Explain how to construct a field of order $343$ not using addition and multiplication tables.
I understand that every finite field has order $p^n$ for some prime $p$. Since $343$ is $7^3$, let $p=7$. I believe I need to find a polynomial of degree 3 which does not factor over $\mathbb{Z}_7$. I have considered the following polynomial $x^3+x+1$ and showed that it is irreducible over $\mathbb{Z}_7$.
I am not sure how to procede from here, any help would be great.
Putting $\,\Bbb F_7:=\Bbb Z/7\Bbb Z\,$ , put $\,\Bbb F_{7^3}=\Bbb F_7[x]/(x^3+x+1)\,$ .
Now, divide any $\,f(x)\in \Bbb F_7[x]\,$ by $\,x^3+x+1\,$ with residue:
$$f(x)=h(x)(x^3+x+1)+r(x)\,\,,\,deg r<3\,\,\,or\,\,\,r(x)=0\,$$
Then,
$$f(x)+(x^3+x+1)=r(x)+(x^3+x+1)\in F_{7^3}$$
Thus, any element in $\,F_{7^3}\,$ as above can be represented by an element in $\,\Bbb F_7[x]\,$ of degree $\,\leq 2\,$ in the quotient ring (field).
Check that there are $\,7^3\,$ elements in in the quotient, and then you multiply and sum modulo $\,7\,$ taking into account that $\,x^3=-x-1=6x+6\,$ in that quotient.