# Prove that for every prime p, there is an irreducible quadratic in Z_p[x]

• Apr 12th 2009, 01:14 PM
Dark Sun
Prove that for every prime p, there is an irreducible quadratic in Z_p[x]
Prove that for every prime p, there is an irreducible quadratic in $\displaystyle Z_p[x]$ (the polynomial ring of the integers mod p).

I have not been able to wrap my brain around this one. Does anyone have any ideas?

I have been looking at $\displaystyle x^2+x+1$ and $\displaystyle x^2+1$, but haven't been able to derive a contradiction.

This is what I have so far: Assume $\displaystyle x^2+1$ is reducible in $\displaystyle Z_p[x]$. Then, $\displaystyle \exists a,b\in Z_p[x]$ s.t. $\displaystyle x^2+1=(x-a)(x-b)=x^2-(a+b)x+ab\Rightarrow a+b=0, ab=1$

Thanks!
• Apr 12th 2009, 02:10 PM
NonCommAlg
Quote:

Originally Posted by Dark Sun
Prove that for every prime p, there is an irreducible quadratic in $\displaystyle Z_p[x]$ (the polynomial ring of the integers mod p).

I have not been able to wrap my brain around this one. Does anyone have any ideas?

I have been looking at $\displaystyle x^2+x+1$ and $\displaystyle x^2+1$, but haven't been able to derive a contradiction.

This is what I have so far: Assume $\displaystyle x^2+1$ is reducible in $\displaystyle Z_p[x]$. Then, $\displaystyle \exists a,b\in Z_p[x]$ s.t. $\displaystyle x^2+1=(x-a)(x-b)=x^2-(a+b)x+ab\Rightarrow a+b=0, ab=1$

Thanks!

if p = 2, consider $\displaystyle x^2 + x + 1.$ for p > 2, choose $\displaystyle a \in \{1, \cdots, p-1 \}$ such that $\displaystyle a$ is not a quadratic residue modulo p. then $\displaystyle x^2 - a$ would be irreducible in $\displaystyle \mathbb{Z}_p[x].$
• Apr 12th 2009, 05:56 PM
Dark Sun
Thanks Dark Knight, this is just the answer I was looking for.

You have helped me a great deal.

I wish you well in your studies!
• Apr 13th 2009, 11:00 AM
ThePerfectHacker
Quote:

Originally Posted by Dark Sun
Prove that for every prime p, there is an irreducible quadratic in $\displaystyle Z_p[x]$ (the polynomial ring of the integers mod p).

Here is another approach but this is not a good approach at all. (Puke)

Let $\displaystyle f(x) = x^{p^2} - x$ and let $\displaystyle K$ be its splitting field.
Then we have that $\displaystyle [K:F]=2$, pick any $\displaystyle a\in K - F$ then the minimal polynomial of $\displaystyle a$ must have degree $\displaystyle 2$.
Thus, you have found an irreducible polynomial of degree 2.