Does the set of all formal Laurent series form an ordered field?

Question

Let $R$ be an ordered field. Consider the set $R[[x]]:=\{\sum_\limits{n \in \mathbb{Z}}a_nx^n:a_n \in R\}$

Addition: $\sum_\limits{n \in \mathbb{Z}}a_nx^n+\sum_\limits{n \in \mathbb{Z}}b_nx^n:= \sum_\limits{n \in \mathbb{Z}}(a_n+b_n)x^n$

Multiplication: $\sum_\limits{n \in \mathbb{Z}}a_nx^n \times \sum_\limits{n \in \mathbb{Z}}b_nx^n:= \sum_\limits{n \in \mathbb{Z}}(a_nb_n)x^n$

Does these operations on $R[[x]]$ makes it a field?

1. Have I correctly defined the multiplication?

2. Is the notation $R[[x]]$, i.e does the notation $R[[x]]$ have some other meaning?
   If yes, is there any standard notation for this set?

3. What order relation can be defined on $R[[x]]$ in order to turn it into an ordered field?

My motive is to show that $R[[x]]$ is Cauchy Complete but is not Order Complete(i.e lub property)

---

Edit: As said by @ the multiplication defined above doesn't even make it an integral domain.

So I am changing the set to $R[[x]]:=\{\sum_\limits{k=-n}a_kx^k: a_k \in R, n \in \mathbb{N} \}$ and multiplication as defined in the answers.

Answer 1

To define an order, it suffices to say which elements are $\geq 0$, since then we can define $a\leq b$ if and only if $(b-a)\geq 0$. (And then one needs to check that this order actually gives us an ordered field.)

One way to define the order is this: a formal Laurent series in $R((x))$, where $R$ is an ordered field, is $\geq 0$ if and only if it is $0$ or $c_n \geq 0$ in $R$, where $n$ is the least power of $x$ with nonzero coefficient $c_n$.

This ordering makes lower powers of $x$ "more significant", so you can think of it as an ordering of $R((x))$ in which $x$ is positive and infinitesimal.

Answer 2

No, it's not a field with the multiplication you defined! The multiplicative identity is $\sum z^n$, hence $z$ is a non-zero element with no multiplicative inverse.

Where did this come from? Offhand I don't see how to define a multiplication that makes $R[[x]]$ into a field. The natural multiplication for formal power series is $$\left(\sum a_nx^n\right)\left(\sum b_n x^n\right)=\sum_n\left(\sum_{j+k=n}a_jb_k\right)x^n,$$but of course that doesn't converge for formal Laurent series...

Answer 3

$\mathbb R[[X]]$ is usually reserved for formal power series. Formal Laurent series are denoted $\mathbb R((X))$. However, they look different than your version. The main difference: you defined multiplication componentwise, essentially making your ring the same as $\mathbb Z^{\mathbb R}$, the ring of real valued maps on the integers (each component $a_nx^n$ of the series corresponds to the fact that $n\mapsto a_n$). This ring is not a field. It's not even an integral domain, since $(X)\cdot(X^2)=0$, for instance, so we have zero-divisors.

The usual way to make the Laurent series into a field (assuming we are considering coefficients from a field, which we do here) is to stipulate that they may only have a finite amount of negative powers and to define multiplication as

$$\left(\sum_{k=-n}^\infty a_k X^k\right)\left(\sum_{k=-m}^\infty b_k X^k\right):=\sum_{k=-(n+m)}^\infty \sum_{i+j=k}a_ib_j X^k.$$

The reason for the finite amount of negative powers is to make the second sum finite, since infinite sums are not well defined in general.

As for a suitable ordering, you can define a positive cone $P\subset\mathbb R((X))$ as the set of all formal Laurent series whose leading coefficient (meaning the one with the smalles index) is nonnegative. With the ordering $f\geq g:\Leftrightarrow f-g\in P$ you get an ordered field.