First recall from linear algebra, that if we have a finite dimensional vector space \(V\) with a non-degenerate bilinear form \(\langle , \rangle \) on \(V\), then \(\{ v_1,\dots ,v_n\} \) a basis of \(V\) if and only if

(This matrix appearing here is called the matrix associated to the pairing \(\langle ,\rangle \) with respect to our chosen basis).

Now, by Proposition 1.7.8 we know that the trace pairing is perfect, which in particular means it is non-degenerate. Moreover, \(\Delta (B)\) is exactly the determinant of the matrix associated to the trace pairing. So, we see that \(B\) consists on linearly independent vectors if and only if \(\Delta (B) \neq 0\).

Let \(T_B=(\operatorname {Tr}_{K/\mathbb {Q}}(b_i b_j))_{i,j}\) be the matrix associated to the trace pairing with respect to \(B\). Then, it is a basic result in linear algebra that if you have the matrix associated to a bilinear pairing and change the basis, then the matrix gets conjugated by the change of basis matrix. This means that \(T_B=P^{t}T_{B'}P\), (where \(()^t\) denotes transpose). Now, taking determinants we get \(\Delta (B)=\det (T_B)=\det (P^t T_{B'}P)=\det (P^t)\det (T_{B'})\det (P)\) which gives the result once you remember that \(\det (P^t)=\det (P)\).

If \(b_i b_k\) is an algebraic integer, then \(\operatorname {Tr}_{K/\mathbb {Q}}(b_i b_j) \in \mathbb {Z}\). Therefore the matrix \(T_B\) has integers coefficients, and therefore the determinant is a integer.

If \(B\) is not an integral basis then we can find some element \(\phi \in \mathcal{O}_K\) such that

with not all the \(y_i\) in \(\mathbb {Z}\). So, let \(N\) be the least common multiple of the denominators of the \(y_i\) (meaning \(Ny_i \in \mathbb {Z}\) for all \(i\)). Now, let \(p\) be a prime factor of \(N\). If we now consider \((N/p)\phi \) then all of the coefficients of \(b_i\) are in \(\frac{1}{p} \mathbb {Z}\) (so they have denominator \(1\) or \(p\).) and at least one of them has denominator \(p\) (since not all the \(y_i\) where in \(\mathbb {Z}\)). So by relabelling, wlog we can assume

with \(y_i \in \frac{1}{p} \mathbb {Z}\)

Now look at

(here \(\lfloor x \rfloor \) denotes the integer part of \(x\)). The both \(\psi \) and \(\phi \) are algebraic integers (as the \(b_i\) are algebraic integers). Therefore, so is \(\theta =\phi -\psi \). By construction, \(\theta \) has coefficients of the for \(\frac{x_i}{p}:=y_i-\lfloor y_i \rfloor \) where \(x_i \in \{ 0,\dots ,p-1\} \) and not all the \(x_i\) are zero (since, again, not all the \(y_i\) were in \(\mathbb {Z}\)). This gives the first part of the lemma.

Now, assume \(x_i \neq 0\), then let us replace \(b_i \in B\) with \(\theta \) to get a new basis \(B'\) which again consists of algebraic integers. Next, we note that the change of basis matrix from \(B\) to \(B'\) is

(here the column of \(x_j/p\)’s is in the \(i\)-th column).

This matrix has determinant \(\frac{x_i}{p}.\) Therefore, by Proposition 2.2.4 we see that \(\Delta (B')=\frac{x_i^2}{p^2}\Delta (B)\). But both \(\Delta (B),\Delta (B')\) are in \(\mathbb {Z}\) by Proposition 2.2.5, therefore \(p^2 \mid \Delta (B)\) giving the result.

The fact that it is finite follows from \(K\) being a finite extension of \(\mathbb {Q}\). Let \(B\) be a basis consisting of algebraic integers, now by repeatedly applying Lemma 2.2.7, we can obtain \(B\) such that \(\mid \Delta (B) \mid \) is as small as possible. This must now be an integral basis, otherwise Lemma 2.2.7 would allow us to find a new basis \(B'\) with \(|\Delta (B')| {\lt} |\Delta (B)|\) contradicting our choice of \(B\).

This follows at once from Lemma 2.2.7.

In this case the change of basis matrix from \(B\) to \(B'\) is an invertible matrix with integer coefficients, so its determinant is \(\pm 1\). Using Proposition 2.2.4, we see that a change of basis matrix won’t alter the discriminant, since the factor of the determinant squared is what appears.