Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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

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:

$$f(x)=h(x)(x^3+x+1)+r(x)\,\,,\,deg r<3\,\,\,or\,\,\,r(x)=0\,$$


$$f(x)+(x^3+x+1)=r(x)+(x^3+x+1)\in F_{7^3}$$

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.