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Hilbert’s Theorem 90.

May 14, 2013

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 k \hookrightarrow F, we can naturally consider F as a k-vector space. We can use this point of view to define a notion of norm in field extensions. For any \alpha \in F, we define the norm of \alpha with respect to this extension as

N_{F/k}(\alpha) = \det(m_{\alpha})

where m_{\alpha} is multiplication by \alpha viewed as a k-linear transformation on F. 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 z = a + bi \in \mathbf{C}, then in the basis (1, i) the application m_z is represented by the matrix \bigl( \begin{smallmatrix} a&-b \\ b & a \end{smallmatrix} \bigr) so we get N_{\mathbf{C}/\mathbf{R}}(z) = a^2 + b^2. A similar argument shows that in a quadratic number field of the form \mathbf{Q}(\sqrt{-d}) for d an integer which is not a square, we find N_{\mathbf{Q}(\sqrt{-d})/\mathbf{Q}}(a+b \sqrt{-d}) = a^2 + b^2d.

Lemma: For a simple algebraic extension k \hookrightarrow k(\alpha), we have N_{k(\alpha)/k}(\alpha) = (-1)^da_0 where a_0 is the constant term of the minimal polynomial of \alpha in k.

Proof: Let f_{\alpha} be the minimal polynomial of \alpha, say of degree d and recall that k(\alpha) \cong k[x]/(f_{\alpha}) can be seen as the field of k-polynomials of degree less than d where multiplication is done modulo (f_{\alpha}). Then \alpha is identified with the class of x. A nice basis for this space is (1, x, x^2, \ldots, x^{d-1}). Here multiplication by \alpha acts on these vectors as 1 \mapsto x, x \mapsto x^2, \ldots x^{d-1} \mapsto x^d (everything modulo (f_{\alpha})). But x^d = -a_0 - a_1x - a_2x^2 - \cdots - a_{d-1}x^{d-1} \mod (f_{\alpha}) so multiplication by \alpha is represented by

\left( \begin{smallmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ && \ddots && \\ 0 & 0 & \cdots & 1 & -a_{d-1} \end{smallmatrix} \right).

This matrix clearly has determinant (-1)^da_0 hence the assertion.

QED

Note that it also follows from the proof that \text{tr}_{k(\alpha)/k}(\alpha) = -a_{d-1}. To get a better description of that norm, we need another lemma:

Lemma: For k \hookrightarrow F a finite extension and \alpha \in F, let [F : k(\alpha)] = r. Then N_{F/k}(\alpha) = (N_{k(\alpha)/k} (\alpha))^r.

Proof: If (e_1, \ldots, e_r) is a basis of F as a vector space over k(\alpha) and (1, x, x^2, \ldots, x^{d-1}) is the canonical basis of k(\alpha) over k, then (e_ix^j) for i=1, \ldots, r and j=1, \ldots, d-1 is a basis for F over k. Ordering this basis by (e_1x^1, e_1x^2, \ldots, e_1x^{d-1}, e_2x^1, \ldots, e_2x^{d-1}, \ldots, e_rx^1, \ldots, e_rx^{d-1}), the matrix of multiplication by \alpha is given by the block diagonal matrix

\left(\begin{smallmatrix} A &&& \\ &A&& \\ & & \ddots & \\ &&&A \end{smallmatrix} \right)

where A is the matrix representing multiplication by \alpha in k(\alpha) as in the first lemma.

QED

The analogous result for the trace is that \text{tr}_{F/k} (\alpha) = r(\text{tr}_{k(\alpha)/k}(\alpha)). These two lemmas let us interpret the norm and trace in a clarifying way:

Proposition: For k \hookrightarrow F a Galois extension and \alpha \in F, we have

N_{F/k}(\alpha) = \displaystyle\prod_{\sigma \in Gal(F/k)} \sigma (\alpha)     and      \text{tr}_{F/k}(\alpha) = \displaystyle\sum_{\sigma \in Gal(F/k)} \sigma (\alpha).

Proof: If \lambda_1, \ldots, \lambda_d are the roots of the minimal polynomial f_{\alpha}(x)  of \alpha whose constant term we denote by a_0 and F is of degree r over k(\alpha), then since (-1)^da_0 = \prod_{i=1}^r \lambda_i, the preceding lemmas tell us that

N_{F/k}(\alpha) = \left(\displaystyle\prod_{i=1}^d \lambda_i \right) ^r.

Now, recall that Gal(F/k) acts transitively on the set of roots of f_{\alpha}(x) since it is irreducible. Denoting by \text{Stab}(\lambda) the stabilizer of a root \lambda for this action, we have \text{Stab}(\lambda) = Gal(F/k(\lambda)) and since there is only one orbit, all stabilizers are conjugate of each others so |Gal(F/k(\lambda))| = |Gal(F/k(\alpha))| = r for all root \lambda. This shows that to each root of f_{\alpha}(x) there corresponds at least r elements of Gal(F/k). Since |Gal(F/k)| = rd, this correspondence is bijective and we have

\displaystyle\prod_{\sigma \in Gal(F/k)} \sigma (\alpha) = \left( \displaystyle\prod_{i=1}^d \lambda_i \right)^r = N_{F/k}(\alpha)

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 k \hookrightarrow F be a cyclic (Galois) extension of degree n, take \sigma a generator for the Galois group and \alpha \in F. Then N_{F/k}(\alpha) = 1 if and only if \alpha = b/\sigma(b) for some b \in F.

