Discriminants and lattices

Let K=\mathbb Q(\alpha) be a quadratic number field. For a,b\in K, recall that the discriminant \Delta(a,b) is defined as

\displaystyle \Delta(a,b):=\left|\begin{matrix} a^{(1)} & a^{(2)}\\ b^{(1)} & b^{(2)}\end{matrix}\right|^2

where a^{(1)}, a^{(2)} are the Galois conjugates of a and b^{(1)}, b^{(2)} are those of b. For any \beta\in K we define its discriminant to be \Delta(\beta):=\Delta(1,\beta).

Write a=a_1+a_2\alpha and b=b_1+b_2\alpha. Then

\left(\begin{matrix} a\\ b\end{matrix}\right)=\underbrace{\left(\begin{matrix} a_1 & a_2\\ b_1 & b_2\end{matrix}\right)}_{A}\left(\begin{matrix} 1\\\alpha\end{matrix}\right)

If \alpha,\bar\alpha are the Galois conjugates of \alpha, then

\Delta(a,b)=\left|\begin{matrix} a_1+a_2\alpha & a_1+a_2\bar\alpha\\ b_1+b_2\alpha & b_1+b_2\bar\alpha\end{matrix}\right|^2=\left|\begin{matrix} a_1 & a_2\\ b_1 & b_2\end{matrix}\right|^2\left|\begin{matrix} 1 & 1\\\alpha & \bar\alpha\end{matrix}\right|^2

\therefore\boxed{\Delta(a,b)=(\det A)^2\Delta(\alpha)}

Now suppose that \mathbb Z[\alpha]=a\mathbb Z+b\mathbb Z. Then \mathbb Z[\alpha] is spanned by \{a,b\}, so there are integers p,q,r,s such that

\underbrace{\left(\begin{matrix} p & q\\ r & s\end{matrix}\right)}_{M}\left(\begin{matrix} a\\ b\end{matrix}\right)=\left(\begin{matrix} 1\\\alpha\end{matrix}\right)

So we have

MA\left(\begin{matrix} 1\\\alpha\end{matrix}\right)=\left(\begin{matrix} 1\\\alpha\end{matrix}\right).

Lemma. If P is a 2\times 2 matrix with integer coefficients and w=(1,\alpha)^T with \alpha\not\in\mathbb Q, then Pw=w if and only if P=I, the 2\times 2 identity matrix.

Proof. This follows from the \mathbb Z-linear independence of \{1,\alpha\}. More concretely,

\underbrace{\left(\begin{matrix} s & t\\ u & v\end{matrix}\right)}_{P}\left(\begin{matrix}1\\\alpha\end{matrix}\right)=\left(\begin{matrix} 1\\\alpha\end{matrix}\right)\Rightarrow\begin{cases}s+t\alpha=1\\ u+v\alpha=\alpha\end{cases}

\therefore s=1,\ t=0,\ u=0,\ v=1\Rightarrow P=I. \square

Thus MA=I, so that \det(M)\det(A)=1. But \det(M) and \det(A) are integers. Hence |\det(M)|=|\det(A)|=1, i.e. \Delta(a,b)=\Delta(\alpha). Thus

Fact. \{a,b\}\subset\mathbb Z[\alpha] spans \mathbb Z[\alpha] if and only if \Delta(a,b)=\Delta(\alpha).

Note that all of the above arguments generalize to arbitrary number fields.

A nice corollary:

Corollary. (a,b) and (c,d) generate \mathbb Z^2 (as a group) if and only if

\left|\begin{matrix} a & b\\ c & d\end{matrix}\right|=\pm 1.

In other words, two bases generate the same lattice only if their fundamental parallelograms have equal areas.


1 Comment

Filed under Geometry, Linear algebra, Number theory

One response to “Discriminants and lattices

  1. Pingback: Pick’s theorem | Samin Riasat's Blog

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