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$R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements

Statement

Let $K = R ((t^{- 1}))$ be the ordered field of $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient, and identify a natural number with its image in $Z$ when it is used as an index. Then:

$n \cdot 1_{K} < t$ for every $n \in N$ ; consequently $K$ is not Archimedean (Archimedean ordered field).
$0_{K} < t^{- (k + 1)} < t^{- k}$ for every $k \in Z$ .
(Countable cofinality.) For every $ε \in K$ with $ε > 0_{K}$ there is $k \in N$ with $0_{K} < t^{- k} < ε$ ; indeed every integer $k > v (ε)$ works.
(The monomials measure the valuation.) For $h \in K$ and $k \in Z$ : if $h (j) = 0$ for every $j \leq k$ then $∣ h ∣ < t^{- k}$ ; and conversely, if $∣ h ∣ < t^{- k}$ then $h (j) = 0$ for every $j < k$ .

Facts & Assumptions

Given: $K$ with its valuation $v$ , leading coefficient $lc$ , monomials $t^{- a}$ and constants $ι (c)$ .

[L1]

For nonzero $h \in K$ , $h (k) = 0$ for $k < v (h)$ and $h (v (h)) = lc (h) \neq 0$ ; $t^{- a}$ is $1$ at index $a$ and $0$ elsewhere, so $t^{- a} \neq 0_{K}$ with $v (t^{- a}) = a$ and $lc (t^{- a}) = 1$ ; and $t = t^{- (- 1)}$ (The formal Laurent series $R ((t^{- 1}))$ : support bounded below, valuation, leading coefficient).

[L2]

$K$ is an ordered field in which $f < g$ holds exactly when $g - f \neq 0_{K}$ and $lc (g - f) > 0$ ; for $f \neq 0_{K}$ one has $∣ f ∣ \neq 0_{K}$ , $v (∣ f ∣) = v (f)$ and $lc (∣ f ∣) > 0$ ; and $n \cdot 1_{K} = ι (n \cdot 1_{R})$ , which for $n \geq 1$ is nonzero with $v = 0$ ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient, Absolute value in an ordered field).

[L3]

For nonzero $f, g \in K$ : $- f \neq 0_{K}$ with $v (- f) = v (f)$ ; and if $v (f) < v (g)$ then $f + g \neq 0_{K}$ with $lc (f + g) = lc (f)$ (Valuation and leading coefficient in $R ((t^{- 1}))$ : $v (f g) = v (f) + v (g)$ , and the behaviour of $v$ under sums).

[L4]

An ordered field $F$ is Archimedean when for every $x \in F$ there is a natural $n$ with $x < n \cdot 1_{F}$ ; and in an ordered field exactly one of $x < y$ , $x = y$ , $y < x$ holds (Archimedean ordered field, Ordered field).

[L5]

The order on $Z$ is total, and every integer $\geq 0$ is the image of a unique natural number; so for every $m \in Z$ there is a natural $k$ whose image exceeds $m$ (The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

Proof

technique · direct

1.1L1L2L3

For every $k \in Z$ the monomial $t^{- k}$ is nonzero with $lc (t^{- k}) = 1 > 0$ , so $t^{- k} > 0_{K}$ by [L2]; and since $v (t^{- k}) = k < k + 1 = v (- t^{- (k + 1)})$ by [L1] and [L3], the difference $t^{- k} - t^{- (k + 1)}$ is nonzero with leading coefficient $lc (t^{- k}) = 1 > 0$ , so $t^{- (k + 1)} < t^{- k}$ .

1.2L1L2L3

Let $n \in N$ . If $n = 0$ then $t - n \cdot 1_{K} = t$ , which is nonzero with $lc (t) = 1 > 0$ . If $n \geq 1$ then $n \cdot 1_{K}$ is nonzero with $v (n \cdot 1_{K}) = 0$ , so $- (n \cdot 1_{K})$ is nonzero with valuation $0$ by [L3], while $v (t) = - 1 < 0$ ; hence $t - n \cdot 1_{K}$ is nonzero with leading coefficient $lc (t) = 1 > 0$ by [L3]. In both cases $n \cdot 1_{K} < t$ by [L2].

