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Please allow me, for the purposes of this question(but only here), to exaggerate matters and state two polemic definitions. Please forget these definitions after answering this question, and pardon my silly nitpicking.

Definition $1$: "Algebraic number theory" is the theory of algebraic numbers. We exclude arithmetic geometry and such.

Definition $2$: "Number theory" is the study of properties of natural numbers.

In the above sense, I seek examples of applications of algebraic number theory to number theory. I mean, those applications which throw light on "numbers" as we know them in primary school. There is of course enlightenment by looking at a bigger picture of so many number rings, but that is not what I mean. I have specifically the application to the down-to-earth integers in mind. What I know are the following:

$1$. The theorem that an odd prime is of the form $a^2 + b^2$ if and only if it is of the form $4n +1$, proved by looking at factorization in the Gaussian ring.

$2$. Pell's equation is solved with Dirichlet's unit theorem.

$3$. von Staudt–Clausen theorem on Bernoulli numbers, proved using cyclotomic theory.

$4$. Certain equations, like the Fermat equation $X^n + Y^n = Z^n$, may "split" in some extension field and thus it makes sense to go to bigger rings, to study diophantine equations. Here I mean the work of Kummer which started ideal theory, algebraic number theory, etc..

I exclude the following:

$5$. Arithmetic geometry can be used together with algebraic geometry, to study diophantine equations. Elliptic curves fall in here, when their geometry is used significantly(such as in the work of Katz-Mazur). That is "arithmetic geometry", for the purposes of this question. I am more interested in hearing about applications of "algebraic number theory", as defined above.

$6$. Again using algebraic geometry and also modular forms, conjectures such as the Ramanujan bound on the tau function can be proved. Here "modular forms" are "analytic", or "transcendental", and also "geometry is involved. So it goes beyond the "algebraic number theory"

$7$. Dirichlet's theorem on arithmetic progressions is "analytic number theory".

So I thus exclude any touch of "analytic number theory" and "arithmetic geometry", from "algebraic number theory" as defined above. But it can include Kummer theory, classfield theory, etc.. I do not know where to put in Dorian Goldfeld's results on the Gauss class number problem. It uses Gross-Zagier, which is significantly geometric, but gives a result expressible in terms of rational integers. Also I do not know whether Iwasawa theory is arithmetic geometry or not. Langlands theory etc., must be excluded, because it is even more abstract. I want only the "first course in algebraic number theory", "basic cyclotomic theory", "classfield theory" etc., in short only those things which are obviously the study of algebraic numbers.

So, question:

Are there other applications of "algebraic number theory" to "study of natural numbers", than the examples 1-4 above?

I tag this question "big-list" because I hope there are indeed quite a few.

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Community wiki? – H. Hasson Feb 24 '10 at 16:01
Catalan's Conjecture, maybe? – Arturo Magidin Feb 24 '10 at 16:12
You haven't given us any motivation for your question. Why is it fruitful to try to separate out algebraic number theory from other deeply intertwined branches of number theory and mathematics? – Pete L. Clark Feb 24 '10 at 16:41
I guess he wants to read a course in algebraic number theory and needs some motivation for his students. I doubt they want to hear motivation for stuff that will never be covered in the course. – darij grinberg Feb 24 '10 at 16:45
Questions that ask for a list of examples cannot, by definition, be answered since no such list is complete. Such questions are generally community wiki. – Qiaochu Yuan Feb 24 '10 at 20:46

A trick I have seen several times: If you want to show that some rational number is an integer (i. e., a divisibility), show that it is an algebraic integer. Technically, it is then an application of commutative algebra (the integral closedness of $\mathbb Z$, together with the properties of integral closure such as: the sum of two algebraic integers is an algebraic integer again), but since you define algebraic number theory as the theory of algebraic numbers, you may be interested in this kind of applications.

Example: Let $p$ be a prime such that $p\neq 2$. Prove that the $p$-th Fibonacci number $F_p$ satisfies $F_p\equiv 5^{\left(p-1\right)/2}\mod p$.

Proof: We can do the $p=5$ case by hand, so let us assume that $p\neq 5$ for now. Then, $p$ is coprime to $5$ in $\mathbb Z$. Let $a=\frac{1+\sqrt5}{2}$ and $b=\frac{1-\sqrt5}{2}$. The Binet formula yields $F_p=\displaystyle\frac{a^p-b^p}{\sqrt5}$. Now, $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ (by the idiot's binomial formula, since $p$ is an odd prime). Note that $p$ is coprime to $5$ in the ring $p\mathbb Z\left[a,b\right]$ (since $p$ is coprime to $5$ in the ring $\mathbb Z$, and thus there exist integers $a$ and $b$ such that $pa+5b=1$). Now,

$\displaystyle F_p=\frac{a^p-b^p}{\sqrt5}\equiv\frac{\left(a-b\right)^p}{\sqrt5}$ (since $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ and since we can divide congruences modulo $p\mathbb Z\left[a,b\right]$ by $\sqrt5$, because $p$ is coprime to $5$ in $p\mathbb Z\left[a,b\right]$)

$\displaystyle =\frac{\left(\sqrt5\right)^p}{\sqrt5}$ (since $a-b=\sqrt5$)

$=5^{\left(p-1\right)/2}\mod p\mathbb Z\left[a,b\right]$.

