence to K[[T]]_{0}. □

2) In elementary analysis the underlying absolute value is archimedean and hence the ordinary proof differs from ours in those places where we use the non-archimedean property of the valuation. But of course, the idea of proof is the same in both cases, archimedean and non-archimedean, and it is possible to give a unified proof.

Corollary 5 shows that every z ∈ K[[T]]_{0} can be uniquely written as

z = T^{k}z_{0}

where k = ord(z) and z_{0} is a unit in K[[T]]_{0}. This implies the relation (1) already stated above.

Proof of Proposition 3: For simplicity we may assume that

f′(0)(0) = 1 .  (8)

If this should not be the case then multiplication of the coefficients of f(Y) with the constant c = f(0)(0)^{−1} will achieve this.

There exists r_{0} > 0 such that all the coefficients a_{0}, a_{1}, …, a_{k} of f(Y) are convergent in the disc |t| ≤ r_{0}. Let us put

M = max_{0≤i≤k} ∥a_{i}∥_{r_{0}} .

It follows from (8) that ∥a_{1}∥_{r_{0}} = ∥f′(0)∥_{r_{0}} ≥ 1; hence M ≥ 1.

In the following proof we choose a radius r < r_{0} which is “sufficiently small”; the precise condition for r will be seen in the course of our arguments. We fix such r and put

I_{r} = { y ∈ K[[T]]_{r} : ∥y∥_{r} < M^{−1} } .

This is an ideal in the ring O_{r} = { y ∈ K[[T]]_{r} : ∥y∥_{r} ≤ 1 } . Observe that I_{r} is closed with respect to the valuation ∥ · ∥_{r}, hence complete.

The proof of Proposition 3 will now be presented as a succession of observations (i)–(iii) with the conclusion (iv).

Observation (i) If y ∈ I_{r} then f(y) ∈ I_{r}. Thus f maps the ideal I_{r} into itself.

To see this we write

f(y) = a_{0} + a_{1}y + a_{2}y^{2} + ⋯ + a_{k}y^{k} .  (9)

According to (3) the power series function a_{0} = a_{0}(t) = f(0)(t) vanishes for t = 0; hence by continuity we have |a_{0}(t)| → 0 if t → 0. Thus if r is sufficiently small we have

∥a_{0}∥_{r} < M^{−1}. ^{3})

3) Explicitly: write a_{0}(T) = ∑_{ν≥1} a_{0,ν}T^{ν} and use the estimate |a_{0,ν}|r^{ν} ≤ Mr_{0}^{−ν} · r^{ν} ≤ Mr_{0}^{−1}r for ν ≥ 1 to see that the condition r < r_{0}M^{−2} does suffice.

As to the other terms in (9) we compute for i ≥ 2

∥a_{i}y^{i}∥_{r} ≤ ∥a_{i}∥_{r}∥y∥_{r}^{i} ≤ ∥a_{i}∥_{r_{0}}∥y∥_{r}^{i} ≤ M · M^{−i} ≤ M^{−1} .

For i = 1 we have a_{1} = f′(0) and the same estimate holds because, due to the next observation, we have ∥f′(0)∥_{r} = 1. Hence the estimate ∥a_{i}∥_{r}∥y∥_{r}^{i} ≤ M^{−1} holds for all i = 0, 1, 2, …, k and so

∥f(y)∥_{r} < M^{−1}

as contended. □

Observation (ii) If y ∈ I_{r} then f′(y)^{−1} ∈ K[[T]]_{r} and ∥f′(y)^{−1}∥_{r} = 1.

To see this we write

f′(y) − 1 = (a_{1} − 1) + 2a_{2}y + ⋯ + ka_{k}y^{k−1} .  (10)

According to (8) the power series function a_{1}(t) − 1 = f′(0)(t) − 1 vanishes for t = 0; hence again by continuity, we have

∥a_{1} − 1∥_{r} < 1

if r is sufficiently small.^{4}) For the remaining terms in (10) we have

∥ia_{i}y^{i−1}∥_{r} ≤ ∥a_{i}∥_{r}∥y∥_{r}^{i−1} ≤ ∥a_{i}∥_{r_{0}}∥y∥_{r}^{i−1} ≤ M · M^{−(i−1)} < 1

since i ≥ 2. It follows

∥f′(y) − 1∥_{r} < 1

and we apply Lemma 4.□

Next we introduce the Newton operator

N (y) = y − f′(y)^{−1}f(y) .  (11)

From observations (i) and (ii) we see that N maps the ideal I_{r} into itself.

