Discriminant of a monic irreducible integer polynomial vs. discriminant of its splitting field

Solution 1:

I think there was some confusion about the splitting field and the field $\mathbb{Q}[x]/(f(x))$, which is isomorphic to the field generated by one root of $f(x)$. (We always assume that $f(x)$ is monic irreducible.)

Let $\alpha$ be a root of $f(x)$, and $L=\mathbb{Q}(\alpha)$ be the field generated by $\alpha$, $\mathbb{Z}[\alpha]$ be the subring of $\mathcal{O}_L$ generated by $\alpha$. The the discriminant of $f(x)$ is the discriminant of the lattice $\mathbb{Z}[\alpha]$. So $\mathrm{Disc}(f)/\mathrm{Disc}(\mathcal{O}_L)$ is the square of $[\mathcal{O}_L: \mathbb{Z}[\alpha]]$. (See III.3 of Lang's "Algebraic number theory").

However, for splitting fields, these things hardly compares. For example, take $f(x)=x^4-x+1$, then the discriminant of $f(x)$ is 229 (a prime, which coincides with the discriminant of the field $L$ in this case) , but the discriminant of the splitting field of $f(x)$ is $229^{12}$ (calculated using Pari), which has 28-digits. (Well, it is not hard to show the discriminant of the splitting field of $f(x)$ share the same prime divisors as the field $L$.)

Sorry about bring up a really old question. It is just I asked myself the same thing today.

Solution 2:

The two are the same if the roots of $f$ form an integral basis of the ring of integers of $\mathbb{Q}[x]/f(x)$ (e.g. if $f$ is a cyclotomic polynomial) because then, well, they're defined by the same determinant (see Wikipedia), but in general they don't. In the general case $\mathbb{Z}[\alpha_1, ... \alpha_n]$ is an order in $\mathcal{O}_K$ so one can write the $\alpha_i$ as an integer linear combination of an integral basis, so the matrices whose determinants define the two discriminants should be related by the square of a matrix with integral entries, hence integral determinant.

In fact if I'm not totally mistaken, the quotient of the two discriminants should be precisely the index of $\mathbb{Z}[\alpha_1, ... \alpha_n]$ in $\mathcal{O}_K$ as lattices, or maybe its square...?

In any case, since the discriminant of the field is defined in terms of $\mathcal{O}_K$ it is the "right" choice for carrying information about, for example, ramification. One can see this even in the quadratic case: if $d \equiv 1 \bmod 4$ then the discriminant of $x^2 - d$ is $4d$ but the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$, and the latter is the "right" choice because $2$ doesn't ramify in $\mathbb{Z} \left[ \frac{1 + \sqrt{d}}{2} \right]$.

Solution 3:

In response to Qiaochu,

$Disc(f)/Disc(\mathcal{O}_K)$ is the square of the index of $\mathbb{Z}[ \alpha _1, \ldots , \alpha _n ]$ in $\mathcal{O}_K$. The index itself is the determinant of the change of basis matrix from $(\alpha _1, \ldots , \alpha _n )$ to an integral basis for $\mathcal{O}_K$. This matrix is squared when taking the discriminant.