Elements of subfield $F_{8}$ of $F_{2}[x]/(x^{6}+x+1)$

Question

I need to find the elements of the subfield $F_{8}$ of $F_{2}[x]/(x^{6}+x+1)$ in their standard representation. I know that $F_{2}[x]/(x^{6}+x+1)$ represents the residu classes of polynomials modulo $(x^{6}+x+1)$ with coefficients either 0 or 1. There are $2^{6}$ of those residu classes because we work with the finite field of 2 elements and the degree of the polynomial is 6. $F_{8}$ is the finite field with 8 elements and I can perfectly construct this finite field. However I have difficulty seeing that $F_{8}$ is a subfield of $F_{2}[x]/(x^{6}+x+1)$. Can someone guide me in the right direction?

Answer 1

There are two kinds of structure that you can take advantage of here.

If you want to use only the multiplicative structure, then you need to know that the multiplicative groups of finite fields are cyclic. Thus $\Bbb{F}_{64}^*\cong C_{63}$ and $\Bbb{F}_8^*\cong C_7$. But a cyclic group $C_{63}$ contains a unique (necessarily also cyclic) subgroup of order seven. It consists, of course, of the ninth powers of elements of $C_{63}$. Therefore the ninth power of any element of $L=\Bbb{F}_{64}$ belongs to the (unique) subfield isomorphic to $K=\Bbb{F}_8$.

Let us abbreviate $\alpha=x+\langle x^6+x+1\rangle$. Thus we know that $\alpha^6+\alpha+1=0$, so $\alpha^6=\alpha+1$. Consequently $\beta=\alpha^9=\alpha^3\cdot\alpha^6=\alpha^4+\alpha^3$. Because $\beta\neq1$, we know that $\beta$ is of order seven, so $\beta^7=1$. We can thus list the elements of the subfield $K$, and they are $$ K=\{0,1,\beta,\beta^2,\beta^3,\ldots,\beta^6\}. $$ Anyway, $K=\Bbb{F}_2[\beta]$. For the purposes of identifying this as a subset of $\Bbb{F}_{64}$ we can write these elements as polynomials evaluated at $\alpha$ by repeatedly using the minimal polynomial of $\alpha$: $$ \begin{aligned} \beta^2&=\alpha^8+\alpha^6=\alpha^8+\alpha+1=\alpha^2\cdot\alpha^6+\alpha+1=\alpha^3+\alpha^2+\alpha+1,\\ \beta^3&=(\alpha^3+\alpha^2+\alpha+1)(\alpha^4+\alpha^3)=(\alpha^3+\alpha^2+\alpha+1)(\alpha+1)\alpha^3=\\ &=(\alpha^4+1)\alpha^3=\alpha^7+\alpha^3=\alpha^3+\alpha^2+\alpha,\\ \end{aligned} $$ At this point we observe that $\beta^3=\beta^2+1$, so $x^3+x^2+1$ (easily seen to be irreducible in $\Bbb{F}_2[x]$ is the minimal polynomial of $\beta$ over the prime field. This gives us another proof for the fact that $$\Bbb{F}_2[\beta]\cong\Bbb{F}_2[y]/\langle y^3+y^2+1\rangle\cong\Bbb{F}_8.$$ Note that I switched to using $y$ as the indeterminate, because otherwise you might confuse it with the $x$ used in your construction of $\Bbb{F}_{64}$. Anyway, this makes it easier for us to continue the list: $$ \begin{aligned} \beta^4&=\beta\cdot\beta^3=\beta(\beta^2+1)=\beta^3+\beta=\beta^2+\beta+1=\\ &=\alpha^4+\alpha^2+\alpha,\\ \beta^5&=\beta(\beta^2+\beta+1)=\beta^3+\beta^2+\beta=\beta+1=\\ &=\alpha^4+\alpha^3+1,\\ \beta^6&=\beta(\beta+1)=\beta^2+\beta=\\ &=\alpha^4+\alpha^2+\alpha+1. \end{aligned} $$ From this it is easy to verify that the subset $\{0,1,\beta,\beta^2,\ldots,\beta^6\}$ is, indeed, also closed under addition as well as multiplication. This follows from the known facts, but I encourage you to check this as an exercise and/or confirmation.

---

Another way of finding elements of the subfield is to use the relative trace map. If $z$ is an arbitrary element of $L$, then we know that $z^{64}=z$. Consequentlye the element $w=tr^L_K(z)=z^8+z$ satisfies the equation $w^8=w$, because $$ w^8=(z^8+z)^8=z^{64}+z^8=z+z^8=w. $$ Therefore $w\in K$. Let's do this for $z=\alpha$. We get $$ tr^L_K(\alpha)=\alpha^8+\alpha=\alpha^2\cdot\alpha^6+\alpha=\alpha^3+\alpha^2+\alpha $$ that we identify as the element $\beta^3$ from above. It is easy to see that $\beta$ and $\beta^3$ both generate the same subfield of $L$. I won't repeat the process of generating the other elements.