2.2.15 Formulae for calculating discriminants
Let us now look at some alternative ways for calculating discriminants.
Let \(K\) be a number field with basis \(B=\{ b_1,\dots ,b_n\} \) and let \(\sigma _1,\dots ,\sigma _n\) be the embeddings of \(K\) into \(\mathbb {C}\). Now let \(M\) be the matrix
Then
By Proposition 1.7.6 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
Now, by the Primitive element theorem 1.5.11, for any number field \(K\) we can find some \(\alpha \) such that \(K=\mathbb {Q}(\alpha )\) and thus \(\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \) is a basis for \(K/\mathbb {Q}\) where \(n=[K:\mathbb {Q}]\). In this case the discriminant is given by:
Let \(K\) be a number field and \(B=\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \) for some \(\alpha \in K\). Then
where \(\sigma _i\) are the embeddings of \(K \) into \(\mathbb {C}\).
First we recall a classical linear algebra result relating to the Vandermonde matrix, which states that
Combining this with Proposition 2.2.16 gives the result.
Lastly, we have probably the more useful formula for computing the discriminant in this case, but before we state it we need the following lemma.
Let \(f\) be a monic irreducible polynomial over a number field \(K\) and let \(\alpha \) be one of its roots in \(\mathbb {C}\). Then
where the product is over the roots of \(f\) different from \(\alpha \).
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.
Let \(K=\mathbb {Q}(\alpha )\) be a number field with \(n=[K:\mathbb {Q}]\) and let \(B=\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \). Then
where \(m_\alpha '\) is the derivative of \(m_\alpha (x)\) (which we recall denotes the minimal polynomial of \(\alpha \)).
By Proposition 2.2.17 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 2.2.18 and Proposition 1.7.6 we see that
which gives the result.
Let \(\alpha \) be a root of \(m_\alpha (x)=x^8-2\) (which is irreducible by Eisensteins criterion 1.2.15). Therefore \(\mathbb {Q}(\alpha )\) has degree \(8\) over \(\mathbb {Q}\) and a basis is \(B:=\{ 1,\alpha ,\dots ,\alpha ^7\} \). From the above we then have
\begin{align*} \Delta (B)= (-1)^{8\cdot 7/2} N_{\mathbb {Q}(\alpha )/\mathbb {Q}} (m_\alpha ’(\alpha ))& = N_{\mathbb {Q}(\alpha )/\mathbb {Q}} (8\alpha ^7)\\ & =N_{\mathbb {Q}(\alpha )/\mathbb {Q}}(8) N_{\mathbb {Q}(\alpha )/\mathbb {Q}}(\alpha ^7)=8^8 \cdot (-2)^7=-2^{31}\end{align*}Here we use that \(N_{K/\mathbb {Q}}(\alpha ^7)=N_{K/\mathbb {Q}}(\alpha )^7\) and that \(N_{K/\mathbb {Q}}(\alpha )=-2\).
Now, by Lemma 2.2.7, if we want to find an integral basis, we need to first pick a basis \(B\) of algebraic integers, and then check for which primes \(p|\Delta (B)\) we have \(p^2| \Delta (B)\). Its at these primes where we might need to modify our basis candidate \(B\).
In order to make this simpler, here is a simple trick.
Let \(K=\mathbb {Q}(\alpha )\)and \(\alpha \) be an algebraic integer such that \(m_\alpha \) satisfies Eisensteins Criterion 1.2.15 for a prime \(p\). Then none of the elements
is an algebraic integer, where \(n=\deg (m_\alpha )\) and \(x_i \in \{ 0,\dots ,p-1\} .\)
We will only prove the case when \(m_\alpha \) is Eisenstein, since the proof of the more general case is identical.
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 2.1.19 says that the norm of an algebraic integer must be an integer. So \(\phi \) couldn’t have been an algebraic integer.
How do we use this? well let me show you one use.
Let \(\alpha \) be a root of \(x^8-2\). Then as we saw in Example 2.2.20 we know that \(B=\{ 1,\alpha ,\dots ,\alpha ^7\} \) has discriminant \(-2^{31}\). So the we only need to check at \(p=2\). But \(m_\alpha \) satisfies Eisensteins criterion 1.2.15 with \(p=2\), therefore the Lemma 2.2.21 tells us that in fact this must be an integral basis, since dividing by \(2\) wont ever give us new algebraic integers.