Try
The monic polynomial f(t)=t³−3t+2 factors as (t−1)²(t+2) over Q. The nearby polynomial g(t)=t³−3t+1 has no repeated root, even though neither a plot nor a list of approximate roots is a proof. We want one exact coefficient calculation that detects this difference.
A polynomial is monic when its leading coefficient is 1. In this lesson f is monic of degree n≥1 over a field F. Choose a splitting field K and list all roots r₁,…,rₙ with their multiplicities. The discriminant is
Disc(f) = product over i<j of (rᵢ−rⱼ)².
A repeated root makes one factor zero. Conversely, in the field K a nonzero product has no zero factors, so a nonzero discriminant means all roots are distinct. The challenge is to calculate this expression in F without already knowing the roots.
Predict
Reordering the root list changes the Vandermonde product by a sign, but squaring removes that sign. More directly, reordering permutes unordered root pairs, and reversing a difference leaves its square unchanged. Thus Disc(f) is a symmetric polynomial in the roots. By the fundamental theorem of symmetric polynomials, its universal integer polynomial can be written as an integer polynomial in the elementary symmetric functions. Those functions are the signed coefficients of the monic polynomial f.
Consequently Disc(f) belongs to F and is independent of both root ordering and the splitting field. This coefficient expression uses no division, so it can be reduced in any characteristic. Predict why that last statement does not permit dividing by 2 or 3 while changing variables in a field where one of those numbers is zero.
The formal derivative f′ is obtained by differentiating powers, with integer coefficients interpreted in F. We will connect the discriminant to f′ without assuming that f′ has degree n−1; its leading coefficient can vanish in positive characteristic.
Experiment
For f=t³+pt+q, work in the three-dimensional F-vector space A=F[t]/(f), with basis 1,t,t². Here A is a quotient algebra, not generally a field. Let T denote multiplication by the class of t. Since t³=−pt−q in this quotient, its matrix has rows (0,0,−q),(1,0,−p),(0,1,0). Multiplication by f′=3t²+p is the operator 3T²+pI.
| Polynomial over Q | p, q | det of multiplication by f′ | Discriminant |
|---|
| t³−3t+2 | −3, 2 | 4(−3)³+27·2²=0 | 0 |
| t³−3t+1 | −3, 1 | −108+27=−81 | 81 |
| t³+t+1 | 1, 1 | 4+27=31 | −31 |
Compute the matrix before using its determinant. It has rows (p,−3q,0), (0,−2p,−3q), (3,0,−2p). Expansion in the first row gives 4p³+27q². In degree three the discriminant is the negative of this determinant, not the determinant itself. The next proof explains that sign and works even for the repeated-root first row.
Understand
The derivative product, including repeated roots
Differentiate f(t)=∏(t−rᵢ) formally over K. Evaluating the product rule at rᵢ gives
f′(rᵢ) = product over j≠i of (rᵢ−rⱼ).
This still holds if roots repeat: all other summands in the product rule contain the factor t−rᵢ indexed by i, and vanish at rᵢ; the remaining product can also be zero. Multiplying these identities over all i pairs the factors for i,j as
(rᵢ−rⱼ)(rⱼ−rᵢ) = −(rᵢ−rⱼ)².
There are n(n−1)/2 pairs. Therefore
Disc(f) = (−1)^(n(n−1)/2) ∏ f′(rᵢ).
Multiplying twice by the same sign returns 1, so this equation has the stated direction. In characteristic two the sign is 1 and the proof remains valid. For n=1 all pair products are empty, f′=1 and the discriminant is 1.
It follows also that Disc(f)≠0 exactly when gcd(f,f′)=1 in F[t]. Indeed a repeated root is a root of f′ by the product formula, whereas a simple root has nonzero derivative there. A nonconstant gcd has a root in a splitting extension; conversely a shared root precludes a Bézout identity equal to 1. Since F[t] is Euclidean, gcd 1 is equivalent to such an identity. If f′=0, the monic gcd is f itself, and since n≥1 the discriminant is zero. In this sense f is separable exactly when its discriminant is nonzero: “separable” here describes this polynomial having distinct roots, not a claim that every polynomial over F does.
