Are there any infinite dimensional division algebras? Appart from the finite dimensional division algebras like $\mathbb{R, C, H, O}$
Are there any infinite dimensional division algebras? (Especially any "exceptional" ones?)
I was thinking maybe the ring over polynomials might be a division algebra if you include negative exponents and allow infinite series. But I'm not sure if every series gives a unique member of the algebra. You might have $(1+x)^{-1} = 1-x+x^2-...$
Well I guess the space of functions is a division algebra since you can add and divide them $f(x)g(x)$ and $f(x)/g(x)$ and has an identity element $1$.
What about the ring over polynomials with rational or irrational exponents?
Or ones based on lattices?
 A: Your question actually describes an infinite dimension division algebra - the field of Formal Laurent series. In particular, the elements of this field are just expressions of the form:
$$a_{-k}x^{-k}+a_{-k+1}x^{-k+1}+\ldots+a_{-1}x^{-1}+a_0+a_1x+a_2x^2+\ldots$$
where the series can go on forever to the right, but may not have infinitely many terms of negative exponent (although can have arbitrarily many). Note that we're not concerning ourselves with convergence or anything like that - these are purely expressions that you manipulate by adding them coefficient wise and multiplying them by taking every pair of terms from the two series, taking their product, then collecting like terms (which is, for each coefficient, a finite process due to the fact that there are only finitely many negative terms included in any series).
It's sort of a pain to write out the exact formula for the multiplicative inverse of an element, but you can do it fairly nicely in two steps: First, note that every non-zero element is of the form
$$c\cdot x^n\cdot F$$
where $c$ is an element of the field we're taking our coefficients from and where $F$ is of the form $F=1+a_1x+a_2x^2+\ldots$. Since we can clearly invert $c$ as it's just a real number (or something like that) and we can invert $x^n$, all we need to do is invert $F$. We can do that by solving
$$(1+a_1x+a_2x^2+\ldots)\cdot (1+b_1x+b_2x^2+\ldots)=1+0x+0x^2+\ldots$$
which gives the equations, for each $n\geq 1$ that
$$\sum_{i=0}^{n}a_ib_{n-i}=0$$
which rearranges to say
$$b_n=-\sum_{i=1}^na_ib_{n-i}$$
after we pull out one term from the sum. We can then inductively figure out the power series inverse to any of the form $1+a_1x+a_2x^2+\ldots$ and extend that as you wish.
It's also worth noting that this construction gives a division ring whenever we take our coefficients from a division ring - so if we want something non-commutative, we could apply this construction to have quaternion coefficients.
A: The rational functions (in one variable) provide an example of such a ring ... after we mod out by the appropriate equivalence relation, namely "equality off a finite set" (so that e.g. "$x+3$" and "${(x-2)(x+3)\over x-2}$" are the same thing). And the rational functions are only the tip of a much larger iceberg of "well-behaved" functions, especially when we work over $\mathbb{C}$ instead of $\mathbb{R}$ (see the notion of analytic function).
However, this idea isn't so straightforward for arbitrary collections of functions. Rational functions are extremely well behaved, and when we lose that good behavior things get quite difficult. For example, let $f(x)=1$ if $x>0$ and $0$ if $x\le 0$, and let $g(x)=1-f(x)$. Then the product of $f$ and $g$ is the always-zero function, but both $f$ and $g$ are nonzero a lot of the time so there's no clear picture of which one "ought to be" the zero element of our algebra. In general, finding a natural equivalence relation making a division algebra out of a given class of functions can be quite hard.
