Construct a finite field of 128 elements.

Question

By using Galois' Theorem I know I need to find an irreducible polynomial of degree 7 in $\mathbb{Z}_2$. However I'm not sure how to find one efficiently, without showing that it is not the product of a polynomial of degree 5 and another of degree 2, or one of degree 3 and another of degree 4.

Answer 1

I’m hoping that you know how to raise something to a ridiculously high power by means of ridiculously few multiplications.

Well, you’re looking for an irreducible septic polynomial $f(x)$ over $\Bbb F_2$, the field with two elements. There are eighteen of them, not so many when you consider how many polynomials over $\Bbb F_2$ of degree seven there are. Nonetheless, I recommend flailing about and trying random polynomials that may work.

How do you tell that your ring $K_f=\Bbb F_2[x]/\bigl(f(x)\bigr)$ really is a field? As long as your chosen $f$ has no repeated factors, this $K_f$ will be a direct sum of fields. And if $f$ is reducible, there will be at least two of these, all rather smaller than $\Bbb F_{128}$, and certainly no element in any such thing will be of order $127$, a property that all the elements of $\Bbb F_{128}$, except for $0$ and $1$, will have.

Thus, if any element of $K_f$ has order $127$, then $K_f$ is a field. For instance, if I take $f(x)=x^7+x^2+1$, then $x^{127}=x^6+x^4+1$. Bad guess! But try Darsen’s guess, from an earlier answer, now deleted, of $f(x)=x^7+x+1$, really the first guess you ought to examine. Lo and behold!, you get $x^{127}=1$, and just to check, $(1+x^2+x^5)^{127}=1$ as well, as must be.

So there you are, with a nice septic irreducible polynomial over $\Bbb F_2$ .

Answer 2

As Lubin explained, a bit of experimenting will take you the distance sooner rather than later. Shaking my bag of tricks and examining what falls out.

We need an irreducible septic $f(x)\in\Bbb{F}_2[x]$. Clearly $f(x)$ must contain the terms $x^7$ and $1$ for otherwise $x\mid f(x)$. The binomial $x^7+1$ is divisible by $x+1$ so we need a trinomial at the very least. Let's try $$ f(x)=x^7+x^a+1 $$ for some $a\in\{1,2,\ldots,6\}$.

1. Such a trinomial has no linear factors for $f(0)=f(1)=1$.
2. Could it have a quadratic factor? The sole irreducible quadratic, $x^2+x+1$, has a primitive third root $\omega$ as a root. We have $$f(\omega)=\omega^7+\omega^2+1=\omega+\omega^a+1.$$ We see that this is zero if and only $a\equiv2\pmod3$. This rules out $a=2$ and $a=5$.
3. But all the other choices for $a$ actually work! This is because from the construction of the field $\Bbb{F}_8$ we know that all the irreducible cubic polynomials of $\Bbb{F}_2[x]$ are factors of $x^8-x=x(x^7+1)$ and hence also of $x^7+1$. So if $p(x)$ is an irreducible cubic, we have $$f(x)\equiv x^a\pmod{p(x)}.$$ This immediately tells us that $p(x)\nmid f(x)$ (any common factor would be a factor of $x^a$ also etc).

So $$ x^7+x+1,x^7+x^3+1, x^7+x^4+1\quad\text{and}\quad x^7+x^6+1 $$ all work.

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A few concluding remarks for extra credit:

• You may have spotted that the irreducible septic trinomials come in reciprocal pairs. If $f(x)$ is an irreducible septic then so is its reciprocal $\tilde{f}(x)=x^7f(\frac1x)$. I invite you to prove this in a more general case as a fun exercise. Basically if $p(x)$ is the minimal polynomial of an element $\alpha\neq0$, algebraic over a field $K$, then $\tilde{p}(x)$ is the minimal polynomial of $1/\alpha$.
• Lubin hinted at repeated squaring. We can do it here quickly for example with the choice $a=1$. So let $\beta$ be a zero of $x^7+x+1$. Therefore $\beta^7+\beta+1=0$, implying $\beta^7=\beta+1$ and thus also $\beta^8=\beta^2+\beta$. Then start squaring. Remember Freshman's dream, $(a+b)^2=a^2+b^2$ when in characteristic two. Behold. $$ \begin{aligned} \beta^{16}&=\beta^4+\beta^2\\ \beta^{32}&=\beta^8+\beta^4=\beta^4+\beta^2+\beta\\ \beta^{64}&=\beta^8+\beta^4+\beta^2=\beta^4+2\beta^2+\beta=\beta^4+\beta\\ \beta^{128}&=\beta^8+\beta^2=2\beta^2+\beta=\beta. \end{aligned} $$ So $\beta$ is an element of $\Bbb{F}_{128}$. As it is not an element of the prime field and the extenstion degree $7$ is a prime, we can deduce that $\Bbb{F}_{128}=\Bbb{F}_2(\beta)$.
• Oh, one last thing occurred to me while looking at that last calculation. Namely, it may remind you of the calculation of writing consecutive powers $\alpha^i$, $i=1,2,\ldots,7$, where $\alpha$ is a zero of $x^3+x+1$, see here. You may have done that as an exercise when constructing the field $\Bbb{F}_8$. Notice how similar it is. The difference is that as we there reduce powers $\alpha^j$ to linear combinations of $\alpha^0,\alpha^1$ and $\alpha^2$, here we, in each step, reduce $\beta^{2^j}$ to the exact same linear combination of $\beta^{2^0}$, $\beta^{2^1}$ and $\beta^{2^2}$. This is no coincidence! It reflects the relation of the polynomial $$m(x)=x^3+x+1=\sum_j b_jx^j$$ and its linearized associate $$L_m(x)=\sum_j b_j x^{2^j}=x^8+x^2+x.$$ This may be beyond the first course in finite fields, but because $3,7$ and $127$ are all Mersenne primes, it follows that we can repeat this process and show that $L_{L_m/x}(x)/x=x^{127}+x+1$ is also irreducible over $\Bbb{F}_2$. See this old answer of mine for more details. This is further developed when viewing extension fields of $\Bbb{F}_2$ as modules over the polynomial ring $\Bbb{F}_2[\tau]$ with the indeterminate $\tau$ acting via the Frobenius automorphism.