1.7 Norm and Trace

Let \(K/F\) be a field extension, we now look at ways of taking an element in \(K\) and obtaining elements in \(F\).

Definition 1.7.1

Let \(K/\mathbb {Q}\) be a number field and \(\alpha \in K\). Then we define the norm of \(\alpha \) by

\[ N_{K/\mathbb {Q}}(\alpha )= \operatorname{Det}(A_\alpha ) \in \mathbb {Q} \]

and the trace of \(\alpha \) by

\[ \operatorname {Tr}_{K/\mathbb {Q}}(\alpha )=\operatorname{Trace}(A_\alpha ) \in \mathbb {Q}. \]

Here \(\operatorname{Trace}(A_\alpha )\) is the sum of the diagonal entries of \(A_\alpha \).

Example 1.7.2

Let \(K=\mathbb {Q}(\sqrt{5})\) and \(\alpha =3+2\sqrt{5}\), then \(A_\alpha =\left(\begin{smallmatrix}3& 10\\ 2& 3\end{smallmatrix}\right)\) and therefore \(\operatorname {Tr}_{K/\mathbb {Q}}(\alpha )=6\) and \(N_{K/\mathbb {Q}}(\alpha )=-11\).

Proposition 1.7.3

Let \(K\) be a number field of degree \(n\) over \(\mathbb {Q}\) and \(\alpha \in K\). Then

\[ C_\alpha (x)=x^n-\operatorname {Tr}_{K/\mathbb {Q}}(\alpha )x^{n-1} +\cdots +(-1)^n N_{K/\mathbb {Q}}(\alpha ). \]

Proof ▼

This is immediate from the definition of \(C_\alpha \) as the characteristic polynomial of \(A_\alpha \).

Specifically, its a basic result in linear algebra, that says if \(M\) is a \(n\times n\) matrix with characteristic polynomial \(C(X)\) then

\[ C(X)=X^n-\operatorname{Trace}(M)X^{n-1}+\dots +(-1)^n\det (M) \]

where \(\operatorname{Trace}(M)\) is the sum of the diagonal entries of \(M\).

Exercise 1.7.4

Let \(K=\mathbb {Q}(\sqrt{d})\) with \(d\) square-free. Show that

\[ \operatorname {Tr}_{K/\mathbb {Q}}(a+b\sqrt{d})=2a \qquad N_{K/\mathbb {Q}}(a+b\sqrt{d})=a^2-db^2. \]

Proposition 1.7.5

Let \(K\) be a number field and \(\alpha ,\beta \in K\). Then

\[ \operatorname {Tr}_{K/\mathbb {Q}}(\alpha +\beta )=\operatorname {Tr}_{K/\mathbb {Q}}(\alpha )+\operatorname {Tr}_{K/\mathbb {Q}}(\beta ) \qquad N_{K/\mathbb {Q}}(\alpha \beta )=N_{K/\mathbb {Q}}(\alpha )N_{K/\mathbb {Q}}(\beta ). \]

Proof ▼

By definition we only need to consider the trace of \(A_\alpha +A_\beta \) and the determinant of \(A_\alpha A_\beta \). But we know from linear algebra that trace is additive and determinant is multiplicative, so the result follows.

Proposition 1.7.6

Let \(K\) be a number field, \(\alpha \in K\) and let \(\sigma _i\) be the embeddings of \(K\) into \(\mathbb {C}\). Then

\[ \operatorname {Tr}_{K/\mathbb {Q}}(\alpha ) =\sum _i \sigma _i(\alpha ) \qquad N_{K/\mathbb {Q}}(\alpha )=\prod _i \sigma _i(\alpha ) \]

Proof ▼

First recall that the norm and trace of a matrix and invariant under conjugation. Therefore, the \(\operatorname {Tr}_{K/\mathbb {Q}}(\alpha )=\operatorname{Trace}(A_\alpha )=\operatorname{Trace}(P^{-1}A_\alpha P)\) similarly for the norm. Now, by Proposition 1.6.7 we see that we can find \(P\) such that \(P^{-1}A_\alpha P\) is diagonal with entries \(\sigma _i(\alpha )\). From this the result follows.

One of the important properties of the trace function is that it gives us a \(\mathbb {Q}\)-bilinear pairing on \(K\), defined as follows:

Definition 1.7.7

Let \(K\) be a number field. We have a pairing

\[ \langle , \rangle : K \times K \to \mathbb {Q} \]

given by

\[ \langle \alpha ,\beta \rangle = \operatorname {Tr}_{K/\mathbb {Q}}(\alpha \beta ) \]

Proposition 1.7.8

The trace pairing is a \(\mathbb {Q}\)-bilinear perfect pairing. In other words if \(\alpha ,\beta \in K\) and \(\lambda \in \mathbb {Q}\) then

\[ \langle \lambda \alpha ,\beta \rangle =\langle \alpha , \lambda \beta \rangle =\lambda \langle \alpha ,\beta \rangle \]

and if \(\alpha \in K\) is such that \(\langle \alpha ,\beta \rangle =0\) for all \(\beta \in K\), then \(\alpha =0\).

Proof ▼

The fact that it is bilinear follows at once from the definition of \(\operatorname {Tr}_{K/\mathbb {Q}}\). To check it is perfect, consider \(\alpha \neq 0\). Then \(\alpha ^{-1} \in K\) and thus \(\operatorname {Tr}_{K/\mathbb {Q}}(\alpha \alpha ^{-1})=\operatorname {Tr}_{K/\mathbb {Q}}(1)=[K:\mathbb {Q}]\) which is non-zero.