# Ring of formal Laurent series: units and defining operations

I'm struggling to see how one could show that the ring of formal Laurent series in $F$ is unital.

I defined addition and multiplication to be, for $p, q \in F((t))$: if $$p = \sum_{i \in X} a_i t^i$$ for some finite set $X \subset \mathbb{Z}$ and likewise for $Y$ in $$q = \sum_{j \in Y} b_j t^j$$ then

$$p+q = \sum_{k \in X \cup Y}(a_k + b_k)t^k$$

and

$$pq = \sum_{l \in X + Y} \sum_{i+j=l} a_i b_j t^l$$

where $X+Y = \{x+y\mid x \in X, y \in Y\}$ and $i \in X$, $j \in Y$.

I seem to have proved all of the necessary ring axioms save for the distributive property (not yet), but I can't figure out how to find units for this ring. The units of the formal power series are defined as the elements such that $a_0$ is an unit in $F$ (and the other coefficients are defined recursively. However, of course, $0$ may not be even in $X \cup Y$.

## Laurent series and Laurent polynomials

Actually, by using finite $$X$$ and $$Y$$, what you've written is only a definition for Laurent polynomial arithmetic, and not Laurent series arithmetic.

The expression you gave carry over directly, but the condition on $$X$$ and $$Y$$ is that they have a least element. Notice that if both have a least element, then so do $$X\cup Y$$ and $$X+Y$$.

## Now for your question

The expression $$p = \sum_{i \in X} a_i t^i$$ is one way to express $$p$$, but it's not the only way to write $$p$$.

If $$0\notin X$$, then notice that it can be added in for free:

$$p = \sum_{i \in X} a_i t^i=0t^0+\sum_{i \in X} a_i t^i=\sum_{i \in X'} a_i t^i$$ where $$X'=X\cup\{0\}$$. You can see that $$X'$$ still has a least element, which is either the least element of $$X$$ or $$0$$ itself. Naturally $$a_0=0$$ in this case. In other cases, you already have $$a_0\neq 0$$ given to you.

So actually you can see that every coefficient in a Laurent series is defined, it's just that there are limitations that make many of the coefficients $$0$$, and hence omissible. Such conditions are necessary because they ensure that each coefficient of the sum and product can be obtained from computations with finitely many nonzero numbers.

## Identity

I'd like to adapt the notation above so it's a little easier to write things. When I wrote $$p_0$$, I mean the coefficient of $$0$$ in $$p$$. Similarly $$q_5$$ is the $$5$$ coefficient of $$q$$. And $$(pq)_{-1}$$ is the $$-1$$ coefficient of the product $$pq$$.

Using this notation, your arithmetic formulas change into:

$$(p+q)_k = p_k + q_k\\ (p\cdot q)_k = \sum_{i+j=k} p_i q_j$$

The identity is the obvious candidate: the coefficent of $$t^0$$ should be $$1$$, and all other coefficients should be $$0$$. I'm going to call that particular series "$$1$$".

Checking the multiplication:

$$(p\cdot 1)_k = \sum_{i+j=k} p_i 1_j=p_k$$ (Can you see how that works?) This means that $$p\cdot 1=p$$, since all of their coefficients match. Thus $$1$$ is the identity.

## Units of $$F[[t]]$$

The units are not defined as you described. What you wrote was more like a description of the units of $$F[[t]]$$ (but not a definition.) Let's make that clear here.

To find all the units for the ring of power series over a ring $$F$$, I'm going to first have you look at units in the power series ring $$F[[t]]$$, which as you know looks just like $$F((t))$$ except coefficients below $$0$$ are always $$0$$.

First, let's look at units in $$F[[x]]$$. By definition, $$p$$ is a unit if there is a $$q$$ such that $$p\cdot q=1$$. Suppose $$pq=1$$. Then the formula for multiplication says that $$p_0q_0=1\in F$$, so we know for sure that $$p_0$$ has to be a unit.

Convesely, consider what $$q$$ must look like to invert $$p$$ if $$p_0$$ is a unit in $$F$$.

We have: $$(p\cdot q)_0=p_0q_0=1$$ and solving for $$q_0$$ we get $$q_0=p_0^{-1}$$. This is possible to write since $$p_0$$ is a unit of $$F$$.

Next $$(p\cdot q)_1=p_1q_0+p_0q_1=0$$, and solving for $$q_1$$ we get $$q_1=-p_1q_0p_0^{-1}$$

Next $$(p\cdot q)_2=p_2q_0+p_1q_1+p_0q_2=0$$, and solving for $$q_2$$ we get $$q_2=-p_0^{-1}(p_2q_0+p_1q_1)$$.

