TITLE: Ring of formal Laurent series: units and defining operations
QUESTION [8 upvotes]: 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 $X \cup Y$.

REPLY [3 votes]: 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 $\sqrt{-1}$.