A determinant proof that does not require distinct roots
Let M be multiplication by the class of f′ on A=F[t]/(f). The division basis 1,t,…,tⁿ⁻¹ makes this a concrete n-by-n matrix over F. Extend coefficients to K; the same matrix represents multiplication on K[t]/(f), and its determinant is unchanged as an element of K.
For j=0,…,n−1 define bⱼ(t) as the product of the first j factors t−rᵢ, with b₀=1. Each bⱼ is monic of degree j. The coefficient-change matrix from 1,t,…,tⁿ⁻¹ to these polynomials is triangular with diagonal entries 1, so their classes form a basis, even when roots repeat.
For j<n−1, we have t bⱼ=bⱼ₊₁+rⱼ₊₁bⱼ. For j=n−1 the same identity uses bₙ=f, whose class is zero. Thus multiplication by t is lower triangular in the ordered basis b₀,…,bₙ₋₁, with diagonal r₁,…,rₙ. Any polynomial in a triangular matrix is triangular with that polynomial evaluated on its diagonal. Multiplication by f′ consequently has diagonal f′(r₁),…,f′(rₙ), and
det M = ∏ f′(rᵢ).
Combining with the signed product proves the determinant formula for Disc(f). We used a triangular basis, not an eigenbasis or evaluation isomorphism to Kⁿ; those latter arguments would fail at repeated roots. The quotient algebra may have nilpotents, but multiplication and its determinant still make sense.
Derive the low-degree formulas
For f=t²+bt+c, multiplication by t in basis 1,t is T=[[0,−c],[1,−b]]. Hence multiplication by f′=2t+b has matrix [[b,−2c],[2,−b]]. Its determinant is 4c−b². The sign for n=2 is −1, giving Disc(f)=b²−4c.
For f=t³+pt+q, the Experiment matrix has determinant 4p³+27q². The sign for n=3 is also −1, giving Disc(f)=−4p³−27q². Both are polynomial identities over the integers and remain valid over every field, with coefficients interpreted there.
Completing the square by a shift involving b/2 requires 2 invertible. Removing the quadratic term from a general cubic t³+at²+bt+c by t=u−a/3 requires 3 invertible. A cubic already in the displayed form needs no shift and its formula is valid in characteristic three too. For example in characteristic three its derivative is p, and its discriminant is −p³. If p=0, this polynomial has zero derivative and repeated roots in a splitting field, whether or not those roots belong to F.
This lesson defines and proves these formulas only for monic polynomials. We have not asserted a formula for arbitrary leading coefficients or invoked a general resultant theorem. Dividing by a leading coefficient and then transferring a discriminant formula would require explicit normalization and scaling arguments.
Apply
Over Q, compare h=t³−12t+16 with k=t³−6t+4. Compute both discriminants and independently decide repeated-root status using the derivative or a factorization. Then examine f=t²+t+1 over F₂ and over F₃: use the same coefficient formula but interpret its value in the correct field.
Check the changed polynomials and characteristics
Disc(h)=−4(−12)³−27·16²=6912−6912=0. In fact h=(t−2)²(t+4), and over Q its monic gcd with h′=3(t²−4) is t−2. For k the discriminant is 864−432=432, nonzero. Its derivative vanishes only where t²=2; substituting that relation into k gives 4−4t. A common root would have t=1 and t²=2, impossible over Q’s splitting extension. Thus gcd(k,k′)=1, independently of the discriminant arithmetic.
For t²+t+1 the discriminant is −3. Over F₂ it is 1; the derivative is 1 and the polynomial is separable, although it has no root in F₂ itself. Over F₃ the discriminant is zero and the polynomial is (t−1)². The formal derivative is 2t+1, which vanishes at 1. The distinction is separability, not whether a root happens to lie in the base field.
For one more boundary check, t³−12t+16 reduces modulo 3 to t³+1=(t+1)³. Its derivative is zero, so the monic gcd with its derivative is the whole polynomial. The characteristic-zero gcd t−2 cannot simply be transported to characteristic three without rechecking the coefficients and derivative.
Continue
Use polynomial gcds and separability to compare coefficient certificates with field-extension questions. The determinant route is useful precisely when explicitly finding the roots would be awkward.
For a new example, record the field, monicity and degree; build the multiplication matrix in a declared basis; include the parity sign; and verify a claimed repeated root or gcd independently. A zero determinant detects repetition but does not by itself give the multiplicities or a splitting field. Those remain separate algebraic questions.
