irreducible polynomials - Constructing a finite field of order 343.

I just wanted to check my logic on this type of exercise. I know for any prime $p$ and a monic irreducible polynomial in $\mathbb{F}_{p}[x]$ of degree $n$ that a ring like $\mathbb{F}_{p}[x]/(p(x))$ is a field of order $p^{n}$.

So I need an irreducible polynomial of degree $3$ in $\mathbb{Z}_{7}[x]$ since $7^{3}=343$. So I chose $p(x)=x^{3}+x+1$ which is irreducible in $\mathbb{Z}[x]$, so irreducible in $\mathbb{Z}_{7}[x]$. Thus the field that can be constructed is

$$\mathbb{Z}_{7}[x]/(x^{3}+x+1)$$

which has order 343.

Is there anything flawed in this argument or anything that I am missing?

Answer

This is almost correct, but you cannot be sure that $p(x)$ is irreducible in $\mathbb{Z}_7[x]$ just because it is irreducible in $\mathbb{Z}[x]$. For instance, $x^2-2$ is irreducible in $\mathbb{Z}[x]$ but it factors as $(x-3)(x+3)$ in $\mathbb{Z}_7[x]$. Instead, you can verify your polynomial is irreducible in $\mathbb{Z}_7[x]$ by just testing that it has no roots in $\mathbb{Z}_7$ (since if it were reducible, it would have to have a linear factor).