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

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


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

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 $X \cup Y$.


2 Answers 2


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.


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.

  • $\begingroup$ 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? $\endgroup$
    – Lost
    Dec 13, 2013 at 13:56
  • $\begingroup$ @Lost I added a bunch of information which hopefully gets you on track. Hope it helps! $\endgroup$
    – rschwieb
    Dec 13, 2013 at 15:11
  • $\begingroup$ 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. $\endgroup$
    – Lost
    Dec 13, 2013 at 15:27
  • $\begingroup$ 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. $\endgroup$
    – Lost
    Dec 13, 2013 at 15:30
  • 1
    $\begingroup$ @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? $\endgroup$
    – 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.

  • 1
    $\begingroup$ 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 $\endgroup$
    – rschwieb
    May 11, 2023 at 14:26
  • 1
    $\begingroup$ 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 $fg = 1$ $\endgroup$
    – math54321
    Jun 21, 2023 at 4:55
  • $\begingroup$ @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! $\endgroup$
    – M.G.
    Jun 21, 2023 at 14:06

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