I’m not going to prove that every ring of integers of a number field is a Dedekind domain here, but I promise to do so in the next post.

Recall that if K is a number field of degree *n*, then there are exactly *n* distinct ring homomorphisms from K to **C**. I’ll denote them by *w*1, *w*2, …, *w*(*n*), where *w*1 is the natural inclusion. For instance, if K = **Q**(2^(1/3)), then *w*1 will send 2^(1/3) to itself, and *w*2 and *w*3 will send 2^(1/3) to *t**2^(1/3) and (*t*^2)*2^(1/3) where *t*^3 = 1 but *t* != 1.

Those of you who have had the misfortune of taking linear algebra in college will know that there exists something called a matrix and a determinant of a matrix. Let A be any *n***n* matrix, with entries in a field, say **C**. It’s convenient to look at the rows or columns of A as vectors in **C**^*n*. Also, let *a*(*ij*) or *a*(*i*, *j*) be the element in row *i* and column *j*.

The determinant of A is one function of the entries of A. If *n* = 1, then |A| is just *a*11, the only entry of A. If *n* > 1, |A| is gotten by expanding along some row or column of A (it doesn’t matter which row or column). For example, taking the first row, it’s equal to *a*11*B11 + *a*12*B12 + … + *a*(1*n*)*B(1*n*). B(*ij*) is equal to the determinant of the (*n*-1)*(*n*-1) matrix we get if we remove the *i*th row and *j*th column from A, and multiply by -1 if *i*+*j* is odd.

More concretely, if A is 2*2, we can write it as [*a*11, *a*12; *a*21, *a*22]. Expanding along the first row, we get |A| = *a*11*B11 + *a*12*B21 = *a*11*|*a*22| + *a*12*(-1)*|*a*21| = *a*11**a*22 – *a*12**a*21.

The determinant has the following properties, none of which I’m going to prove. If A’ is gotten from A by switching any two rows, or switching any two columns, then |A’| = -|A|. If A’ is gotten from A by adding a multiple of one row to another row, or a multiple of one column to another column, then |A’| = |A|. If A’ is gotten from A by multiplying a single column or a single row by a constant *k*, then |A’| = *k*|A|. |A| = 0 iff the row vectors of A are linearly dependent over a field containing all *a*(*ij*)’s, which is true iff the column vectors of A are linearly dependent over such a field.

Going back to algebraic number theory, given any list of *n* elements of K, (*x*1, *x*2, …, *x*(*n*)), we define their discriminant to be |A|^2, where the element *a*(*ij*) of A is given by *w*(*i*)(*x*(*j*)). |A|^2 is clearly invariant under switching rows or switching columns, so it doesn’t matter how we order the *x*(*j*)’s or the *w*(*i*)’s. Also, it’s sometimes useful to talk about disc(*x*) = disc(1, *x*, *x*^2, …, *x*^(*n*-1)).

The discriminant satisfies the following properties:

- It is always rational. It’s clearly contained in K’, the smallest number field containing all conjugates of K. Every ring homomorphism on K’, say F, will permute the
*w*(*i*)’s, so it will act on A by switching around rows. Since |A|^2 is invariant under switching rows, we get |A|^2 = |F(A)|^2 = F(|A|^2) for all F. In a way it turns |A|^2 into a symmetric polynomial in the conjugates of*a*(where K =**Q**(*a*)), so it will be rational. - If the
*x*(*j*)’s are algebraic integers, then |A|^2 is an integer, because it’s the sum of products of algebraic integers. - |A|^2 can be negative; that is, |A| can be purely imaginary. If K =
**Q**(*i*), then disc(*i*) = |A|^2, A = [1,*i*; 1, –*i*], so |A| = -2*i*and |A|^2 = -4. - |A| = 0 if the
*x*(*j*)’s are linearly dependent over**Q**. If*a*1*x*1 +*a*2*x*2 + … +*a*(*n*)*x*(*n*) = 0, then for every ring homomorphism*w*,*w*(1) = 1, so by addition and division,*w*(*p*/*q*) =*p*/*q*for integers*p*and*q*; so 0 =*w*(*a*1*x*1 +*a*2*x*2 + … +*a*(*n*)*x*(*n*)) =*a*1*w*(*x*1) +*a*2*w*(*x*2) + … +*a*(*n*)*w*(*x*(*n*)). Then letting*c*(*j*) be the*j*th column, we get that*a*1*c*1 +*a*2*c*2 + … +*a*(*n*)*c*(*n*) = 0 and the*a*(*n*)’s are not all 0. - |A| = 0 only if the
*x*(*j*)’s are linearly dependent over**Q**. If*a*1*c*1 +*a*2*c*2 + … +*a*(*n*)*c*(*n*) = 0, we want to show that the*a*(*j*)’s are in**Q**. But they’re invariant under every ring homomorphism of K’, which will just switch around rows, so they must be in**Q**. - From 5, if K =
**Q**(*a*), then disc(*a*) is not zero.