Finite field with 8 elements

Question

I am trying to work out exercise 61 from Rotman's Galois Theory, second edition:

61. Give the addition and multiplication tables of a field having eight elements. (Hint: Factor $x^8 -x$ over $\mathbb{Z}_2$.)

I am able to work out the previous exercise, which asks for a field with four elements by adjoining a suitable root of $x^4 - x$ to $\mathbb{Z}_2$. I suspect a similar strategy should be used for this exercise, but I haven't succeeded in working it out.

I understand that this field may be expressed as $\mathbb{Z}_2[x]/(p(x))$, where $p(x)$ is an irreducible polynomial of degree 3. This would allow me to work out the multiplication and addition tables. However, I don't think this is intended here, since the hint is not used.

When I try to use the hint and factor $x^8 - x$, I get as far as $$ x^8 - x = x(x - 1)(x^6 + x^5 + x^4 +x^3 +x^2 + x + 1) $$

I know that when I adjoin a root $\xi$ with $\xi^7 = -1$ I should be able to factor $$ x^6 + x^5 + x^4 +x^3 +x^2 + x + 1 = (x-\xi)(x-\xi^2)(x-\xi^3)(x-\xi^4)(x-\xi^5)(x-\xi^6) $$

But I cannot justify this with a calculation (when I try to apply polynomial factoring, the equations get quite involved and don't work out). Moreover, the field $\mathbb{Z}_2 / (x^6+x^5+x^4+x^3+x^2+x+1)$ resulting from adjoining this root has $2^6 = 64$ elements, not 8.

I think the proof of the following theorem should be used (quoted verbatim from the textbook):

Theorem 33 (Galois). For every prime $p$ and every positive integer $n$, there exists a field having exactly $p^n$ elements.

Proof. If there were a field $k$ with $|K| = p^n =q$, then $K^\# = K - \{ 0 \}$ would be a multiplicative group of order $q - 1$; by Lagrange's theorem, $a^{q - 1} = 1$ for all $a \in K^\#$. It follows that every element of $K$ would be a root of the polynomial $$ g(x) = x^q - x .$$ We can now begin the construction. By Kronecker's theorem, there is a field $E$ containing $\mathbb{Z}_p$ over which $g(x)$ splits. Define $F = \{ \alpha \in E : g(\alpha) = 0 \}$; that is, the set of all the roots of $g(x)$. Since the derivative $g'(x) = qx^{q - 1} - 1 = -1$ (because $q = p^n$ and $E$ has characteristic $p$), Lemma 32 shows that the $\gcd(g, g') = 1$, and so $g(x)$ has no repeated roots; that is, $|F| = q = p^n$.

We claim that $F$ is a field, which will complete the proof. If $a, b \in F$, then $a^q = a$ and $b^q = b$. Therefore, $(ab)^q = a^qb^q = ab$, and $ab \in F$. By Lemma 32(iii), replacing $b$ by $-b$, we have $(a - b)^q = a^q - b^q = a - b$, so that $a - b \in F$. Finally, if $a \not= 0$, then $a^{q-1} = 1$ so that $a^{-1} = a^{q - 2} \in F$ (because $F$ is closed under multiplication). $\ \bullet$

This theorem actually shows that the roots of $x^8 - x$ form a field when working in $\mathbb{Z}_2$. I understand the multiplicative structure of these roots, but I haven't succeeded in working out the splitting fields, and I'm having trouble working out the additive structure. I also haven't used yet that we are working in $\mathbb{Z}_2$.

Answer 1

Based on Jyrki Lahtonen's comment I am able to work out the answer.

My mistake was that I didn't see an obvious factorization of $p(x) = x^6+x^5+x^4+x^3+x^2+x+1$. Polynomials of this kind (cyclotomic polynomials) are irreducible in $\mathbb{Q}[x]$ -- but not necessarily in $\mathbb{Z}_2$!

Indeed we can just try dividing by some different polynomials as suggest in the comments and answers on this question. Since $p(x)$ is of degree 6, it must either be prime or have a divisor of degree three or smaller. We can also observe that any divisor must have 1 as the constant term.

Trying to divide $p(x)$ by $x + 1$, $x^2 + 1$, $x^2 + x + 1$, $x^3 + 1$ does not result into a factorization. However, when we try to divide by $x^3 + x + 1$ (and not forgetting to do arithmetic modulo 2!), we obtain $$ x^8 - x = x(x - 1)(x^3 + x + 1)(x^3 + x^2 + 1) $$

Now we adjoin a root $\xi$ of $x^3 + x + 1$. The field that is formed this way is isomorphic to $\mathbb{Z}_2/(x^3 + x + 1)$. That is, our field is isomorphic to polynomials of order < 3 (since we have $\xi^3 = \xi + 1$) with coefficients in $\mathbb{Z}_2$. So we have the elements $0, 1, \xi, \xi^2, \xi^3 = \xi + 1, \xi^4 = \xi^2 + \xi, \xi^5 = \xi^2 + \xi + 1, \xi^6 = \xi^2 + 1$.

Now working out the multiplication and addition tables is trivial. As Jyrki Lahtonen has noted, it is easier to use the form $\xi^k$ for multiplication, and to use the form $a \xi^2 + b \xi + c$ for addition. Of course, we have $\xi^a + \xi^b = \xi^{a + b}$ (where $a + b$ might need to be reduced modulo 7, since $\xi ^ 7 = 1$).

The addition table is as follows (I left out the zero polynomial to keep the size of the table manageable):

$$\begin{matrix} + & 1 & \xi & \xi^2 & \xi + 1 & \xi^2 + \xi & \xi^2 + \xi + 1 & \xi^2 + 1\\ 1 & 0 & \xi + 1 & \xi^2 + 1 & \xi & \xi^2 + \xi + 1 & \xi^2 + \xi & \xi^2 \\ \xi & \xi + 1 & 0 & \xi^2 + \xi & 1 & \xi^2 & \xi^2 + 1 & \xi^2 + \xi + 1 \\ \xi^2 & \xi^2 + 1 & \xi^2 + \xi & 0 & \xi^2 + \xi + 1 & \xi & \xi + 1 & 1 \\ \xi + 1 & \xi & 1 & \xi^2 + \xi + 1 & 0 & \xi^2 + 1 & \xi^2 & \xi^2 + \xi \\ \xi^2 + \xi & \xi^2 + \xi + 1 & \xi^2 & \xi & \xi^2 + 1 & 0 & 1 & \xi + 1 \\ \xi^2 + \xi + 1 & \xi^2 + \xi & \xi^2 + 1 & \xi + 1 & \xi^2 & 1 & 0 & \xi \\ \xi^2 + 1 & \xi^2 & \xi^2 + \xi + 1 & 1 & \xi^2 + \xi & \xi + 1 & \xi & 0 \\ \end{matrix}$$