In other words, the number $F_p-5^{\left(p-1\right)/2}$ is divisible by $p$ in the ring $\mathbb Z\left[a,b\right]$. Hence, $\frac{F_p-5^{\left(p-1\right)/2}}{p}$ is an algebraic integer. But it is also a rational number. Thus, it is an integer, so that $p\mid F_p-5^{\left(p-1\right)/2}$ and thus $F_p\equiv 5^{\left(p-1\right)/2}\mod p$, qed.

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"Finding the period of the Fibonacci sequence modulo a prime" itself counts as an application, I'd say. – Qiaochu Yuan Feb 24 '10 at 17:25

The truncated exponential polynomial $1 + x + x^2/2! + ... + x^n/n!$ is irreducible for all positive integers $n$. This result is due to Schur and the proof uses prime ideal factorizations in the number field generated by one root of this polynomial.

Schur actually proved a more general theorem about irreducibility of polynomials of the form $$1 + a_1x + (a_2/2!)x^2 + ... \pm (1/n!)x^n,$$ where the $a_i$'s are all integers. The special case where all $a_i$'s are 1 is comprehensible on its own and already interesting as a problem on rational polynomials whose proof uses algebraic number theory.

This doesn't directly shed light on properties of "numbers" in primary school, as expressed in the original question, but unless the intention is to try to teach algebraic number theory to primary school students, an application of the subject to polynomials seems worthwhile for motivation.

For that matter, algebraic number theory provides a conceptual understanding of the Eisenstein irreducibility criterion (it is closely related to totally ramified primes) and explains why most polynomials $f(X)$ in ${\mathbf Z}[X]$ that are irreducible will not have an "Eisenstein translate" $f(X+c)$ for any integer $c$.

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Proofs for the two results cited here can be found at and – KConrad Feb 28 '10 at 2:36

You've listed quite a few restrictions, but hopefully this example will be "admissible".

Fix an integer $n>0$. Algebraic number theory sheds light on which rational primes are of the form $x^2+ny^2$.

Specifically, consider the order $\mathbb{Z}[\sqrt{-n}]$ of $\mathbb{Q}(\sqrt{-n})$. The relevant theorem is that a prime $p$ is represented by the quadratic form $x^2+ny^2$ if and only if it splits completely in the Ring Class Field of $\mathbb{Z}[\sqrt{-n}]$.

Here is the idea of the proof (for simplicity I'll restrict to the case in which $\mathbb{Z}[\sqrt{-n}]$ is the maximal order of $K=\mathbb{Q}(\sqrt{-n})$:

Direction 1: If $p=x^2 + ny^2$ for a prime $p\nmid 2n$, then $p\mathcal{O}_K=(x + \sqrt{-n}y)(x − \sqrt{-n}y)$. The ideal $(x+\sqrt{-n}y)$ is both prime and principal, so $p$ splits completely in the Hilbert Class Field of $K$.

Direction 2: Suppose conversely that $p$ splits completely in the Hilbert Class Field of $K$. Then we must have $p\mathcal{O}_K=\mathfrak{p}\mathfrak{q}$. By assumption both $\mathfrak{p},\mathfrak{q}$ split completely in the Hilbert Class field of $K$, so they must be principal. Say $\mathfrak{p} =(x+\sqrt{-n} y)\mathcal{O}_K$ and $\mathfrak{q} =(x-\sqrt{-n} y)\mathcal{O}_K$. Then $p=x^2+ny^2$.

Note that it is not difficult, using the Dedekind-Kummer theorem, to show that this is equivalent to the statement:

There is a monic irreducible polynomial $f_n(x)\in \mathbb{Z}[x]$ such that if an odd prime $p$ divides neither n nor the discriminant of $f_n(x)$ then $p$ is represented by $x^2+ny^2$ if and only if $(\frac{-n}{p})=1$ and $f_n(x)\equiv 0\pmod{p}$ has an integral solution.

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The Frobenius and Cebatarov density theorems -- the factorization of an integer polynomial modulo various primes is controlled by the Galois group of the corresponding extensions. Even if you are really hard core about excluding analytic results, to the point that you won't accept statements about Dirichlet density, surely it is worth understanding why $x^3+2 x^2-x-1$ always factors into (linear)(linear)(linear) or stays irreducible, but is never (linear)(quadratic).

