Let p be a prime. Determine the number of irreducible quadratic polynomials over Z_p

(Worried)

Printable View

- Dec 11th 2008, 01:02 PMmandy123determine irreducible quadratic polynomials
Let p be a prime. Determine the number of irreducible quadratic polynomials over Z_p

(Worried) - Dec 11th 2008, 08:07 PMThePerfectHacker
The polynomial $\displaystyle x^{p^2} - x$ factors into a product of monic irreducible polynomials of order dividing $\displaystyle 2$.

There are $\displaystyle p$ linear monic polynomials. Let $\displaystyle n$ be the number of monic irreducible quadradic polynomials.

Then by counting degrees of polynomials in $\displaystyle x^{p^2} - x = \prod_{\deg p(x) | 2}p(x)$

We see that $\displaystyle p + 2N = p^2 \implies N = \tfrac{1}{2}(p^2 - p)$.

(This formula can be generalized to polynomials of degree $\displaystyle m$ by applying Mobius inversion formula) - Dec 12th 2008, 07:22 AMmandy123
So then what if p is not a prime, what if we are told to

Determine the number of irreducible quadratic polynomials over Z_p?

how would that change the answer? - Dec 12th 2008, 08:28 AMThePerfectHacker
Finite fields have orders of power of primes. Thus, your question would be to count the number of irreducible quadradics over $\displaystyle \mathbb{F}_q$ where $\displaystyle q=p^n$ (a power of prime). In this case the same theorem applies i.e. $\displaystyle x^{q^2} - x$ factors into monic irreducible polynomials having order dividing $\displaystyle 2$.