Why is the discriminant of a number field well-defined?

Question

Let $K$ be a number field of degree $n$ and $z_1,\dots,z_n$ a $\mathbb{Z}$-basis for $\mathcal{O}_K$. It is clear that we can find another basis $y_1,\dots,y_n$ so, why $\Delta_K$ is independent of our choice of basis?

Answer 1

This is a nice exercise in linear algebra: let $\langle,\rangle$ be a bilinear pairing on a finite dimensional vector space $V$, let $v_1,\ldots,v_n$ and $v_1',\ldots, v_n'$ be two bases for $V$, let $M$ be the change of basis matrix, and define the matrices $A=(\langle v_i,v_j\rangle)_{i,j}$ and $A'=(\langle v_i',v_j'\rangle)_{i,j}$. Then show that $A' = M^{tr}AM$. Taking determinants, you get ${\rm det}\;A' = ({\rm det}\;M)^2\cdot{\rm det}\;A$. So in your situation, if the two bases generate the same lattice over $\mathbb{Z}$, then $M$ is invertible over the integers, hence its determinant is invertible in the integers, i.e. is $\pm 1$, and $A$ and $A'$ have the same determinant.

Depending on how you defined the discriminant of a number field, you might have to do Exercise 2: define a pairing on $K$ such that if $v_1,\ldots,v_n$ is an integral basis, the above matrix $A$ is the discriminant of $K$.