Basis over GF(2^5) with sum of its elements = 1

Question

I need to work over $GF(2^5)$ starting from $GF(2)$ and I would need a basis $\{a_0, a_1, \cdots , a_4\}$ over $GF(2^5)$ such that $a_0+a_1+\cdots +a_4=1$. Do you know how could i proceed?

Thank you!

Answer 1

As already discussed in comments, normal bases always have this property. If $\alpha\in GF(2^n)$ generates a normal basis, then $\{\alpha,\alpha^2,\alpha^4,\alpha^8,\ldots,\alpha^{2^{n-1}}\}$ is linearly independent over the prime field. But, the sum of those Galois conjugates is (the trace of $\alpha$) invariant under the Frobenius automorphism, hence an element of the prime field $GF(2)$, that is, either $0$ or $1$. The former case immediately shows that the conjugates don't form a linearly independent set, so the sum of the elements of any normal basis is automatically $=1$.

The following special property of the extension degree $n$ adds an IMHO interesting angle to this question. We shall see that the assumptions hold when $n=5$.

Proposition. Assume that the extension degree $n$ has the following properties:

• $n$ is an odd prime, and
• $2$ is a generator of the multiplicative group $\Bbb{Z}_n^*$. Then an element $\alpha\in GF(2^n)\setminus GF(2)$ generates a normal basis if and only if $$ tr(\alpha)=\alpha+\alpha^2+\alpha^4+\cdots+\alpha^{2^{n-1}}=1. $$

Proof. Let's view $GF(2^n)$ as a module over the polynomial ring $GF(2)[\tau]$, where the indeterminate $\tau$ acts via the Frobenius automorphism, i.e. $\tau\cdot x= x^2$ for all $x\in GF(2^n)$. Because $x^{2^n}=x$ for all $x\in GF(2^n)$, we see that all the elements are annihilated by $\tau^n-1$. If $x\in GF(2^n)$ we denote its annihilator by $$ \mathrm{Ann}(x)=\{f(\tau)\in GF(2)[\tau]\mid f(\tau)\cdot x=0\}. $$ The annihilator is always an ideal of $GF(2)[\tau]$. As the polynomial ring is a PID, the annihilator is generated by its lowest degree non-zero polynomial. In view of the above the generator must be a factor of $\tau^n-1$.

If $\alpha$ does not generate a normal basis, then there exists a non-trivial linear dependency relation involving the elements $\tau^i\cdot\alpha$, $i=0,1,\ldots,n-1$. This happens if and only if $Ann(\alpha)$ contains a non-zero polynomial of degree $<n$. In other words,

$\alpha\in GF(2^n)$ generates a normal basis if and only if $\mathrm{Ann}(\alpha)$ is generated by $\tau^n-1$.

This means that the question whether a proper factor of $\tau^n-1$ may be in the annihilator. In general the factorization of $\tau^n-1$, while well understood, can be quite complicated. My point is that the listed assumptions about $n$ are equivalent to the factorization $$ (\tau^n-1)=(\tau-1)(\tau^{n-1}+\tau^{n-2}\cdots+\tau+1) $$ having irreducible factors. If $n$ is even, then $\tau^n-1=(\tau^{n/2}-1)^2$ has other factors. If $n$ is not a prime, then the cyclotomic polynomial $\Phi_n(\tau)$ is a proper factor of the degree $n-1$ factor. And, last but not least, if $n$ is an odd prime, that last factor $\Phi_n(\tau)$ is irreducible if and only if $2$ is a generator of $\Bbb{Z}_n^*$.

All this means that in the present case $\alpha\in GF(2^n)$ generates a normal basis if and only if $(\tau-1)\cdot\alpha\neq0$ and $\phi_n(\tau)\cdot\alpha\neq0$. Let's check what these mean.

• We have $(\tau-1)\cdot\alpha=0$ if and only if $\alpha^2-\alpha=0$ if and only if $\alpha\in GF(2)$.
• We have $(\tau^{n-1}+\tau^{n-2}\cdots+\tau+1)\cdot\alpha=0$ if and only if $$\alpha^{2^{n-1}}+\alpha^{2^{n-2}}+\cdots+\alpha^2+\alpha=0$$ if and only if $tr(\alpha)=0$.

So for these special values of $n$ we only need to throw out elements of the prime field $Gf(2)$ and element with trace zero. The rest of them will yield normal bases. QED.

Examples.

• If $n=5$, then $2$ generates $\Bbb{Z}_5^*$ because $2^0=1$, $2^1=2$, $2^2=4$, $2^3=8\equiv3$ and all the non-zero residue classes modulo $5$ are powers of two.
• A similar calculation shows that the proposition also applies, when $n=11$ or $n=13$, but does not apply, when $n=7$ or $n=17$.
• It is widely believed that the condition is satisfied by infinitely many primes. This would be implied by Artin's conjecture on primite roots.

Answer 2

Find a basis $1,b_1,\ldots,b_4$ which contains $1$. Then let $a_0=1+b_1+\cdots+b_4$, $a_1=b_1,\ldots,a_4=b_4$.

Answer 3

The sums of the roots of the primitive polynomials $$x^5+x^4+x^3+x^2+1, \quad x^5+x^4+x^3+x+1, \quad x^5+x^4+x^2+x+1\tag{1}$$ in $\mathbb F_2[x]$ all equal $1$ (the coefficient of $x^4$ is $1$ in each case) and the five roots are a normal basis of $\mathbb F_{32}$ over $\mathbb F_2$. If you are using $x^5+x^2+1$ for constructing $\mathbb F_{32}$ and denoting the roots as $\alpha, \alpha^2, \alpha^4, \alpha^8, \alpha^{16}$, then these roots sum to $0$ as you have discovered. The roots of the polynomials in $(1)$ above are respectively

$x^5+x^4+x^3+x^2+1\colon \quad\alpha^3, \alpha^6, \alpha^{12}, \alpha^{24}, \alpha^{17}$

$x^5+x^4+x^3+x+1\colon \quad\alpha^{11}, \alpha^{22}, \alpha^{13}, \alpha^{26}, \alpha^{21}$

$x^5+x^4+x^2+x+1\colon \quad\alpha^5, \alpha^{10}, \alpha^{20}, \alpha^{9}, \alpha^{18}$

and each set is a basis that sums to $1$ as desired. The first set listed above is the one that the OP found for himself, as mentioned in his comment on his own question.