Continuing inductively, we produce an infinite series of $$q_i$$ with the property that their Laurent series $$q$$ satisfies $$p\cdot q=1$$. We have now established that units of $$F[[t]]$$ are exactly the series with a unit of $$F$$ in first coefficient.

## Units of $$F((t))$$

(Note: as observed in the comments and other solution, this is insufficient when $$F$$ is not a domain. When the coefficient ring has zero divisors, there can be "more" units than this analysis expects. It should be fine for at least domains, and in particular, $$F$$ a field.)

Finally, we tackle your unit question. First, notice that $$t$$ is a unit in this ring since it contains the series $$t^{-1}$$ which is $$1$$ in the $$-1$$ spot and $$0$$ elsewhere. Multiplying you'll find that $$t\cdot t^{-1}=1$$.

Now let $$p$$ be any nonzero Laurent series. Obviously $$p=t^{z}p'$$ where $$p'$$ is a power series with a nonzero first coefficient. Since $$t^z$$ is obviously a unit, $$p$$ is a unit of $$F((t))$$ iff $$p'$$ is a unit of $$F[[t]]$$, but we already know that means that the lowest nonzero coefficient of $$p'$$ is a unit of $$F$$.

So there you have it: the units of $$F((t))$$ are the ones with lowest nonzero coefficient a unit of $$F$$. If $$F$$ is a field, then actually $$F((t))$$ is a field too. It turns out to be the ring of fractions of $$F[[t]]$$, in that case.

• I was thinking that by saying $X$ was finite would be the same thing as (implicitly) saying it had a least element, but I'll make sure to add that extra assumption, thanks. How should I go about finding units?
– Lost
Dec 13, 2013 at 13:56
• @Lost I added a bunch of information which hopefully gets you on track. Hope it helps! Dec 13, 2013 at 15:11
• Thanks a lot, this really helps out. When I looked around for about the units, I couldn't find anything that didn't look convoluted. The fact that $F((t))$ is the ring of fractions of $F[[t]]$ is what I'll be proving right afterward.
– Lost
Dec 13, 2013 at 15:27
• I hadn't thought about $p = t^zp'$, I tried defining $p^{-1}$ to be the series for which the indices belonged to $-X$, the set consisting of the additive inverses of $X$ and you know how well that turned out.
– Lost
Dec 13, 2013 at 15:30
• @rschwieb: the second part of the statement "$p$ is a unit of $F((T))$ iff $p'$ is a unit of $F[[t]]$" needs some justification. Certainly, at first we only get from $p=t^zp'$ that the power series $p'$ is a unit of $F((t))$. But why is it immediately clear that $p'$ then must be a unit in $F[[t]]$ too?
– M.G.
Jun 19, 2018 at 18:45

I am late to the game, but after some reflection upon my remark to rschwieb's answer, I came to the conclusion that his answer is in general incorrect when $$F$$ is not an integral domain. Here is an explicit counter-example:

Let $$F:=\mathbb{C}[X]/(X^2)$$ and let $$\varepsilon$$ be the image of $$X$$ in $$F$$, i.e. $$F=\mathbb{C}[\varepsilon]$$ with $$\varepsilon^2=0$$. Consider $$f(t):=\frac{\varepsilon}{t^2}-\frac{i}{t}\in F[t,1/t]\subseteq F((t))$$ and $$g(t):=\varepsilon+it\in F[t]\subseteq F[[t]]$$ Then $$f(t)g(t)=1$$ showing that the lowest non-zero coefficient (of $$f$$) need not be a unit in $$F$$.

The cool thing about this example is that it is minimal in that it uses non-trivial nilpotent element of lowest possible order, which in turn - turns out - precludes constructing a $$f$$ with lowest degree $$(-1)$$, so the term $$t^{-2}$$ is actually necessary, and requires a pair of numbers that are inverse to each other with respect to both addition and multiplication, which forces introducing a $$\sqrt{-1}$$.

EDIT: As pointed out by user math54321 in the comments, the existence of $$\sqrt{-1}$$ is not necessary to realize a non-trivial example of minimal degree.

• Hi, I somehow missed your comment 5 years ago. Today someone upvoted and I was drawn to take another look. Thanks for the careful reading. I've added a note along the lines of the problem you've brought forward. +1 May 11, 2023 at 14:26
• There's no need to introduce $\sqrt{-1}$: if $R$ is any non-reduced ring, then there exists $0 \ne a \in R$ with $a^2 = 0$, and taking $f := a + t, g := -\frac{a}{t^2} + \frac{1}{t} \in R((t))$ gives $fg = 1$ Jun 21, 2023 at 4:55
• @math54321: Oh, you are completely right! I must have missed that case. Of course one "conjugate" the nilpotent element instead! Thanks for noticing and for correcting me!
– M.G.
Jun 21, 2023 at 14:06