I am mainly interested in the fields $\mathrm{GF}(2^n)$, but the question can be asked for any prime.

We can write out each element $x\in\mathrm{GF}(2^n)$ in base $2$ and note that its additive group combined with multiplication by elements of $\mathrm{GF}(2)$ is isomorphic to the vector space $\left(\mathbb{Z}/(2\mathbb{Z})\right)^n$. Let $v:\mathrm{GF}(2^n)\to\left(\mathbb{Z}/(2\mathbb{Z})\right)^n$ stand for this "vectorisation" operation.

Linear maps on $\left(\mathbb{Z}/(2\mathbb{Z})\right)^n$ may be represented by $n\times n$, $\{0,1\}$-valued matrices.

Since field multiplication is linear for any $x\in\mathrm{GF}(2^n)$ there is a matrix $M_x$ such that for all $y\in\mathrm{GF}(2^n)$ \begin{align} M_x v(y) = v(x\cdot y), \end{align}

There are, however $2^{n\times n}$ matrices and only $2^{n}$ field elements, so the question is what can we say about the structure of the set of matrices $\{M_x \mid x\in \mathrm{GF}(2^n)\}$ as a subset of the full set of matrices?

Loosely speaking - if I give you a matrix then how can you tell if it represents a field element?

  • $\begingroup$ The matrix associated to a multiplication in a field is invertible (unless you multiply by $0$). $\endgroup$ – Severin Schraven Jul 8 '20 at 12:09
  • $\begingroup$ I wonder if it's easier if instead of asking which matrices are of some special class, you ask what the matrix looks like for a member of that special class? $\endgroup$ – Michael Hardy Jul 8 '20 at 14:38

We note that any finite field $GF(p^n)$ can be presented in the form $GF(p^n) = \Bbb Z_p[x]/\langle q(x)\rangle $, where $\Bbb Z_p = \Bbb Z/p\Bbb Z$ and $q$ is an irreducible polynomial of degree $n$. Relative to the basis $\{1,x,\dots,x^{n-1}\}$, we find that the matrix $M_x$ corresponding to multiplication by (the distinguished indeterminate) $x$ is given by the companion matrix $C_q$ of $q$. It follows that a matrix $M$ corresponds to a field element if and only if there exists there exists a polynomial $f$ for which $M = f(C_q)$.

Because the matrix $C_q$ is non-derogatory, it turns out that there is such a polynomial $f$ if and only if $C_q M = MC_q$ (cf. Horn and Johnson Matrix Analysis theorem

We can get another perspective on this if we take the elements of the matrix to themselves be elements of $GF(p^n)$. Any irreducible polynomial over $\Bbb Z_p$ splits into distinct linear factors over its splitting field. It follows the polynomial $q$ splits into linear factors with $$ q(x) = (x - a_1)\cdots (x - a_n), \quad a_i \in GF(p^n). $$ It follows that the matrix $C_q$ is diagonalizable $GF(p^n)$. A matrix $M$ will commute with $\operatorname{diag}(a_1,\dots,a_n)$ if and only if it is also diagonal. So, given a matrix, it suffices to change bases and check whether the transformed matrix has the required block structure.

More specifically, the eigenvectors of $C_p$ correspond to the polynomials of multiplication by $x$, which are $$ q_i(x) = q(x)/(x - a_i), i = 1,\dots,n. $$ In particular, we can see that $xq_i(x) = a_i q_i(x)$, modulo $q(x)$. $M$ will correspond to multiplication by an element of $GF(p^n)$ if and only if $p_i(x)$ is an eigenvector of (the operator over $\Bbb Z_p/q(x)$ corresponding to) $M$ for all $i$.

  • $\begingroup$ @J Thanks!${}{}{}$ $\endgroup$ – Ben Grossmann Jul 8 '20 at 12:34
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    $\begingroup$ I'm not sure if it's possible for an irreducible polynomial over a finite field to have repeated roots (which is a possibility that I account for in my answer). If anybody has an example of such a polynomial or knows that no such polynomial exists, please let me know. $\endgroup$ – Ben Grossmann Jul 8 '20 at 12:46
  • $\begingroup$ That's a really interesting point. I know it sounds crazy, but you could ask that as a question. :) $\endgroup$ – John Hughes Jul 8 '20 at 13:05
  • $\begingroup$ @J Fair point.${}$ $\endgroup$ – Ben Grossmann Jul 8 '20 at 13:24
  • $\begingroup$ @J Found this in the suggestions as I started typing the question; updating accordingly. $\endgroup$ – Ben Grossmann Jul 8 '20 at 13:28

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