# Hilbert’s Theorem 90.

Hilbert’s theorem 90 is the 90’th theorem in Hilbert’s *Zahlbericht* (meaning number report according to google translate), which is a famous report on the state of algebraic number theory at the end of the nineteenth century. It is a basic theorem shedding some light on cyclic field extensions (a Galois extension with a cyclic Galois group) and it seems to lead to some really big ideas in algebraic number theory. As a corollary of this, we directly get a parametrization of the rational points on the circle, hence of integral pythagorean triples.

Given a finite extension , we can naturally consider as a -vector space. We can use this point of view to define a notion of norm in field extensions. For any , we define the **norm** of with respect to this extension as

where is multiplication by viewed as a -linear transformation on . It also makes sense to speak of the trace of an element with respect to a field extension and we actually get an analogous “additive” Hilbert 90 without further difficulty.

Take for example the complex numbers as an extension of the reals. If , then in the basis the application is represented by the matrix so we get . A similar argument shows that in a quadratic number field of the form for an integer which is not a square, we find .

**Lemma: **For a simple algebraic extension , we have where is the constant term of the minimal polynomial of in .

**Proof: **Let be the minimal polynomial of , say of degree and recall that can be seen as the field of -polynomials of degree less than where multiplication is done modulo . Then is identified with the class of . A nice basis for this space is . Here multiplication by acts on these vectors as (everything modulo ). But so multiplication by is represented by

.

This matrix clearly has determinant hence the assertion.

QED

Note that it also follows from the proof that . To get a better description of that norm, we need another lemma:

**Lemma:** For a finite extension and , let . Then .

**Proof:** If is a basis of as a vector space over and is the canonical basis of over , then for and is a basis for over . Ordering this basis by , the matrix of multiplication by is given by the block diagonal matrix

where is the matrix representing multiplication by in as in the first lemma.

QED

The analogous result for the trace is that . These two lemmas let us interpret the norm and trace in a clarifying way:

**Proposition:** For a Galois extension and , we have

and .

**Proof:** If are the roots of the minimal polynomial of whose constant term we denote by and is of degree over , then since , the preceding lemmas tell us that

.

Now, recall that acts transitively on the set of roots of since it is irreducible. Denoting by the stabilizer of a root for this action, we have and since there is only one orbit, all stabilizers are conjugate of each others so for all root . This shows that to each root of there corresponds at least elements of . Since , this correspondence is bijective and we have

as we wanted.

QED

We are now in a position to state and prove our main theorem, which characterizes unit norm elements in a cyclic extension:

**Theorem (Hilbert 90):** Let be a cyclic (Galois) extension of degree , take a generator for the Galois group and . Then if and only if for some .

One direction is immediate. Indeed, for any and , we have

(where the products are all taken over ). For the other direction, we need a lemma:

**Lemma:** If are pairwise distinct -automorphisms of , then they are linearly independant over .

**Proof of the lemma:** Let be a minimal linear relation among the automorphisms. Since they are pairwise distinct, there exists such that . Then for all , since , we have also that . We obtain in this way that for all ,

and

so subtracting give us which is a non-trivial relation (since we took so that ). But this relation is shorter than our initial relation, which we supposed to be minimal. This is a contradiction and the lemma is proven.

**Proof of the theorem:** Suppose and define as . Then it suffices to find a fixed point of . It is easy to see that for , we have

.

In particular, so if , we have for all . In particular, the order of is . Since moreover is -linear over , we get a representation by . From this point of view, it is easy to determine if has a fixed point. Indeed, for any group and any representation , recall that

is the projection of onto the trivial summand of the representation. So to show that a group action has a fixed point, it suffices to show that this map is not identically . But in our case we have

and since is a generator for the Galois group, the automorphisms are linearly independent by the lemma so is not identically zero because that would give us a non-trivial linear relation between them. This shows that for some and that .

QED

Note that the analogous result for the trace is that in a cyclic extension, is and only if . As a cool application of this theorem, we can give a parametrization of pythagorean triples. Indeed, consider the equation for not a square. Recall that the norm of is

(this is easy to see with the finer interpretation of the norm as a product over the Galois group). So rational solutions of our equation correspond exactly to the elements of unit norm in . By Hilbert’s theorem 90 and since conjugation generates the Galois group, these elements are those of the form

.

Since multiplying and by some rational number does not change the expression, we can take and as integers. Therefore, the rational solutions to the equation are

where . Letting , we find all the rational points on the unit circle, which give us the integer pythagorean triples by the correspondence .