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

1. ## 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 $Z_p[x]$ (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 $x^2+x+1$ and $x^2+1$, but haven't been able to derive a contradiction.

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

Thanks!

2. Originally Posted by Dark Sun
Prove that for every prime p, there is an irreducible quadratic in $Z_p[x]$ (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 $x^2+x+1$ and $x^2+1$, but haven't been able to derive a contradiction.

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

Thanks!
if p = 2, consider $x^2 + x + 1.$ for p > 2, choose $a \in \{1, \cdots, p-1 \}$ such that $a$ is not a quadratic residue modulo p. then $x^2 - a$ would be irreducible in $\mathbb{Z}_p[x].$

3. 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!

4. Originally Posted by Dark Sun
Prove that for every prime p, there is an irreducible quadratic in $Z_p[x]$ (the polynomial ring of the integers mod p).
Here is another approach but this is not a good approach at all.

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