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?

  • $\begingroup$ The term "exceptional" usually refers to some kind of classification. Here there is no such thing, so there is no real place for anything exceptional. $\endgroup$ Aug 20, 2019 at 15:58

2 Answers 2


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.

  • $\begingroup$ Ah yes, rational functions make it more easy than considering infinite series. $\endgroup$
    – zooby
    Aug 19, 2019 at 23:58
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    $\begingroup$ If you want it to be non-commutative you can look at $\Bbb{H}\otimes_{\Bbb{R}} \Bbb{R}(x)= \Bbb{R}(x)+\Bbb{R}(x)i+\Bbb{R}(x)j+\Bbb{R}(x)k$ $\endgroup$
    – reuns
    Aug 20, 2019 at 0:03
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    $\begingroup$ Analytic real-valued functions (on a connected real manifold) are okay, too. This amounts to the observation that a nonzero analytic function has a discrete set of zeroes. This falls apart for disconnected manifolds/topological fields. $\endgroup$
    – tomasz
    Aug 20, 2019 at 13:02

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.

  • $\begingroup$ That's a good one. I notice when you multiply power you have $x^n\times x^m=x^{n+m}$. I wonder if you could get a ring if you changed the rule to $x^n\times x^m = x^{(n+m)\mod p}$ ? Well that wouldn't be infinite. But maybe you could make $p$ very large. I wonder if it would still be a division ring. $\endgroup$
    – zooby
    Aug 20, 2019 at 2:08
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    $\begingroup$ @zooby Note that $x^{(n+m) \mod p}$ is the same as $x^{n+m} \mod x^p-1$. As $x^p-1 = (x-1)(x^{p-1}+...+1)$ is obviously not irreducible for $p>1$, I think the answer to your follow-up question is no, but I am a bit rusty on that stuff. $\endgroup$
    – mlk
    Aug 20, 2019 at 9:13
  • $\begingroup$ @zooby As mlk points out, that wouldn't be a division ring since $(x-1)\cdot (x^{p-1}+\ldots + 1)=0$, which implies that $(x-1)$ lacks an inverse. You can mess with the rule for powers in other ways - you can do things like allow rational exponents, as long as the denominators stay bounded (called Puiseux series) or even allow any well-ordered subset of $\mathbb R$ to be a set of exponents (called Hahn series). You could also play around with non-commutative rules for multiplication of monomials, but this leads to considerable difficulties. $\endgroup$ Aug 20, 2019 at 12:43

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