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In the same spirit: there exist polynomials irreducible over Q which factor mod p for every p. – David Zureick-Brown Feb 24 '10 at 18:29
Are there any completely algebraic proofs of Chebotarev (the standard proof uses L-functions and essentially contains the proof of Dirchlet's theorem inside of it)? – Ben Linowitz Feb 24 '10 at 18:44
I'm not sure exactly what you want. Since the statement contains the notion of density (either Dirichlet or otherwise), you need to do some analysis somewhere. There are proofs of the Dirichlet density version which involve no complex analysis or Tauberian theorems; just elementary sums and estimates on the real line. – David Speyer Feb 24 '10 at 18:53
On the other hand, there is the point I am making above. Without any analysis, you can prove that the factorization of f must correspond to the action of one of the elements of G. You need the analysis to know that every element of G occurs. – David Speyer Feb 24 '10 at 18:55
K. Conrad has a paper ( in which he asserts that the only cases of Dirichlet's theorem that can be proven in a "Euclidean" manner (i.e. imitating Euclid's proof of the infinitude of the primes) are for arithmetic progressions of the form a mod m where a^2 = 1 mod m, and there is some evidence that there is no essentially algebraic proof of any other cases. – Qiaochu Yuan Feb 24 '10 at 20:49

There are great proofs of quadratic reciprocity which go through the ideal theory of cyclotomic fields.

And it is basically impossible to discuss cubic and biquadratic reciprocity without algebraic number theory.

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Thanks. +1. Darij Grinberg also mentioned this in comments. – Regenbogen Feb 24 '10 at 17:49

I don't think that the following example is terribly important, but it certainly goes to show that very simple-minded properties of integers can only be understood in terms of quite deep relations. Here is one about Mersenne numbers. Actually even the Lucas-Lehmer test ist best understood in terms of algebraic numbers (quadratic number fields).

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The number field sieve fits your requirements perfectly, I think.

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Yes it does fit. However this is more like a CS-style application? I was having more pure math style in mind. Somehow I got brainwashed into thinking that blind (almost mechanized) computation is a sort of baser mental activity, compared to stuff involving ideas in the domain of thought. – Regenbogen Feb 24 '10 at 17:45
So suddenly prime factorization is not a mathematical activity? Whether the computation is a baser mental activity or not, the design of algorithms is a nontrivial mathematical endeavor. – Qiaochu Yuan Feb 24 '10 at 17:49
Well, the design of algorithms is for really smart chaps. If I were to take up that subject for example, I might get lost into inconsequential stuff. Whereas in pure mathematics, certain subjects are so abstract that it is harder to get lost. – Regenbogen Feb 24 '10 at 17:54
That's an interesting statement. Most people I know would find it much easier to get lost in abstract stuff. When designing an algorithm it is easy not to get lost: you are on the right track if the algorithm is fast and on the wrong track if the algorithm is slow. In any case, I don't see the point of making all these unnecessary divisions between subjects. Wouldn't you prefer that your students get a taste of many kinds of mathematics and of their unity? – Qiaochu Yuan Feb 24 '10 at 18:01
Here is a nice reason to mention the number field sieve at least in passing (if its details are too involved for the class). When you first learn about the ring of integers O_K in a number field K, they all look like Z[a]. Not every O_K has this special form, so from an abstract point of view you might start to look down on the rings Z[a]. However, computations in a ring Z[a] are far more efficient than in O_K (to start, try finding a Z-basis in each). For computational purposes, the number field sieve makes essential use of rings Z[a] in preference to O_K itself. – KConrad Feb 24 '10 at 19:37

The Ramanujan constant $e^{\pi\sqrt{163}}$ is almost a natural number, or more generally, the so-called Heegner numbers are very close to being natural numbers. The explanation can be found here, just to cite a web-ready reference, and is based on the theory of modular functions and complex multiplication (which I consider under the umbrella of algebraic number theory).

This may be not the type of examples you are looking for (they really explain why a transcendental number is almost a natural number, not why a natural number is almost a transcendental number) but it perhaps brings home a better point: algebraic number theory can explain not only properties of natural numbers, but also properties of transcendental numbers (think of values of zeta and L-functions, etc).

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Well, $e$ and $\pi$ are not quite about natural numbers, but I agree that this is a great motivation for a course in algebraic NT (though I doubt Regenbogen will ever come to this motivation in his lecture). – darij grinberg Feb 24 '10 at 22:05
Well, let us forget about my lecture and concentrate on the question at hand! – Regenbogen Feb 24 '10 at 22:13

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