Irreducible polynomials of degree $2$ over a finite field

Question

By consider the nonquadratic residues, one has an irreducible polynomial of degree $2$ over ${\Bbb F}_p$ for $p$ being odd: $x^2+r$. Also we know that $x^2+x+1$ is irreducible over ${\Bbb{F}_2}$. How can irreducible polynomials of degree $2$ over ${\Bbb F}_{p^n}$ in general look like?

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For case when $p$ is odd, I'm wondering if we still have $x^2+r$ irreducible over $K:={\Bbb F}_{p^n}$ for some $r\in K$. But I don't see how to go on with this idea.

Answer 1

I just found that one can use the map $f:K\to K$ with $f(x)=x^2$ when $p$ is odd and $f(x)=x^2+x$ when $p=2$. By using the fact that on a finite set, a function is injective iff surjective, we can tell that in both cases, $f$ is not surjective since $$ f(1)=f(-1)\quad p\ \hbox{odd} $$ and $$ f(0)=(1). $$

Thus there exists irreducible polynomial of the form $x^2-r$ when $p$ is odd and $x^2+x+r$ when $p=2$ where $r\in K$ not in the image of $f$.

Answer 2

a polynomial of degree 2 is irreducible iff has a root (if p is odd )iff $b^2-4ac$ has a square root.so P is irreducible over $Z_p$ iff $b^2-4ac$ is a quadratic residue mod p.$F_{p^2} \subset F_{p^2k}$ and every degree 2 polynomial is reducible over $F_{p^2}$ so over $F_{p^{2k}}$. by the tower law you can see if a degree 2 polynomial is irreducible over$z_p$ it is irreducible over $F_{p^{2k+1}}$(because $2\not|2k+1$)so P is irreducible over $F_{p^{2k+1}}$ iff $b^2-4ac$ is a quadratic residue mod p

Answer 3

Let $p$ be odd. Then any element $a$ of a finite field $\mathbb{F}_{p^n}$ is a zero of a unique quadratic polynomial $X^2 - b$ with $b \in \mathbb{F}_{p^n}$. Since any such polynomial (except the one for $b=0$) has two zeroes that are distinct (the two zeroes are negatives of eachother and therefore distinct because $p$ is odd), there will be polynomials left that do not have any zeroes. Any such polynomial is irreducible.

A similar argument can be given for $p=2$, considering polynomials of the form $X^2 - X - b$, $b \in \mathbb{F}_{2^n}$. Such a polynomial has two distinct zeroes (its zeroes have sum $1$), and since an element of $\mathbb{F}_{2^n}$ is a root of at most one such a polynomial, there are irreducible polynomials of this form.