One direction is immediate. Indeed, for any \beta \in F and \sigma \in Gal(F/k), we have

N_{F/k}(\beta/\sigma(\beta)) = \prod (\tau(\beta)/\tau \circ \sigma (\beta)) = \prod \tau(\beta) / \prod \tau'(\beta) = 1

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

Lemma: If \varphi_1, \ldots, \varphi_m are pairwise distinct k-automorphisms of F, then they are linearly independant over F.

Proof of the lemma: Let \varphi = \sum_{j=1}^rc_j \varphi_{i_j} = 0 be a minimal linear relation among the automorphisms. Since they are pairwise distinct, there exists \lambda \in F such that \varphi_{i_1}(\alpha) \neq \varphi_{i_2}(\alpha). Then for all x \in F, since \varphi(x) = 0, we have also that \varphi(\lambda x) = 0. We obtain in this way that for all x \in F,

0 = \varphi(\lambda x) = \displaystyle\sum_{j=1}^r c_j\varphi_{i_j}(\lambda) \varphi_{i_j}(x)

and

0 = \varphi_{i_1}(\lambda)\varphi(x) = \displaystyle\sum_{j=1}^rc_j \varphi_{i_1}(\lambda)\varphi_{i_j}(x)

so subtracting give us 0 = \sum_{j=2}^rc_j(\varphi_{i_j}(\lambda) - \varphi_{i_1}(\lambda)) \varphi_{i_j}(x) which is a non-trivial relation (since we took \lambda so that \varphi_{i_2}(\lambda) \neq \varphi_{i_1}(\lambda)). 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 N_{F/k}(\alpha) = 1 and define \tau : F \to F as \tau(b) = \alpha \sigma (b). Then it suffices to find a fixed point of \tau. It is easy to see that for k \in \{1, \ldots, n\}, we have

\tau^k(b) = \alpha \sigma(\alpha)\sigma^2(\alpha) \cdots \sigma^{k-1}\sigma^k(b).

In particular, \tau^n(b) = \prod_{i=1}^n\sigma^n(\alpha)\sigma^n(b) = N_{F/k}(\alpha)b so if N_{F/k}(\alpha) = 1, we have \tau^n(b) = b for all b \in F. In particular, the order of \tau is n. Since moreover \tau is k-linear over F, we get a representation \rho : \mathbf{Z}/n\mathbf{Z} \to GL(F) by \rho(m) = \tau^m. From this point of view, it is easy to determine if \tau has a fixed point. Indeed, for any group G and any representation \phi : G \to GL(V), recall that

p := \dfrac1{|G|} \displaystyle\sum_{g \in G} \phi(g)

is the projection of V 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 0. But in our case we have

\begin{aligned} p &= \frac1n(1_F + \tau + \tau^2 + \cdots + \tau^{n-1}) \\ &= \frac1n(1_F + \alpha \sigma+ \alpha \sigma(\alpha)\sigma^2 + \cdots + \alpha \sigma(\alpha)\cdots \sigma^{n-2}(\alpha)\sigma^{n-1}) \end{aligned}

and since \sigma is a generator for the Galois group, the automorphisms 1_F, \sigma, \sigma^2, \ldots, \sigma^{n-1} are linearly independent by the lemma so p is not identically zero because that would give us a non-trivial linear relation between them. This shows that \tau(b) = b for some b \in F and that \alpha = b/\sigma(b).

QED

Note that the analogous result for the trace is that in a cyclic extension, Tr(\alpha) = 0 is and only if \alpha = \beta - \sigma(\beta). As a cool application of this theorem, we can give a parametrization of pythagorean triples. Indeed, consider the equation x^2 + dy^2 = 1 for d not a square. Recall that the norm of \alpha = a + b\sqrt{-d} \in \mathbf{Q}(\sqrt{-d}) is

N_{\mathbf{Q}(\sqrt{-d})/\mathbf {Q}}(\alpha) = a^2 + db^2

(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 \mathbf{Q}(\sqrt{-d}). By Hilbert’s theorem 90 and since conjugation generates the Galois group, these elements are those of the form

\alpha = \dfrac{a+b\sqrt{-d}}{a-b\sqrt {-d}} = \dfrac{a^2 - db^2}{a^2 + b^2} + \dfrac{2ab}{a^2 + b^2}\sqrt{-d}.

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

(x,y) = \left(\dfrac{a^2 - db^2}{a^2 + b^2}, \dfrac{2ab}{a^2 + b^2}\right)

where (a,b) \in \mathbf{N}^2. Letting d=1, we find all the rational points on the unit circle, which give us the integer pythagorean triples by the correspondence (a/c, b/c) \mapsto (a,b,c).

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