Observation (iii) If y ∈ I_{r} then ∥f(N y)∥_{r} ≤ M · ∥f(y)∥_{r}^{2} .

To obtain this estimate we use Taylor’s formula which we write in the form

f(Y + Z) = f(Y) + f′(Y)Z + f^{(2)}(Y)Z^{2} + ⋯ + f^{(k)}(Y)Z^{k} (12)

where Y, Z are independent variables. The polynomials f^{(ν)}(Y) are the modified higher derivatives of f(Y) in the sense of Hasse and F.K.Schmidt; they are defined by the formula

f^{(ν)}(Y) = ∑_{ν≤i≤k} \binom{i}{ν}a_{i}Y^{i−ν} .  (13)

4) Similar explicit verification as in (i). This time the condition r < r_{0}M^{−1} will suffice.

For the coefficients we have the estimate

∥\binom{i}{ν}a_{i}∥_{r} ≤ ∥a_{i}∥_{r} ≤ ∥a_{i}∥_{r_{0}} ≤ M

and hence for y ∈ I_{r} (since ∥y∥_{r} < M^{−1} < 1)

∥f^{(ν)}(y)∥_{r} ≤ M .  (14)

Now, in Taylor’s formula we substitute Y ↦ y and Z ↦ −f′(y)^{−1}f(y) and observe that, due to the definition of the Newton operator, the first two terms cancel and Taylor’s formula starts with the quadratic term:

f(N y) = f^{(2)}(y) (f′(y)^{−1}f(y))^{2} + ⋯ + f^{(k)}(y) (f′(y)^{−1}f(y))^{k} .

For each term on the right hand side, we deduce using (14) and observation (ii):

∥f^{(ν)}(y) (f′(y)^{−1}f(y))^{ν}∥_{r} ≤ M · ∥f(y)∥_{r}^{ν} ≤ M · ∥f(y)∥_{r}^{2}

the last inequality because ν ≥ 2 and ∥f(y)∥_{r} < 1 by (i). □

Conclusion (iv):

Consider the Newton sequence y_{n} defined inductively by

y_{n+1} = N (y_{n})

as in (5), with the initial term y_{0} = 0. From observation (iii) we obtain by induction

∥f(y_{n})∥_{r} ≤ M^{2^{n}−1}∥f(y_{0})∥_{r}^{2^{n}} (n = 1, 2, 3, …)

Moreover the definition (5) of the Newton sequence shows that

∥y_{n+1} − y_{n}∥_{r} = ∥f′(y_{n})^{−1}f(y_{n})∥_{r} ≤ ∥f(y_{n})∥_{r} ≤ M^{2^{n}−1}∥f(y_{0})∥_{r}^{2^{n}}

where we have used observation (ii). Since M∥f(y_{0})∥_{r} < 1 by observation (i), these formulas show us, firstly, that the Newton sequence y_{n} is a Cauchy sequence with respect to the r-norm ∥ · ∥_{r}; hence the limit y = lim_{n→∞} y_{n} exists in I_{r}. Secondly, we see that f(y_{n}) converges to 0; hence f(y) = 0.

Since y_{0} = 0 we see that f(y_{0})(0) = f(0)(0) = 0 due to the hypothesis (3). From (5) we conclude y_{1}(0) = 0. By induction one verifies that

f(y_{n})(0) = f(0)(0) = 0 and y_{n+1}(0) = 0 (n = 1, 2, 3, …)

Hence for the limit y = lim_{n→∞} y_{n} we also have y(0) = 0 .

Proposition 3 is proved. □

Accordingly, we now know that K((T))_{0} is henselian. Since it is dense in K((T)), it is well known that the regularity of K((T)) over K((T))_{0} is necessary and sufficient for K((T))_{0} to be existentially closed in K((T)). See e.g., F. V. Kuhlmann’s thesis [K]. It remains to prove:

Proposition 6 Let p = char(K) > 0. The field K((T)) is a separable extension of K((T))_{0}. This means that K((T)) is linearly disjoint to (K((T))_{0})^{1/p} over K((T))_{0}.

Let us remark that

(K[[T]])^{1/p} = K^{1/p}[[T^{1/p}]] .

If a power series z ∈ K[[T]] converges in a disc with radius r > 0 then its p-th root z^{1/p} ∈ K^{1/p}[[T^{1/p}]], as a power series in the variable T^{1/p}, converges in the disc with radius r^{1/p}. And conversely. Thus

(K[[T]]_{r})^{1/p} = K^{1/p}[[T^{1/p}]]_{r^{1/p}} .

