By Proposition 2.1 we know that \(\operatorname {Tr}_{K/\mathbb {Q}}(b_i b_j)= \sum _k \sigma _k(b_i)\sigma _k(b_j)\) which is the same as the \((i,j)\) entry of \(M^t M\). Therefore

First we recall a classical linear algebra result relating to the Vandermonde matrix, which states that

Combining this with Proposition 2.5 gives the result.

We first write \(f(x)=(x-\alpha )g(x)\) which we can do (over \(\mathbb {C}\)) as \(\alpha \) is a root of \(f\), where now \(g(x)=\prod _{\beta \neq \alpha } (x-\beta )\). Differentiating we get

If we now evaluate at \(\alpha \) we get the result.

By Proposition 2.6 we have \(\Delta (B)=\prod _{i {\lt} j}(\alpha _i-\alpha _j)^2\) where \(\alpha _k:=\sigma _k(\alpha )\). Next, we note that the number of terms in this product is \(1+2+\cdots +(n-1)=\frac{n(n-1)}{2}\). So if we write each term as \((\alpha _i-\alpha _j)^2=-(\alpha _i-\alpha _j)(\alpha _j-\alpha _i)\) we get

Now, by lemma 2.7 and Proposition 2.1 we see that

which gives the result.

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 2.9 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 2.11, therefore \(p^2 \mid \Delta (B)\) giving the result.

Suppose for contradiction that \(\phi \in \mathcal{O}_K\) and let \(x_d\) be the first non-zero coefficient, so

Now, rewrite this as \(\phi =\frac{1}{p}(x_d\alpha ^d +\alpha ^{d+1}\beta )\) for some \(\beta \in \mathcal{O}_K\). Next, multiply through by \(\alpha ^{n-1-d}\), then we have

Now, since \(m_\alpha \) satisfies Eisenstein at \(p\), we see that \(\alpha ^n=pf(\alpha )\) for some \(f \in \mathbb {Z}[x]\) and therefore the above gives us that

and thus

Lets now calculate the norm of this:

By Eisenstein the constant coefficient of \(m_\alpha \) is divisible by \(p\) but not \(p^2\), so since the constant coefficient of \(m_\alpha \) is \(N_{K/\mathbb {Q}}(\alpha )\) we see that \(N_{K/\mathbb {Q}}(\alpha )=p a\) where \(p \nmid a\). Therefore we have

But this cant be in \(\mathbb {Z}\) since \(p\) doesn’t divide \(x_d\) or \(a\), and this gives us a contradiction since Proposition 2.10 says that the norm of an algebraic integer must be an integer. So \(\phi \) couldn’t have been an algebraic integer.