Irreducible polynomial of degree 2 over a $\mathbb{Z}_{7}$

Question

I am not that great at abstract algebra, so I need some feedback on whether my way of thinking is right or wrong. I have this one exercise that reads:

Find an irreducible polynomial of degree $2$ over $\mathbb{Z}_{7}$

Which sounds simple. Additionally, class materials contain the following theorem:

Let $w$ be an element of order $n$ in a finite Abelian field $\mathbb{F}_{q}$ of characteristic $p$, and let $m$ be the order of $p$ in $\mathbb{Z}^{*}_{n}$. Then the coefficients of the $m$th degree polynomial $f(x) = \displaystyle\prod_{i=0}^{m-1} \left(x-{w^{p}}^{i}\right)$ are field integers. Furthermore, $f$ is irreducible in $\mathbb{Z}_{p}[x]$.

Now I wasn't really sure how to go about this thing, but from what I can tell, both order and characteristic are $7$ for the case of $\mathbb{Z}_{7}$, and since we are looking for polynomial of degree $2$, $m=2$ so the product formula will take the following form:

$$\displaystyle\prod_{i=0}^{1} \left(x-{w^{7}}^{i}\right)$$

So for some element, say 3, I would get:

$$(x-1)(x-3^7) \equiv (x-1)(x-3)\mod 7$$

Which equals to $x^2 - 4x + 3$. Which I guess is a irreducible polynomial of degree $2$ over $\mathbb{Z}_{7}$.

My question here is whether or not I am right, or rather just how wrong I am and which part did I misunderstand. It would be much appreciated if someone could explain this a bit to me, since I am still confused by the materials I went through and that I found online.

Answer 1

The easiest way is this: Irreducible polynomials of degree 2 or 3 are irreducible iff they have no roots (this is true over any field). So just find some $x^2+a$ with no roots.

Answer 2

\begin{align} & 1^2 \equiv (-1)^2 \equiv 1 \pmod 7 \\ & 2^2 \equiv (-2)^2 \equiv 4 \\ & 3^2 \equiv (-3)^2 \equiv 2 \end{align} There are no nonzero congruence classes modulo $7$ other than those of $\pm1,\pm2,\pm3.$ Therefore there are no nonzero squares modulo $7$ other than $1,2,4.$ Thus the following equations have no solutions in $\mathbb Z_7~{:}$ \begin{align} & x^2 - 3 = 0 \\ & x^2 - 5 = 0 \\ & x^2 - 6 = 0 \end{align}

Generally, modulo a prime number, exactly half of the nonzero congruence classes are squares because the squares of $\pm k$ are congruent to each other, so $k\mapsto k^2$ is a two-to-one mapping.