For r → 0 we obtain

(K[[T]]_{0})^{1/p} = K^{1/p}[[T^{1/p}]]_{0} .

The same relation holds for the respective quotient fields, i.e., we may replace the double square brackets by double ordinary brackets.

Now consider the following diagram:

K^{1/p}((T^{1/p})) K((T^{1/p})) K((T^{1/p})) |||||| K((T^{1/p}))_{0} K((T^{1/p}))_{0} K^{1/p}((T^{1/p}))_{0} K^{1/p}((T^{1/p}))_{0} |||||| K^{1/p}((T^{1/p})) K((T^{1/p})) K((T)) K((T)) |||||| K((T))_{0} K((T))_{0} K((T^{1/p}))_{0}

The next lemma shows that linear disjointness holds in the left lower portion of the diagram:

Lemma 7 K((T)) and K((T^{1/p}))_{0} are linearly disjoint over K((T))_{0} . In fact, the p elements T^{j/p}, 0 ≤ j ≤ p − 1, form a basis of K((T^{1/p})) over K((T)), and also a basis of K((T^{1/p}))_{0} over K((T))_{0}.

Proof : According to the euclidean algorithm each integer n has a unique representation of the form

n = j + pm with 0 ≤ j ≤ p − 1  (15)

and m integer. Hence T^{n/p} = T^{j/p}T^{m} and we obtain the direct sum decomposition

K[[T^{1/p}]] = ∑_{0≤j≤p−1} T^{j/p}K[[T]] ,

which shows that the T^{j/p} form a basis of K[[T^{1/p}]] over K[[T]], and hence also for the respective quotient fields. Explicitly, if

z = ∑_{n} b_{n}T^{n/p}

is a power series in T^{1/p} with coefficients b_{n} ∈ K, and if we express each exponent n as in (15) then

z = ∑_{0≤j≤p−1} T^{j/p}z_{j}

where

z_{j} = ∑_{m} c_{j,m}T^{m} with c_{j,m} = b_{pm+j} .

If z ∈ K[[T^{1/p}]]_{r} then lim_{n→∞} |b_{n}|r^{n/p} = 0. For each fixed j, the subsequence |b_{pm+j}|r^{m+(j/p)} also tends to 0; multiplication by r^{−j/p} shows lim_{m→∞} |c_{j,m}|r^{m} = 0, hence z_{j} converges in the disc |t| ≤ r, which is to say that z_{j} ∈ K[[T]]_{r}. The arguments can be reversed, and therefore

K[[T^{1/p}]]_{r} = ∑_{0≤j≤p−1} T^{j/p}K[[T]]_{r} .

For r → 0 we conclude

K[[T^{1/p}]]_{0} = ∑_{0≤j≤p−1} T^{j/p}K[[T]]_{0} .

Thus 1, T^{1/p}, …, T^{(p−1)/p} is a basis of K[[T^{1/p}]]_{0} over K[[T]]_{0}, and the same holds for the respective quotient fields. □

It remains to show linear disjointness for the right upper portion of the above diagram. That portion refers to T^{1/p} as indeterminate. For the purpose of the following proof, let us write T again instead of T^{1/p} in order to comply with the notations used elsewhere in this manuscript. Thus we have to consider the following diagram:

K^{1/p}((T)) K((T)) K((T)) |||||| K((T))_{0} K((T))_{0} K^{1/p}((T))_{0} K^{1/p}((T))_{0} |||||| K^{1/p}((T))

In this situation we claim:

Lemma 8 The fields K((T)) and K^{1/p}((T))_{0} are linearly disjoint over K((T))_{0}. In fact: Let F ⊂ K^{1/p}((T))_{0} be a subextension of finite degree over K((T))_{0}; let u_{1}, …, u_{m} be a K-basis of the residue field L of F and ũ_{1}, …, ũ_{m} be foreimages of the u_{i} in F. Then the ũ_{i} form a basis of F over K((T))_{0} which at the same time is a basis of K((T)) · F over K((T)).

K^{1/p}((T)) K((T)) · F K((T)) · F ||||||F F K^{1/p}((T))_{0} K^{1/p}((T))_{0} |||||| K^{1/p}((T)) K((T)) · F K((T)) K((T)) |||||| K((T))_{0} K((T))_{0}  F

Remark: In the above lemma the notion of “residue field” refers, of course, to the initial degree valuation of the power series field. The residue field of K((T)) is K, and K is also the residue field of K((T))_{0}.

