Question

If char K≠2 and f∈K[x] is a cubic whose discriminant is a square in K, then f is either irreducible or factors completely in K.

Answer 1

If a cubic polynomial has a discriminant that is a square in K, then the polynomial is either irreducible or factors completely in K.

Step-by-step explanation:

In this question, we are given that the discriminant of a cubic polynomial, f, is a square in K. We need to show that f is either irreducible or factors completely in K.

Let's use a proof by contradiction. Assume that f is not irreducible and does not factor completely in K. This means that f can be factored into two non-constant polynomials, f = gh, where g and h are polynomials of degree at least 1.

Now, using the fact that the discriminant of f is a square in K, we can rewrite the discriminant as a square of a polynomial in K, say d = (ax² + bx + c)², where a, b, and c are elements of K.

Expanding d, we get d = a²x⁴ + 2abx³ + (2ac + b²)x² + 2bcx + c².

Comparing the coefficients of d with that of the discriminant of f, we get the following system of equations:

a² = 1, 2ab = 0, 2ac + b² = 0, 2bc = 0, c² = 0.

Solving these equations, we find that a = ±1, b = 0, c = 0.

Since a ≠ 2, we have a contradiction. Therefore, our assumption that f is not irreducible and does not factor completely in K is incorrect. Hence, f is either irreducible or factors completely in K.