# Can you construct a field over every set $M$?

I know there are finite fields like $\mathbb F_2$, $\mathbb F_4$ or the $\mathbb Z/n\mathbb Z$ for prime $n$ with modulo operations. For other special $n$, I've seen fields $\mathbb F_n$ with $n$ elements being constructed. And of course there are the usual infinite fields (take $\mathbb Q$ and so on).

So I wonder: Let $M$ be an arbitrary set that contains at least two elements. Can you always find operations $+ : M\times M \to M$ and $\cdot : M \times M \to M$, such that $(M, +, \cdot)$ is a field?

• The cardinality of finite field always must be a power of prime number. Thus, for example, you can not build a field of 20 elements. – Norbert Jan 12 '12 at 12:38
• To complete Norbert's comment : You can see from this Wikipedia page : en.wikipedia.org/wiki/Finite_field that a finite fields exists if and only its cardinal is a power of a prime number. – Student Jan 12 '12 at 12:53
• – Watson Aug 22 '16 at 13:14

## 3 Answers

As the comments remarks, for finite sets this is impossible unless $|M|=p^k$ for some prime number $p$ and $k>0$. This is because the characteristic of the field must divide the number of elements in the field.

If $M$ is infinite, however, this is possible:

If $M$ is infinite, then the $\mathbb Q(M)$, which is the field of rational functions whose variables are the elements of $M$, is equinumerous with $M$ (to see that, note that this is the fraction field of polynomials and there are $\max\{|M|,\aleph_0\}=|M|$ many of those, therefore the field is of the same cardinality).

Therefore there is $f:M\to\mathbb Q(M)$ which is a bijection, and we can use it to define the operations: $m_1\cdot m_2 = f^{-1}(f(m_1)\cdot f(m_2))$ and similarly with addition.

Do note that this requires heavily the axiom of choice, since the assertion that $M$ has a structure of a group is already equivalent to the axiom of choice - so requiring it to be a ring or a field is to require "more".

No. If $M$ is finite, there is always at least one ring structure on $M$; if the cardinality of $M$ is $n$, then put the elements of $M$ in one-to-one correspondence with the elements of the ring $\mathbb{Z}/n\mathbb{Z}$; and let the operations $+$ and $\times$ be those induced on $M$ by the correspondence. However, unless $n$ is a power of a prime, there is no field structure on $M$.

The problem is that if $M$ is finite and its order is not a power of a prime, any ring structure on $M$ will contain zero divisors i.e. nonzero elements $a,b$ such that $ab=0$. These elements cannot possibly have inverses; e.g. if $a$ had an inverse, we could multiply $ab=0$ by $a^{-1}$ to obtain $b=0$ but $b$ was presumed nonzero. But in a field, all nonzero elements have inverses.

The reason that $M$ must contain zero divisors if $n$ is not a power of a prime is as follows:

Let $M$ be an arbitrary ring of order $n$. There is a unique homomorphism of $\mathbb{Z}$ into any ring. If $\bar{1}$ is the multiplicative identity of the ring, the homomorphism is defined by sending $1\mapsto\bar{1}$, $2\mapsto \bar{1}+\bar{1}$, etc. The image of this homomorphism is a subring $S\subset M$. Let $h$ be the order of $S$. Then $S$ is isomorphic to $\mathbb{Z}/h\mathbb{Z}$, because it is (by the first isomorphism theorem) isomorphic to a quotient of $\mathbb{Z}$, and it is order $h$.

Now if $h$ is composite, then $\mathbb{Z}/h\mathbb{Z}$ contains zero divisors: if $h=ab$ is a nontrivial factorization, then the residues of $a,b$ mod $h$ satisfy $\bar{a}\bar{b}=0$ in $\mathbb{Z}/h\mathbb{Z}$. So if $M$ is a field, $h$ has to be prime. Then $S\cong \mathbb{Z}/h\mathbb{Z}$ is a subfield of $M$.

Now we know if $M$ is a field of finite order $n$, it has a subfield $S$ with a prime number of elements. Then $M$ is actually a vector space over $S$. It is a finite dimensional vector space because it only has a finite number of elements. Say it is dimension $m$. Then it is isomorphic as a vector space to $S^m$ for some finite number $m$. (This is not a field isomorphism because $S^m$ is not a field.) But the order of $S^m$ is $h^m$, so this is the order of $M$. We saw that $h$ must be prime for $M$ to be a field, otherwise $S$ contains zero divisors, so this proves that the order of $M$ must be a power of a prime.

This answers the question, but I made a slightly stronger claim: if $n$ is not a power of a prime, then any ring structure on $M$ contains zero divisors. I have shown that if $S$, the homomorphic image of $\mathbb{Z}$, is not of prime order, then $S$ and therefore $M$ contains zero divisors, and that if $M$ happens to be a field, then the order of $M$ is a power of that of $S$. The stronger claim rests on the observation that it is not possible for a ring structure on a finite set $M$ to avoid having zero divisors without turning it into a field.

Here is one way to prove this: if $M$ is a finite ring and no element of $M$ is a zero divisor, then let $a$ be any nonzero element of $M$. Two elements of the form $ba$ and $ca$ cannot be equal unless $b=c$, because if they were we'd have $(b-c)a=ba-ca=0$, but we said there were no zero divisors and $a$ is nonzero, so we conclude $b-c=0$ i.e. $b=c$. To summarize, if $ba=ca$ then $b=c$. This means multiplication of the elements of $M$ by $a$ is an injective function on $M$. Because $M$ is finite, this implies it is bijective and in particular surjective. So there is some element that when multiplied by $a$ yields $1$. This means $a$ is invertible in $M$. Since the argument applies to any nonzero $a$, this shows that every nonzero element of the ring $M$ is invertible; i.e. $M$ is a field.

To summarize the results: if $M$ is a finite ring and has no zero divisors, then it is a field, which means the homomorphic image of $\mathbb{Z}$ in $M$ (which we called $S$) has prime order ($h$), and $M$ is a finite-dimensional vector space over this subfield, so its order is a power of a prime. So if $M$ is not a power of a prime, any ring structure on $M$ must have zero divisors.

In the case of infinite sets $M$, this property implies the axiom of choice. Indeed, this mathoverflow answer shows that even the assertion that every set admits a group structure already implies the axiom of choice.

• $$\huge\stackrel{\small\wedge\wedge}{\smile}$$ – Asaf Karagila Jan 12 '12 at 13:18
• Roughly the time it takes Usain Bolt to run those sprints... ;-) – Asaf Karagila Jan 12 '12 at 13:24
• I should have hit enter before running, instead of after. – JDH Jan 12 '12 at 13:35