If K is of finite inseparability degree then [K^{1/p} : K] is finite, hence in the above Lemma we can take F = K^{1/p}((T)) and L = K^{1/p}. Moreover, we can take ũ_{i} = u_{i}. In this case the following proof is identical with the proof given in our earlier version of this manuscript. It has been Kuhlmann’s idea how to modify that proof in order to deal also with fields of infinite degree of inseparability.

Proof of Lemma 8: Since the residues u_{i} of the ũ_{i} are linearly independent over K by construction, it follows that the ũ_{i} are linearly independent over K((T)). In fact, for every linear combination

z = ∑_{1≤i≤m} z_{i}ũ_{i}

with z_{i} ∈ K((T)) we have

ord(z) = min_{1≤i≤m} ord(z_{i})  (16)

and this implies the linear independency of the ũ_{i} over K((T)). Hence it remains to show that the ũ_{i} form a basis of F over K((T))_{0} ^{5}) which is to say that

[F : K((T))_{0}] = [L : K] .

In general, the residue field extension is of degree ≤ than the degree of the original field extension, and equality implies that the field extension has no defect.

Thus we have to show that the valued field F, as an extension of K((T))_{0}, has no defect.^{6})

We proceed by induction with respect to the field degree [F : K((T))_{0}]. If the degree is 1 then there is nothing to prove. Now we assume that F is already known to be defectless over K((T)), and we consider a proper simple extension F′ = F (z) ⊂ K^{1/p}((T))_{0}. We can assume that z ∈ K^{1/p}[[T]]_{0}, i.e., that z is a convergent power series with coefficients in K^{1/p}.

The p-th power z^{p} is in K[[T]]_{0} and hence in F. Thus [F′ : F] = p. For the residue fields we have [L′ : L] ≤ p. Since L′|L is purely inseparable we conclude that [L′ : L] = p or = 1. In the first case F′ is defectless over F. By induction assumption F is defectless over K((T))_{0} and it follows that F′ is defectless over K((T))_{0}.

In the second case F′ has the same residue field as F, hence F′ is an immediate extension of F. We have to show that this case does not occur.

As an immediate extension, F′ is contained in the maximal immediate extension of F, i.e., in the completion F̂ of F. ^{7}) This completion contains K((T)) which is the completion of K((T))_{0}. Thus

F̂ ⊃ K((T)) · F = ∑_{1≤i≤m} K((T))ũ_{i}

(where we have used the induction assumption). Now the right hand side, being a finite dimensional vector space over the complete field K((T)), is also complete and therefore

F̂ = ∑_{1≤i≤m} K((T))ũ_{i} .

Consequently, since z ∈ F′ ⊂ F̂ we conclude that there is a linear represen-

5) In the terminology of [K] the ũ_{i} form a valuation basis of F over K((T))_{0}.

6) It seems that by slight modification of our arguments it would be possible to show that any finite extension F of K((T))_{0} is defectless, i.e., that K((T))_{0} is a defectless field.

7) Observe that the valuation of F is discrete, being induced by the initial degree valuation of K^{1/p}((T)).

tation of the form

z = ∑_{1≤i≤m} a_{i}ũ_{i}  (17)

with coefficients a_{i} ∈ K((T)). From (16) we infer that a_{i} ∈ K[[T]].

Since z is convergent this implies, we claim, that the coefficients a_{i} are also convergent, hence they are contained in K[[T]]_{0} and therefore z ∈ F, a contradiction.

Thus we have to prove the following lemma which, actually, is the heart of Kuhlmann’s proof of Proposition 6. For clarity, let us review the situation of the lemma.

K a complete valued field of characteristic p
K′ a valued algebraic field extension of K ^{8})

u_{i} for 1 ≤ i ≤ m, finitely many elements in K′ which are linearly

independent over K

ũ_{i} = ũ_{i}(T) power series in K′[[T]] such that ũ_{i}(0) = u_{i}

z = z(T) a power series in K′[[T]] which can be written as a linear combination of the form (17) with coefficients a_{i} ∈ K[[T]]

In this situation we have:

Lemma 9 If the power series z(T) and the ũ_{i}(T) are convergent, then the coefficients a_{i}(T) are convergent too.

