# 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.