Polynomial discriminant

For a monic polynomial with roots $r_i$, its discriminant is ${\prod }_{i<j}(r_i-r_j{)}^2$ and vanishes exactly for a repeated root.

For a separable polynomial $f\in F[T]$ in characteristic two, the Vandermonde determinant of its roots is fixed by every Galois automorphism: root permutations change it only by a sign, and $-1=1$ in $F$. It therefore lies in $F$. Its square, multiplied by the square of the appropriate leading-coefficient power for a nonmonic polynomial, is the polynomial discriminant. Thus being a square does not detect containment of the Galois group in the alternating group. For example $T^2+T+1$ over ${\mathbb{F}}_2$ has discriminant one and Galois group $C_2$, while $A_2$ is trivial. The discriminant criterion for an alternating Galois group requires characteristic different from two.

A squarefree polynomial discriminant does not ensure that the polynomial defines a field of its stated degree. For example, $x^3-2x-1=(x+1)(x^2-x-1)$ has discriminant five. Its roots have degrees one or two, so no root generates a cubic number field. In particular $1,\alpha ,{\alpha }^2$ cannot be a field basis for any of its roots. The squarefree discriminant certifies an integral basis argument assumes that the polynomial is the minimal polynomial: then the discriminant of its power basis is the field discriminant times the square of the order index. For a reducible squarefree polynomial the analogous rank belongs instead to its quotient algebra, which is a product of fields, not to one root field.

For a depressed quartic $x^4+p{x}^2+qx+r$ with roots ${\alpha }_1,\dots ,{\alpha }_4$, the three pair-partition quantities ${\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$ and its two analogues are the roots ofThe Galois action on these roots is the quotient of its action on the four quartic roots by the Klein four-group.

Let a separable polynomial $f$ of degree four be irreducible over a field $F$. If all three pair sums $r_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$, $r_2={\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4$, and $r_3={\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3$ lie in $F$, its Galois group is the Klein four-group. Indeed $r_1-r_2=({\alpha }_1-{\alpha }_4)({\alpha }_2-{\alpha }_3)\ne 0$, and similarly the other pair sums differ. Every Galois automorphism therefore fixes each of the three partitions into pairs. The kernel of this action of $S_4$ on three partitions is $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Irreducibility gives a transitive group action on the four roots, so the orbit-stabilizer theorem forces $4$ to divide the group order. Hence the Galois group equals $V_4$. The argument also works in characteristic two.

If an irreducible quartic over a field of characteristic zero has a cubic resolvent of a quartic whose Galois group of a polynomial is $S_3$, then the quartic has Galois group $S_4$. Irreducibility places the quartic group among the transitive subgroups of the symmetric group on four points, while the action on its three root pairings is surjective onto $S_3$. Of the possibilities $C_4,V_4,D_8,A_4,S_4$, only $S_4$ has this image.

The Vandermonde product ${\prod }_{i<j}(r_i-r_j)$ is alternating, and its square is the discriminant of the monic polynomial with roots $r_i$.

For a nonnegative weight $w$ and monic polynomials $p_j$ of degree $j$, this determinant is ${\prod }_i\sqrt{w(x_i)}{\prod }_{i<j}(x_j-x_i)$. Its square is the product of the weights times the squared Vandermonde determinant. For normalized weighted polynomials it differs by the product of their norm factors.

Evaluating arbitrary monic polynomials of successive degrees $0,\dots ,m-1$ gives the same determinant as evaluating the monomials of those degrees. The basis-change matrix is triangular with diagonal entries one, so its determinant is one. This permits an orthogonal polynomial basis to replace the monomials in eigenvalue densities.

The left side is an alternating polynomial of total degree one more than $\Delta (x)$. Dividing by the Vandermonde determinant gives a symmetric homogeneous polynomial of degree one, necessarily $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$; differentiation in $t$ and the Euler theorem for homogeneous functions give $b=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Thus the identity holds as a polynomial identity, even at repeated coordinates. At shifted partition coordinates with $t=-1$, it proves the removable-corner recurrence in the hook-length formula.

For distinct ${\ell }_i$ and $c_j$, expand the exponential entries in powers of $t$. Nonzero determinant terms first occur at the distinct powers $0,1,\dots ,m-1$. The coefficient is the product of their two Vandermonde determinants divided by the factorial product. Ratios of such alternants yield the Weyl dimension formula by taking a torus character to the identity.

A Vandermonde matrix has entries $V_{ij}={\lambda }_i^{j-1}$. Distinct nodes give determinant ${\prod }_{i<j}({\lambda }_j-{\lambda }_i)\ne 0$, explaining independence of polynomial evaluations and distinct-eigenvalue Krylov subspaces.

A generalized Vandermonde matrix evaluates a chosen list of functions ${\varphi }_0,\dots ,{\varphi }_n$ at interpolation nodes $x_0,\dots ,x_n$. For a basis of polynomials of polynomial degree at most $n$ and distinct nodes, it is an invertible matrix by uniqueness of polynomial interpolation. Choosing monomials recovers the ordinary Vandermonde matrix.

Collect the distinct nonzero values among the ${\lambda }_j$ as ${\mu }_1,\dots ,{\mu }_r$ with positive multiplicities $m_j$. The first $r$ power sums imply ${\sum }_j{m}_j{\mu }_j^k=0$ for $1\le k\le r$. The coefficient matrix is an invertible weighted Vandermonde matrix, so all $m_j$ vanish, a contradiction unless $r=0$. Its invertibility also follows because a polynomial ${\sum }_{k=1}^r{a}_k{z}^k$ vanishing at zero and all the ${\mu }_j$ has $r+1$ distinct roots and degree at most $r$.

The depressed cubic $x^3+ax+b$ has discriminant $-4a^3-27b^2$.

For a monic separable polynomial over a field of characteristic other than two, its Galois group is contained in the alternating group exactly when its discriminant is a square in the base field. The Galois action multiplies the Vandermonde product by the sign of the induced root permutation.