The formal Laurent series $\mathbb{R}((t^{-1}))$: support bounded below, valuation, leading coefficient

Throughout, $\mathbb{R}$ is the field of real numbers with its order (The real numbers, The reals form a totally ordered field) and $\mathbb{Z}$ is the totally ordered commutative ring of integers (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, Order on the integers, The integers form a totally ordered ring).

For a function $f : \mathbb{Z} \to \mathbb{R}$ write

$\operatorname{supp} f \;:=\; \{\, k \in \mathbb{Z} : f(k) \ne 0 \,\},$

and say that $\operatorname{supp} f$ is bounded below when there is $m \in \mathbb{Z}$ with $f(k) = 0$ for every $k < m$. The set of formal Laurent series in $t^{-1}$ over $\mathbb{R}$ is

$K \;=\; \mathbb{R}((t^{-1})) \;:=\; \{\, f : \mathbb{Z} \to \mathbb{R} \;\mid\; \operatorname{supp} f \text{ is bounded below} \,\},$

equipped with

$(f + g)(k) := f(k) + g(k), \qquad (fg)(k) := \sum_{i + j = k} f(i)\,g(j),$

where the product sum ranges over the pairs $(i,j) \in \mathbb{Z} \times \mathbb{Z}$ with $i + j = k$ and $f(i)g(j) \ne 0$. That set of pairs is finite for every $k$, and $f + g$ and $fg$ again lie in $K$: this is $\mathbb{R}((t^{-1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below ↗, which also proves that $K$ with these operations is a commutative ring whose zero $0_K$ is the constant function $0$ and whose identity $1_K$ is the function taking the value $1$ at $0$ and $0$ elsewhere.

Distinguished elements. For $n \in \mathbb{Z}$ let $t^{-n} \in K$ be the function taking the value $1$ at $n$ and $0$ at every other index; so $t^{0} = 1_K$, and $t := t^{-(-1)}$ is the function taking the value $1$ at $-1$. For $c \in \mathbb{R}$ let $\iota(c) \in K$ be the function taking the value $c$ at $0$ and $0$ elsewhere. The notation $t^{-n}$ is defined here as a name; that it is consistent with the ring multiplication, $t^{-m} \, t^{-n} = t^{-(m+n)}$, is proved in $\mathbb{R}((t^{-1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below ↗.

Series notation. Because $\operatorname{supp} f$ is bounded below, say by $m$, one writes

$f \;=\; \sum_{k \ge m} f(k)\, t^{-k},$

a purely notational device: the object is the function $f$, and no convergence of any kind is asserted or used.

Valuation and leading coefficient. Let $f \in K$ with $f \ne 0_K$. Then $\operatorname{supp} f$ is nonempty and bounded below, so it has a least element ($\mathbb{R}((t^{-1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below ↗). Define

$v(f) \;:=\; \min \operatorname{supp} f \in \mathbb{Z}, \qquad \operatorname{lc}(f) \;:=\; f(v(f)) \in \mathbb{R} \setminus \{0\}.$

$v(f)$ is the valuation and $\operatorname{lc}(f)$ the leading coefficient of $f$. Neither is defined at $f = 0_K$, whose support is empty; every statement about $v$ or $\operatorname{lc}$ in this library carries the hypothesis $f \ne 0_K$ explicitly.

Order. The positive cone of $K$ is

$P \;:=\; \{\, f \in K : f \ne 0_K \text{ and } \operatorname{lc}(f) > 0 \,\},$

that is, a nonzero series is positive exactly when its lowest-index nonzero coefficient is a positive real. That $(K, P)$ is an ordered field (Ordered field, Field) is $\mathbb{R}((t^{-1}))$ is an ordered field, ordered by the sign of the leading coefficient ↗, and that every nonzero element of $K$ is invertible is $\mathbb{R}((t^{-1}))$ is a field: every nonzero formal Laurent series is invertible ↗. As in any ordered field, $f < g$ means $g - f \in P$.

• Why the support must be bounded below. It is exactly what makes the product a finite sum. If arbitrary functions $\mathbb{Z} \to \mathbb{R}$ were admitted, the defining sum for $(fg)(k)$ would range over an infinite set of pairs and would denote nothing, since $K$ carries no notion of convergence. The condition is preserved by both operations, which is the content of $\mathbb{R}((t^{-1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below ↗.

• Indices run over all of $\mathbb{Z}$, and the edge cases are real. The zero series has empty support and no valuation. A nonzero constant series $\iota(c)$ has $v(\iota(c)) = 0$ and $\operatorname{lc}(\iota(c)) = c$, so the index $k = 0$ is an ordinary index and not a boundary. Negative indices are admitted, and they are what makes $t = t^{-(-1)}$, whose support is $\{-1\}$, an element of $K$; a series may have finitely many terms of negative index but never infinitely many.

• The order is not the coefficientwise order. Two series are compared by their lowest differing coefficient, not by all of them at once, and this is what makes $t^{-1}$ smaller than every positive real constant while $t$ is larger than every real constant. The consequences are drawn in $\mathbb{R}((t^{-1}))$ is non-Archimedean, and the monomials $t^{-k}$ are cofinal below its positive elements.

• Relation to the rational functions. The ordered field $\mathbb{R}(t)$ of Not every ordered field is Archimedean, ordered so that $f > 0$ exactly when $f(x) > 0$ for all sufficiently large real $x$, is the standard first example of a non-Archimedean ordered field, and standard treatments identify it with a subfield of $K$ by expanding each rational function at infinity. This page neither constructs that identification nor uses it, and no item here may be cited for it: everything proved about $K$ below is proved from the definition above and nothing else. What the two objects share, and all that is used here, is the idea of ordering by behaviour at infinity.