$\alpha^{2^{m} + 1}$ is a generator of $\operatorname{GF}(2^m)$?

Question

This is from a paper (Partitions in the S-Box of Streebog and Kuznyechik) about S-Boxes:

Let $\operatorname{GF}(2^{2m}) = \mathbb{F}_2[X]/p(X)$ be a finite field of even degree defined by a primitive polynomial $p$. The multiplicative subgroup $\operatorname{GF}(2^{2m})^*$ is cyclic and generated by $\alpha$ which is such that $p(\alpha) = 0$.

In this context, $\alpha^{2^{m} + 1}$ is a generator of the multiplicative subgroup of the subfield $\operatorname{GF}(2^m)$

I don't understand why it holds that $\alpha^{2^{m} + 1}$ is a generator of the subfield?

score 6 · Accepted Answer · answered Jun 23 '19 at 16:25

Since you haven't heard of splitting fields, here is an argument based on more elementary group theory.

Let $G$ denote a cyclic group. Then, it is easy to show (or you might already know it) that if $\alpha \in G$ is an element of order $n$ (we write $\mathsf{ord}(\alpha)=n$ to denote this), then $$\mathsf{ord}(\alpha^k)= \frac{\mathsf{ord}(\alpha)}{\gcd(\mathsf{ord}(\alpha),k)} = \frac{n}{\gcd(n,k)}.$$ Choose $G = \text{GF}(2^{2m})^*$ whose generator $\alpha$ is of order $n = 2^{2m}-1$, observe that $2^{2m}-1 = (2^m-1)(2^m+1)$, and deduce that $$\mathsf{ord}(\alpha^{2^{m}+1}) = \frac{2^{2m}-1}{\gcd(2^{2m}-1,2^{m}+1)} = \frac{(2^m-1)(2^m+1)}{2^{m}+1} = 2^m-1.$$ Thus, $\alpha^{2^{m}+1}$ generates the unique cyclic subgroup (of order $2^m-1$) of $\text{GF}(2^{2m})^*$ which itself is of order $2^{2m}-1$. Now, $\text{GF}(2^{m})$ is a subfield of $\text{GF}(2^{2m})$ and so this subgroup must be $\text{GF}(2^{m})^*$.

Beautiful explanation, thanks! Could you please also explain to me why $\operatorname{GF}(2^{2m}) \to \operatorname{GF}(2^{m}): \operatorname{Tr}_m(x) = x^{2^{m}} \oplus x$ holds? — winklerrr, Jun 24 '19 at 20:39
@winklerrr That is the definition of the relative trace from $GF(2^{2m})$ to $GF(2^m)$. No other explanation is needed. — Jyrki Lahtonen, Jul 12 '19 at 03:29

score 0 · Answer 2 · edited Jun 23 '19 at 16:26

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Hint: Remember that $\operatorname{GF}(p^m)$ is the splitting field of $x^{p^m} - x$ over $\operatorname{GF}(p)$. If $m \mid n$ so that $\operatorname{GF}(p^n)$ is a field extension of $\operatorname{GF}(p^m)$, then for $u \in \operatorname{GF}(p^n)$, we have $u \in \operatorname{GF}(p^m)$ if and only if $u$ is a root of $x^{p^m} - x$.

Can you use these to show that $\alpha^{2^m + 1}$ is a unit in $\operatorname{GF}(p^k)$ and has maximal order $2^m - 1$?

edited Jun 23 '19 at 16:26

Dilip Sarwate

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answered Jun 21 '19 at 23:10

Squeamish Ossifrage

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Never heard of "splitting field" and I don't understand what $x^{p^m} - x$ means and what it has to do with the generator? Does it hold that $x^{p^m} = x$? So $x^{p^m} - x = 0$? But why over $\operatorname{GF}(p)$? Could you please show me why $\alpha^{2m + 1}$ is a unit in $\operatorname{GF}(p^k)$? – winklerrr Jun 22 '19 at 09:41

$\alpha^{2^{m} + 1}$ is a generator of $\operatorname{GF}(2^m)$?

2 Answers2