Reference request: relationship between discriminant and smoothness of a conic over arbitrary fields I'm looking for an (ideally modern) reference of the relationship between the discriminant and smoothness of projective conics over arbitrary fields (including those of characteristic 2). Let $k$ be a field. Let $q(x,y,z)$ be the polynomial
$$q(x,y,z) = ax^2 + by^2 + cz^2 + dxy + exz + fyz\in k(a,b,c,d,e,f)[x,y,z]$$
where $a,b,c,d,e,f$ are independent transcendentals over $k$. Then $b(v,w) := q(v+w) - q(v) - q(w)$ defines a bilinear form on $k(a,b,c,d,e,f)^3$. Let $M_b$ be the associated Gram matrix, then $\det(M_b)$ is divisible by 2 as an element of $\mathbb{Z}[a,b,c,d,e,f]$, and hence we may define:
$$d_q := \frac{1}{2}\det(M_b)\in\mathbb{Z}[a,b,c,d,e,f]$$
Now let
$$Q(x,y,z) = Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz\in k[x,y,z]$$
then define $D_Q := d_q(A,B,C,D,E,F)$ viewed as an element of $k/k^2$.
Let $\overline{k}$ be an algebraic closure of $k$. I believe the following statement is true:
For any field $k$, the subscheme of $\mathbb{P}^2_{\overline{k}}$ given by $Q(x,y,z)$ is integral (reduced and irreducible) if and only if $D_Q\ne 0$.
Does anyone know of a reference (ideally modern and in english) for this? (assuming it's true).
 A: abx's comment gives a French reference "Résultant, discriminant" by Michel Demazure.
Here's a rough digest of the relevant parts:
Let $k$ be an arbitrary field. Let $f\in k[X_1,\ldots,X_n]$ be a homogeneous polynomial of degree $d\ge 2$. Demazure describes the discriminant $\text{disc}(f)$ of $f$, which is itself a polynomial in the coefficients of $f$, hence an element of $k$, such that $\text{disc}(f) = 0$ if and only if the hypersurface in $\mathbb{P}_k^{n-1}$ defined by $f$ is smooth.
For smoothness, he uses the Jacobian criterion, which amounts to saying that the only common zero (in an algebraic closure $\overline{k}$) of the polynomials $f,D_1f,\ldots,D_nf$ is the origin $(X_1,\ldots,X_n) = (0,\ldots,0)$, where $D_i := \frac{\partial}{\partial X_i}$.
He characterizes this "nonexistence of nontrivial zeroes" in terms of the resultant of the family $(D_1f,\ldots,D_nf)$. First, let
$$a(n,d) := \frac{(d-1)^n - (-1)^n}{d}\qquad (\text{Demazure, p360, discussion after Lemme 9})$$
Then the universal discriminant $\text{disc}$ (for homogeneous polynomials of degree $d$) is defined by the rule
$$d^{a(n,d)}\text{disc} = \text{res}(D_1P_{n,d},\ldots,D_nP_{n,d})\qquad(\text{Demazure, Definition 4, p362})$$
where $\text{res}$ is the resultant (see below), and $P_{n,d}$ is the ``universal homogeneous polynomial of degree $d$ in $n$ variables'' (e.g., $P_{3,2}$ is just $q(x,y,z)$ in the OP). Precisely, $P_{n,d} = \sum_{\alpha} T_\alpha X^\alpha$ where the sum ranges over all $\alpha\in\mathbb{Z}_{\ge 0}^n$ satisfying $\sum_{i=1}^n\alpha_i = d$, $X^\alpha = \prod_{i=1}^n X_i^{\alpha_i}$, and the $T_\alpha$'s are independent transcendentals.
The resultant is somewhat complicated in general, but if $g_1,\ldots,g_n$ is a family of linear polynomials in $k[X_1,\ldots,X_n]$, where $g_i = \sum_{j = 1}^n g_{ij}X^j$, then the resultant is just
$$\text{res}(g_1,\ldots,g_n) = \det(g_{ij})\qquad\text{(Demazure, Exemple 1, p349)}$$
Thus, the universal discriminant for homogeneous polynomials of degree $d$ is a polynomial with coefficients in $k$ in the variables $T_\alpha$, and given a homogeneous degree $d$ polynomial $f = \sum_{\alpha}c_\alpha X^\alpha$, its discriminant is
$$\text{disc}(f) = \text{disc}(\{c_\alpha\})\qquad(\text{Demazure, Definition 4, p362, see $\S1$ for notation.})$$
(ie, this is the element of $k$ obtained by replacing the $T_\alpha$'s in the universal discriminant with the actual coefficients $c_\alpha$ of $f$).
Finally, he shows that this discriminant vanishes if and only if $f$ and its $n$ partial derivatives have no nontrivial zero in any field extension (Proposition 12), which is equivalent to the smoothness of the associated hypersurface by taking affine charts and applying the jacobian criterion for smoothness (also see p335 in the introduction).
In the case $(n,d) = (3,2)$ (ternary quadratic forms), we find $a(3,2) = 1$, and hence $\text{disc}(f)$ is precisely as given in the OP.
