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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.

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This is analogous to constructing the complex numbers over the reals, based on the irreducible polynomial $x^2+1$. Think how this is used to define the complex numbers. Then do something similar in your case. –  GEdgar Nov 25 '12 at 14:10
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Your field will be $\mathbb{Z}_7[x]/\langle x^3 + x + 1 \rangle$. That means you can identify elements of the field with polynomials of degree $\le 2$, and the laws are given by adding or multiplying elements and then reducing them modulo $x^3 + x + 1$ (that is computing the remainder in the division). –  Joel Cohen Nov 25 '12 at 14:16
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1 Answer

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.

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