Modern Algebra: Fields/Polynomials/Irreducible

[lifeonfire]

Homework Statement

Prove that if F is a finite field then there is a quadratic polynomial in F[x] that is irreducible over F

Homework Equations

The Attempt at a Solution

To Prove:At least 1 Quadratic polynomial:Ax^2 +Bx +C not = P(x)Q(x)
I know that if P(x) is a polynomial of degree n over any field F, then P(x) = 0 has at most n distinct solutions.
Therefore a quadratic polynomial will have at least 2 zeros in any F.
But I don't what to do after that...

 

[mighty2000]

To prove that there is a quadratic polynomial that is irreducible over F, we need to show that it cannot be factored into two polynomials of lower degree over F. We can do this by contradiction.

Assume that all quadratic polynomials over F can be factored into two polynomials of lower degree. Then, for any quadratic polynomial P(x) = Ax^2 + Bx + C, there exists two polynomials Q(x) and R(x) such that P(x) = Q(x)R(x).

This means that P(x) = 0 has at least two distinct solutions over F, since if P(x) = Q(x)R(x), then either Q(x) or R(x) must be equal to 0. However, as you mentioned, a polynomial of degree n over F can have at most n distinct solutions. Since n = 2 for a quadratic polynomial, this means that P(x) = 0 can have at most 2 distinct solutions over F.

But we assumed that P(x) = 0 has at least two distinct solutions, which contradicts our previous statement. Therefore, our assumption that all quadratic polynomials over F can be factored into two polynomials of lower degree must be false. This means that there must exist at least one quadratic polynomial in F[x] that is irreducible over F, since it cannot be factored into two polynomials of lower degree.

In conclusion, if F is a finite field, there must exist at least one quadratic polynomial in F[x] that is irreducible over F.