1.3L1L2L3L4

Conversely, let $h \in K$ and $k \in Z$ with $∣ h ∣ < t^{- k}$ , and suppose $h \neq 0_{K}$ with $v (h) < k$ . Then $v (∣ h ∣) = v (h) < k = v (t^{- k})$ and $lc (∣ h ∣) > 0$ by [L2], so $∣ h ∣ - t^{- k}$ is nonzero with leading coefficient $lc (∣ h ∣) > 0$ by [L3], giving $t^{- k} < ∣ h ∣$ and contradicting $∣ h ∣ < t^{- k}$ by the trichotomy of [L4]. Hence $h = 0_{K}$ or $v (h) \geq k$ , and in either case $h (j) = 0$ for every $j < k$ by [L1].

2.1step 1.1L1L2L3

Let $h \in K$ and $k \in Z$ with $h (j) = 0$ for every $j \leq k$ . If $h = 0_{K}$ then $∣ h ∣ = 0_{K} < t^{- k}$ by [step 1.1]. Otherwise $h \neq 0_{K}$ with $v (h) > k$ , so $∣ h ∣ \neq 0_{K}$ with $v (∣ h ∣) = v (h) > k = v (t^{- k})$ by [L1] and [L2]; then $t^{- k} - ∣ h ∣$ is nonzero with leading coefficient $lc (t^{- k}) = 1 > 0$ by [L3], so $∣ h ∣ < t^{- k}$ by [L2].

2.2step 1.1L1L2L3L5

Let $ε \in K$ with $ε > 0_{K}$ , so $ε \neq 0_{K}$ and $lc (ε) > 0$ by [L2]; put $m : = v (ε)$ and use [L5] to fix a natural $k$ with $k > m$ . Then $v (ε) = m < k = v (t^{- k}) = v (- t^{- k})$ by [L1] and [L3], so $ε - t^{- k}$ is nonzero with leading coefficient $lc (ε) > 0$ , that is $t^{- k} < ε$ ; and $t^{- k} > 0_{K}$ by [step 1.1]. The same computation applies to every integer $k > m$ .

2.3step 1.2L4

By [step 1.2], $n \cdot 1_{K} < t$ for every natural $n$ ; by the trichotomy of [L4] no natural $n$ can then satisfy $t < n \cdot 1_{K}$ , so the defining condition of [L4] fails at $x = t$ and $K$ is not Archimedean.

3.1step 1.1step 2.1step 1.3step 2.2step 2.3∎

Clause 1 is [step 1.2] with [step 2.3], clause 2 is [step 1.1], clause 3 is [step 2.2], and clause 4 is [step 2.1] together with [step 1.3].

Remarks

Why clause 3 is the pivotal one. The valuation takes its values in $Z$ , which has countable cofinality, and clause 3 is the translation of that fact into the order of $K$ : a countable family, the monomials $t^{- k}$ with $k \in N$ , already gets below every positive element. This is what makes the sequential Cauchy condition in $K$ testable against countably many thresholds, and it is the reason a sequence indexed by $N$ suffices to reach a limit in Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete. Nothing like it would hold if the exponents were allowed to range over a group of uncountable cofinality.
Non-Archimedean here is a statement about $t$ , not about the constants. The canonical naturals of $K$ are the constant series $n \cdot 1_{K} = ι (n \cdot 1_{R})$ (clause 3 of $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient), all of valuation $0$ , and what bounds them above is $t$ , of valuation $- 1$ . The computation in step 1.2 uses nothing about $t$ beyond that: every positive element of negative valuation exceeds every canonical natural, because a strict inequality between valuations decides the comparison outright, whatever the coefficients are.

$R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

R((t−1))R((t−1)) is non-Archimedean, and the monomials t−kt−k are cofinal below its positive elements

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

$R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements