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Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't?

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The discriminant of $K$ is equal to the gcd of $disc(f)$ as $f$ ranges over irred. polys. of all algebraic integers in $K$ which are primitive elements (i.e. which generate $K$ over $\mathbb Q$). – Matt E Oct 30 '10 at 3:18
up vote 6 down vote accepted

Edit: Every time I say "splitting field" in the following answer I really mean $\mathbb{Q}[x]/f(x)$.

The two are the same if the roots of $f$ form an integral basis of the ring of integers of the splitting field (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 $\mathbb{Z} \left[ \frac{1 + \sqrt{d}}{2} \right]$.

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It hardly ever happens when the roots of $f(x)$ forms the the integral basis of the splitting field. For example, take $f(x)=x^3-2$, then the splitting field has degree $6$ over $\mathbb{Q}$, but $f(x)$ has only 3 roots. – Jiangwei Xue May 7 '11 at 17:19

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.

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Thanks for the correction, Jiangwei. – Qiaochu Yuan May 7 '11 at 19:26

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.

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