This must be wellknown to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d1} x^{d1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over)?

10$\begingroup$ Yes, this is true: the discriminant is geometrically irreducible. This is proved, in higher generality, on page 15 (Ch. 1.1. B) of Gelfand, Kapranov and Zelevinsky's book, Discriminants, Resultants and Multidimensional Determinants. $\endgroup$ – Vesselin Dimitrov Oct 24 '15 at 22:35

2$\begingroup$ @VesselinDimitrov Thanks! I will look, but is there some conceptual reason why something like this should be true (other than it would be shocking otherwise)? $\endgroup$ – Igor Rivin Oct 24 '15 at 22:56

$\begingroup$ The conceptual reason is geometric. I wrote an answer outlining it. $\endgroup$ – Vesselin Dimitrov Oct 24 '15 at 23:35
The discriminant locus has the following geometric interpretation, given in the introductory chapter of [Gelfand, Kapranov, Zelevinsky: Discriminants, Resultants and Multidimensional Determinants].
Let $X \subset \mathbb{P}^d$ be the rational normal curve, $[u:v] \mapsto [u^d:u^{d1}v:\cdots : v^d]$. We may view a linear form $l \in \Gamma(\mathcal{O}_{\mathbb{P}^d}(1))$ as filling in coefficients for a degree$d$ polynomial, viz. restricting $l$ to $X$. This identifies the space of nonzero polynomials (modulo scalars) of degree $\leq d$ as the projective linear dual $P^{\vee}$ of $P := \mathbb{P}^d$. The roots of a polynomial are the (preimages on $\mathbb{P}^1 \hookrightarrow \mathbb{P}^d$ of) the intersection points of $X$ with the hyperplane $\{l = 0\}$. The discriminant locus is then the closure $X^{\vee} \subset P^{\vee}$ of the set of all the hyperplanes of $P$ that are tangent to $X$.
This gives the conceptual reason, since the following straightforward geometric argument (Prop. 1.3 in [GKZ]) proves in general that $X^{\vee} \subset P^{\vee}$ is irreducible whenever $X \subset P = \mathbb{P}^d$ is irreducible. Assuming WLOG that $X$ is regular, the incidence variety $W \subset P \times P^{\vee}$ of pairs $(x,H)$ with $x \in X$ and $H$ a hyperplane tangent to $X$ at $x$, is a projective bundle over the irreducible variety $X$, and hence irreducible. It follows that $X^{\vee}$, being the image $W \hookrightarrow P \times P^{\vee} \to P^{\vee}$ of an irreducible variety, is itself irreducible.
The geometric interpretation applies to more general discriminants, studied in [GKZ].
(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.)
The discriminant is defined by the property that, when $$ p(x) = x^d  s_1 x^{d1} + \cdots + (1)^ds_d = (xt_1)(xt_2)\cdots(xt_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_it_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_2(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_it_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_it_j) $$ for some constants $c'_1$ and $c'_2$.
However, when the characteristic of the field is not $2$, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_it_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$. Thus, $D$ is irreducible when the characteristic of the field is not $2$.
As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression $$ F = \prod_{i<j} (t_it_j) = \prod_{i<j} (t_i+t_j) $$ is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$. One then has $D = F^2$, so $D$ is a square. The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$ (up to constant multiples).

$\begingroup$ As Ofir's answer implicitly suggests, this argument doesn't seem to work as it stands in characteristic 2. $\endgroup$ – Geoff Robinson Oct 25 '15 at 12:51

2$\begingroup$ Yes, of course, obviously the last step in my argument assumes that the characteristic is not $2$, but, then as Ofir's comment points out, $D$ can, in fact, be reducible in characteristic $2$, so I don't regard that as a flaw, but as a feature. The argument shows exactly where characteristic $2$ is needed. (In fact, $D$ is reducible over characteristic $2$ when $d=2$ and $d=3$, and maybe in all degrees.) $\endgroup$ – Robert Bryant Oct 25 '15 at 12:56

$\begingroup$ This is pretty close to the sort of thing I was thinking of (that is, a more invariant theoretic proof  the factors must be invariants of the alternating group). $\endgroup$ – Igor Rivin Oct 25 '15 at 13:00

3$\begingroup$ In characteristic $2$ the polynomial $\prod_{i<j} (t_it_j)=\prod_{i<j} (t_i+t_j)$ is symmetric in $t$'s, hence a polynomial in $s$'s (the fundamental theorem of symmetric polynomials holds in all commutative rings), thus the discriminant is a square of an irreducible polynomial. $\endgroup$ – Jarek Kuben Oct 25 '15 at 13:17

2$\begingroup$ I didn't suggest it was a flaw, but felt that the exception should be noted. $\endgroup$ – Geoff Robinson Oct 25 '15 at 13:53
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 $q^{d1}$ monic nonsquarefree 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 $D(a_0,\cdots,a_{d1})=0$ is of size $q^{d1}$. On the other hand, WeilLang bounds express the size of the zero locus of $D=0$ over $\mathbb{F}_{q_0}$ as $q_0^{d1} \cdot (1+O(\frac{1}{\sqrt{q_0}}))$ times the number of geometrically irreducible components of $D=0$ of dimension $d1$. 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.


$\begingroup$ @IgorRivin Thank you. It is not original though  many functionfield number theorists know it well, but I've never seen it written down. $\endgroup$ – Ofir Gorodetsky Oct 25 '15 at 13:11

$\begingroup$ By the way, I assume that the formula for discriminant (as in Robert Bryant's answer) takes care of the proper power possibility. $\endgroup$ – Igor Rivin Oct 25 '15 at 20:25