Is Every Set Fieldable?
This is the first in a hopefully long series on some of the most absolutely stupid questions I came across while studying mathematics. The idea is presenting a stupid toy problem, the context behind it and how I came across a solution. My expectation is that none of the questions posed in this series will be of any significance. The first question we will explore is set-theoretic in nature (because of course): can every infinite set be endowed with the structure of a field? In other words, given an infinite cardinal number , are there fields of cardinality ?
Notice that the statement is clearly false for finite sets. It is a well known fact that all finite fields are finite extensions of the fields of prime order . In particular, the order of a finite field must be a prime power.
I first asked myself a similar question while trying to apply a result from Group Theory to an unrelated problem. Specifically, I unfortunately found myself in the position of trying to apply a theorem about groups to a set which did not posses a natural group structure. The question then becomes natural: does have any group structure at all? Using Cantor’s normal form I was able to show that is in bijection with the additive group of a certain commutative ring . If we fix a bijection we can thus endow with the structure of an Abelian group by declaring that this bijection is actually a group isomorphism.
In fact, the same argument goes to show that every infinite set can be endowed with the structure of a commutative ring! Unfortunately this argument falls short of a proof of our statement about fields, since the ring I constructed turned out to contain divisors of zero. The question then came: can I somehow improve this proof to account for fields? Coming up with an answer required me to develop a new method, which I would now like to share. Explicitly, we will show that, given a cardinal , the field
of rational functions in variables is in bijection with .
To be clear, the fact that there are fields of cardinality is an immediate corollary of the Löwenheim–Skolem theorem, but we will alternatively present a constructive proof. As it turns out, much of the delicacies of this proof come down to one of the many pathologies of cardinal arithmetics. Specifically, throughout this article we will use the the fact that, given two infinite cardinals and , . Our proof has two primary steps, the first of which consists of showing that…
Taking the polynomial ring doesn’t change the cardinality
Denote by the ring of rational polynomials in variables , . We want to establish . Clearly . Indeed, there is a natural injection
The challenge thus is showing that is no larger than itself. We start by bounding the cardinality of the set of monomials in the variables . Given a non-negative integer , there is a canonical surjection
Hence . Since the union of countably many sets of size has size no larger than , this implies
By the same token, it follows from the fact is countable that , the set of rational combinations of the monomials in the variables , has cardinality . Next we show that formally inverting polynomials does not change the cardinality of our rings. In other words…
Taking the field of fractions doesn’t change the cardinality
Let be the field of rational functions in the variables , . We want to establish that . Once again, it is already clear that . Indeed, there is a canonical injection
given by the composition of the canonical injection with the inclusion .
On the other hand, there is a surjection
and therefore . It thus follows from the Cantor-Bernstein theorem that is precisely .
This concludes our proof. I would like to point out that this proof also works for constructing fields of positive characteristic. Indeed, if we take instead of , the exact same argument show that . In addition, I am strongly convinced that similar methods can be used to show that every infinite set can be endowed with the structure of a algebraically closed field — of arbitrary characteristic!
On the other hand, I should note that most of what we discussed is nonsense unless we work under the assumption of the axiom of choice, for if the axiom of choice does not hold we can find a set which does not have a cardinal. I’m not entirely sure of whether or not I expect our result holds without the assumption of the axiom of choice, but I am inclined towards trying to disprove it. I’ll leave that for another post though. See you then!