First, an easier way of proving that the discriminant is the norm of the different is by considering the discriminant matrix A, whose (*i*, *j*)th entry is *w*(*i*)(*x*(*j*)) where the elements *x*(*j*) form an integral basis for O(K) and the functions *w*(*i*) are the conjugates of K. d(K) = |A|^2. Now, taking the transpose of A, A^T, and multiplying it by A gives a matrix whose (*i*, *j*)th entry is tr(*x*(*i*)*x*(*j*)), and whose determinant is |A|^2 = d(K).

Now, instead of taking (A^T)A, we can take (B^T)A, where B represents any integral basis of a fractional ideal I. The determinant of B is, up to a sign, N(I)*SQRT(d(K)), so |(B^T)A| = (+/-)N(I)d(K). To show that N(D) = d, it’s enough to show that |(B^T)A| is a unit when B represents the inverse different.

Since every element of (B^T)A is of the form tr(*xy*) where *y* is in O(K) and *x* is in the inverse different, |(B^T)A| is an integer. Conversely, if it’s not a unit, then apply column operations, which are equivalent to column operations on A, or row operations, which are equivalent to row operations on B^T, to make one row divisible by some integer *k*.

Then we can divide the trace element *x* corresponding to that row by *k* and stay in the inverse different. But that’s a contradiction, since in every integral basis, no element can be divisible by an integer, since replacing *kx* by *x* will yield a basis of smaller discriminant. Note that this proof does not use local techniques for the discriminant.

Second, Sirix commented on the other different post, but the spam filter swallowed the comment and I accidentally deleted it.

What is the advantage of taking (instead of O(K)) a

subring Bof O(K) generated overZbysomebasis for K over Q before defining inverse element (that is, one defines “inverse element” as all a in K such that tr(ab) is inZwhenever b is in B)? (J.S. Milne does this in his free notes on algebraic number theory, Remark 2.32 on page 30, but doesn’t explain what is it useful for)

That’s just a more general definition of dual modules or fractional ideals. Given any fractional ideal J, we can define its dual J* to be the set of all elements *x* in K such that tr(*xy*) is in **Z** whenever *y* is in J. Clearly, O(K)* is the inverse different, I. Almost as clearly, J* = (J^(-1))I and J** = J.

And third, the original definition of the different had nothing to do with trace forms and everything to do with differentials. Dedekind defined the different as the ideal generated by all elements of the form *m*‘(*a*)(*a*), i.e. the derivative of the minimal polynomial of *a*, evaluated at *a*, subject to the restriction that K = **Q**(*a*) (or else the different will be unit ideal, since *m*‘(1)(1) = 1). The two definitions are the same, although I won’t prove it here.

Is it known why the spam filter blocked me?

Because it’s insane. It fried my own comments at one point.