# How Many Number Fields Are There?

Modern results on a problem from the 1800s.

Sometimes I glance through the current *Annals of Mathematics* to see what types of new results are being recognized as important. If you’re unaware, *Annals* is widely considered to be the best journal of math research (obviously, others are in contention and I mean no disrespect to them).

If your research gets accepted there, it has something important to it: new ideas, old problems, widespread interest, etc.

I ran across an article entitled “Enumerating Number Fields” by Jean-Marc Couveignes this year. We won’t get into any of the technical details, but this is a concept most people with just high school algebra can appreciate.

Have no fear if you’ve never heard these terms before. We’ll go through lots of examples you’re already familiar with to get a handle on what’s going on.

# Number Fields and Terminology

A ** number field** is a finite (algebraic) extension of the rational numbers ℚ.

If you’ve seen the complex (imaginary) numbers, then you’re already familiar with the concept of adjoining new numbers to known number systems. You get imaginary numbers by adjoining *i* to ℝ.

We’ll notate this ℝ(*i*). This is everything of the form a+b*i* where a and b are real numbers. These are called the complex numbers ℂ=ℝ(*i*). Recall that *i*=√-1, so it has the property *i*²=-1.

Now, you might be thinking, where did *i* come from? Did we make it up at random?

Before, we said that a number field is a finite “algebraic” extension of ℚ. The term ** algebraic** here means we can only adjoin new numbers if they solve a polynomial equation with integer coefficients.

So, in the case of *i*, we adjoined it because it solves x²+1=0, and no other real number solves it.

Now ℂ isn’t a number field because you can’t adjoin finitely many new algebraic numbers to ℚ to get ℂ. For example, π is in ℂ, but it cannot be obtained by adjoining algebraic numbers to ℚ.

An example of a number field would be ℚ(√2). This is just the collection a+b√2, where a and b are rational numbers. We’re allowed to adjoin √2 because it solves x²-2=0.

Adjoining a number ℚ(*s*) isn’t as simple as making a+b*s*, though. It just happened to be that simple in the above two examples (because they solved a quadratic, i.e. degree 2, equation).

If we adjoin ∛2, then we make a+b∛2+c(∛2)². This is because adjoining a number means taking all combinations of sums and products (and differences and quotients) of elements of ℚ and the new number.

Other examples are ℚ(*i*) or ℚ(√2, √3). As you see, we don’t have to limit ourselves to only adjoining one. As long as we adjoin finitely many, it will still be a number field.

A surprising thing is that the *Primitive Element Theorem* says you can always find one number to adjoin to get the whole thing, even if you were to adjoin 100 new numbers.

For example, ℚ(√2, √3)=ℚ(√2+√3). It’s a common textbook exercise to prove this if you want to get more of a feel for what’s going on.

This means that for the rest of this article, you can assume a * number field* L=ℚ(

*s*) for some

*s*∈ℂ.

Now you might start to see some of the reasons number fields are the fundamental object of study in algebraic number theory. There’s an intimate connection between these and factoring/solving polynomial equations.

It would take an entire course in Galois theory to get even a surface understanding of just how deep this connection goes.

# How Many Are There?

Since number fields are so important, the first question might be: how many are there?

The answer isn’t very interesting, because there are infinitely many.

This isn’t as obvious as it sounds. Mathematicians only want to consider two number fields to be “different” if you can’t turn one into the other while preserving the addition and multiplication structure.

This is called an ** isomorphism**. We won’t get into the technical definition here.

So, if you wanted to prove there are infinitely many number fields, you’d need to come up with an infinite collection of number fields and then somehow prove they are all mutual non-isomorphic.

There are plenty of ways to do this, and it’s simple enough to be an exercise in a first course on the topic, but it’s also a good cautionary tale that even “obvious sounding” things require some work to prove when dealing with this topic.

This gives us motivation for the next two definitions: degree and discriminant.

## Degree and Discriminant

Recall that the ** degree** of a polynomial aₙxⁿ+…+a₁x+a₀ is just n. It’s the highest exponent that occurs.

A ** quadratic equation** ax²+bx+c has degree 2.

The * degree* of a number field L=ℚ(

*s*) is just the degree of the smallest degree polynomial, p(x), (with rational coefficients) such that p(

*s*)=0.

We know that such a polynomial exists, because that was part of the definition of a number field (recall, we’re only adjoining *algebraic* numbers).

The degree of ℚ(√2) and ℚ(*i*) are both 2 because of x²-2 and x²+1 respectively. The degree of ℚ(√2, √3) is 4.

So, the next obvious question is: are there finitely many number fields of degree 2 (or any fixed n)?

No, again!

The slickest proof I know of for this is that if s and t are squarefree, then ℚ(√s) and ℚ(√t) are not isomorphic because they have different discriminants.

This gives us some motivation for why we might want to define discriminants.

You might have heard this term when learning the quadratic formula. The solutions to ax²+bx+c=0 are given by:

The part under the square root: b²-4ac, is called the ** discriminant**. You can look up the discriminant of a general polynomial, but it’s quite complicated. The discriminant basically tells you if there are no solutions, a repeated solution, or if all the solutions are already in ℚ.

I had to give the warning earlier about obvious things running into subtleties. The definition of the discriminant of a number field is quite complicated (way beyond the scope of this article).

For our purposes, you can think of the * discriminant* of L=ℚ(

*s*) as the discriminant of the smallest degree polynomial, p(x), so that p(

*s*)=0.

So, the discriminant of ℚ(√2) is the discriminant of x²-2, which is 0²-4(1)(-2)=8.

The discriminant of ℚ(i) is the discriminant of x²+1, which is 0²-4(1)(1)=-4.

I’ll reiterate that this isn’t actually the definition. It matches sometimes (including the two examples I gave), and it doesn’t match other times.

Thinking of it this way gives you a feel for the definition, though, and why it might exist.

Discriminants show up all over the place, including in the definition of Dedekind zeta functions, which are a generalization of the Riemann zeta function. You might know about this from the Riemann Hypothesis.

# Counting Number Fields

From now on, we’ll notate the discriminant of L as Δ(L).

A theorem of Hermite and Minkowski, going all the way back to the late 1800's, says that for any integer N>0, there are only finitely many number fields, L, with |Δ(L)|<N.

(The bars indicate ** absolute value**).

In other words, we finally can get some finiteness results by bounding the discriminant. This is the key idea behind all the modern results.

**New Question:** If we fix the degree, d, and bound the discriminant, how many number fields are there?

Let’s go back to our trusty examples with square roots. If we fix d=2, then we know the number fields look like L=ℚ(√m), where m is square-free.

If we consider x²-m=0 as the minimal polynomial, then we see the discriminant Δ(L)=4m (though, again, I lied earlier, and sometimes this is actually just m).

So, one rudimentary estimate is that there are less than N number fields of degree 2 and discriminant bounded by N.

We did it! We got a result that helps us understand how many number fields there are.

Minkowski gave a very good bound in general, and that’s what people have been trying to improve on ever since. There was work in the 1970’s by Odlyzko that made improvements in the techniques used.

The main conjecture is that there a constant c (depending on the fixed degree d) so that the number of number fields of discriminant bounded by N is proportional to cN as N→∞.

Ellenberg and Venkatesh had a 2006 paper in *Annals* on results in the “relative” case (so not just over ℚ). Obviously, I’ve skipped a bunch of history, and more of it can be found there.

And finally, Jean-Marc Couveignes has the 2020 *Annals* paper I mentioned in the introduction on even more results of this form.

So you see, classical math problems from the 1800’s can still have quite a bit of active current research using modern tools.