TITLE: irreducibility of discriminant
QUESTION [34 upvotes]: This must be well-known to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d-1} x^{d-1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d-1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over)?

REPLY [17 votes]: Some remarks:

Exception: In some cases it is actually reducible: Over a finite field of characteristic 2, the discriminant of $x^2+a_1x+a_0$ is $D(a_0,a_1)=a_1^2$.
(Almost) Irreducibility over finite fields: Standard $\zeta$-function manipulations show that over $\mathbb{F}_q[T]$ there are exactly $q^{d-1}$ monic non-squarefree polynomials (hint: the Dirichlet series of $\mu^2(f)=1_{f\text{ squarefree}}$ is $\frac{\zeta(u)}{\zeta(u^2)}$ where $\zeta(u) = \sum_{f \text{ monic}} u^{\deg f}$). 
In other words, the zero locus of $D(a_0,\cdots,a_{d-1})=0$ is of size $q^{d-1}$. On the other hand, Weil-Lang bounds express the size of the zero locus of $D=0$ over $\mathbb{F}_{q_0}$ as $q_0^{d-1} \cdot (1+O(\frac{1}{\sqrt{q_0}}))$ times the number of geometrically irreducible components of $D=0$ of dimension $d-1$. So using the information from the zeta functions for $q=q_0^k, k \to \infty$, we find that $D$ is a power of an irreducible polynomial. 
(Almost) Irreducibility over $\overline{\mathbb{Q}}$: This reduces to irreducibility over number fields, which reduces to irreducibility over rings of integers (Gauss lemma) which follows from irreducibility over finite fields by working modulo some prime ideal.