Proof : (i) The proof will based on the following fact concerning the vector space

L = ∑_{1≤i≤m} Ku_{i}

spanned by the u_{i}. For any x ∈ L we write x = x_{1}u_{1} + ⋯ + x_{m}u_{m} with x_{i} ∈ K and put

μ(x) = max(|x_{1}|, …, |x_{m}|) .

The function x → μ(x) is a vector space norm of L over K, i.e., it satisfies the relations

μ(x + x′) ≤ max(μ(x), μ(x′))
μ(cx) = |c| · μ(x)  for c ∈ K
μ(x) = 0 ⇔ μ = 0

The valuation x → |x| is also a vector space norm of L. Now since K is complete, it is well known that any two vector space norms on L define the

8) For the following lemma it is not necessary that K′ is purely inseparable over K.

same topology. Explicitly, there is a constant M > 0 (depending on the given basis u_{i} of L) such that

μ(x) ≤ M · |x| .  (18)

See e.g. our old paper [R].

(ii) This being said, we now start with the proof of Lemma 9. After replacing z by z − z(0) we may assume that z(0) = 0. Since the power series z(T) converges in some neighborhood of zero in K′ we conclude by continuity that lim_{r→0} ∥z∥_{r} = 0. Hence for r sufficiently small we have that

∥z∥_{r} ≤ M^{−1}.  (19)

For each i we write

ũ_{i} = u_{i} + z_{i}(T)

where z_{i}(0) = 0. Again, since z_{i}(T) converges we have

∥z_{i}∥_{r} ≤ M^{−1}  (1 ≤ i ≤ m)  (20)

if r is sufficiently small.

With r chosen this way, we claim that each ∥a_{i}∥_{r} exists and ∥a_{i}∥_{r} ≤ 1. To see this, let us write

z(T) = ∑_{n>0} c_{n}T^{n}
z_{i}(T) = ∑_{n>0} c_{i,n}T^{n}
a_{i}(T) = ∑_{n≥0} a_{i,n}T^{n}.

with coefficients c_{n}, c_{i,n} ∈ K′ and a_{i,n} ∈ K.

Observe that a_{i,0} = a_{i}(0) = 0; this follows from

0 = z(0) = ∑_{1≤i≤m} a_{i}(0)u_{i}

and the linear independency of the u_{i} over K. Our contention is that

|a_{i,n}|r^{n} ≤ 1  (1 ≤ i ≤ m)  (21)

if r is sufficiently small in the sense of (19), (20). This can be seen by induction. For n = 0 we have a_{i,n} = 0 and there is nothing to prove. Now suppose n > 0. Comparing the coefficients of T^{n} in (17) we obtain

c_{n} = ∑_{1≤i≤m} a_{i,n}u_{i} + ∑_{1≤i≤m} ∑_{0≤ν<n} a_{i,ν}c_{i,n−ν} .

In the second sum on the right hand side only those coefficients a_{i,ν} do appear for which the induction assumption applies. We conclude

| ∑_{1≤i≤m} a_{i,n}u_{i}| r^{n} ≤ max (|c_{n}|r^{n}, max_{1≤i≤m} max_{0≤ν<n}(|a_{i,ν}|r^{ν}|c_{i,n−ν}|r^{n−ν}))

≤ max (∥z∥_{r}, max_{1≤i≤m}(1 · ∥z_{i}|_{r}))
≤ M^{−1}

where we have used (19) and (20), besides of the induction assumption. On the other hand, in view of (18) we have for each i:

|a_{i,n}| ≤ M · | ∑_{1≤i≤m} a_{i,n}u_{i}|

and multiplication with r^{n} yields (21) for the exponent n.

We have seen that for sufficiently small radius r > 0 we have

∥a_{i}∥_{r} = max_{n≥0} |a_{i,n}|r^{n} ≤ 1 (1 ≤ i ≤ m) .

We choose one such radius, say r_{0}. Then for r < r_{0} (and r in the value group of K) we compute as usual

|a_{i,n}|r^{n} ≤ ∥a_{i}∥_{r_{0}} · (\frac{r}{r_{0}} )^{n} ≤ (\frac{r}{r_{0}} )^{n}

and this converges to 0 for n → ∞. In other words: If r < r_{0} then the power series a_{i}(T) converges in the disc |t| ≤ r.

Lemma 9 is proved, and so is Theorem 1.

References:

[K] F.V. Kuhlmann, Henselian function fields, Thesis Heidelberg (1989)

[R] P. Roquette, On the prolongation of valuations, Trans. AMS 88 (1958), 42-56