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The smallest non-trivial finite field of characteristic two is $$ \Bbb{F}_4=\{0,1,\beta,\beta+1=\beta^2\}, $$ where $\beta$ and $\beta+1$ are primitive cubic roots of unity, and zeros of the polynomial $x^2+x+1$. Here the multiplication table is given once we know how to write the non-zero elements as powers of $\beta$. Extend the idea to fields of eight and sixteen elements.

Those fields can be constructed as $$ \Bbb{F}_8=\Bbb{F}_2[\alpha], \quad\text{and}\quad \Bbb{F}_{16}=\Bbb{F}_2[\gamma], $$ where $\alpha$ has minimal polynomial $x^3+x+1$, and $\gamma$ has minimal polynomial $x^4+x+1$, both irreducible in $\Bbb{F}_2[x]$.

Task:

Calculate the tables for base $\alpha$ discrete logarithm of $\Bbb{F}_8$ and base $\gamma$ discrete logarithm of $\Bbb{F}_{16}$.

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  • $\begingroup$ A related meta-thread. It may be best to discuss eventual objections to this thread there. $\endgroup$ – Jyrki Lahtonen Dec 3 '13 at 19:01
  • $\begingroup$ All: The purpose of this Q/A pair is to give an on-site explanation of the use of discrete logarithms in certain finite fields of charactertistic two, and examples of their use. I refer to this question in the tag wiki. The earlier versions of using discrete log are buried in answers to coding theoretical questions and such. My hope is that a standalone Q/A will be more accessible to the interested readers. $\endgroup$ – Jyrki Lahtonen Dec 4 '13 at 12:31
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    $\begingroup$ Jyrki, have you seen this? link.springer.com/chapter/10.1007%2F978-3-642-55220-5_1 $\endgroup$ – David Holden Oct 10 '14 at 14:58
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A (base-$g$) discrete logarithm of a finite field $\Bbb{F}_q$, is a function $$ \log_g:\Bbb{F}_q^*\to\Bbb{Z}_{q-1} $$ defined via the equivalence $g^j=x\Leftrightarrow \log_g(x)=j$. For this to be well-defined it is imperative that $g$ is a primitive element, i.e. a generator of $\Bbb{F}_q^*$, and that the domain of $\log_g$ is the residue class ring of integer modulo $q-1$, as $g^{q-1}=g^0=1$.

It immediately follows that the discrete logarithm satisfies the familiar rules $$ \begin{aligned} \log_g(x\cdot y)&=\log_g(x)+\log_g(y),\\ \log_g(x^n)&=n\cdot\log_g(x) \end{aligned} $$ for all elements $x,y\in \Bbb{F}_q^*$ and all integers $n$. The arithmetic on the r.h.s. is that of the ring $\Bbb{Z}_{q-1}$.


It is known that when $q=8$, a zero $\alpha$ of $x^3+x+1$ generates $\Bbb{F}_8^*$. This is proven by the following calculation, where we repeatedly use the fact that we are working in characteristic two, and that we have the relation $\alpha^3=\alpha+1$. $$ \eqalign{ \alpha^0&=&&=1,\\ \alpha^1&=&&=\alpha,\\ \alpha^2&=&&=\alpha^2,\\ \alpha^3&=&&=1+\alpha,\\ \alpha^4&=&\alpha\cdot\alpha^3=\alpha(1+\alpha)&=\alpha+\alpha^2,\\ \alpha^5&=&\alpha\cdot\alpha^4=\alpha(\alpha+\alpha^2)=\alpha^2+\alpha^3=\alpha^2+(1+\alpha)&=1+\alpha+\alpha^2,\\ \alpha^6&=&\alpha\cdot\alpha^5=\alpha(1+\alpha+\alpha^2)=\alpha+\alpha^2+\alpha^3= \alpha+\alpha^2+(1+\alpha)&=1+\alpha^2,\\ \alpha^7&=&\alpha\cdot\alpha^6=\alpha(1+\alpha^2)=\alpha+\alpha^3=\alpha+(1+\alpha)&=1. }$$

We see from the end results in the last column that all the non-zero quadratic polynomials evaluated at $\alpha$ appear. This is yet another confirmation of the fact that $\alpha$ is a primitive element.

The discrete logarithm is used to replace the cumbersome multiplication (and raising to an integer power) of the field with more familiar integer arithmetic. Exactly like the old-timers used logarithm tables to replace the error-prone multiplication with the easier addition.

For example $$ (1+\alpha)(1+\alpha+\alpha^2)=\alpha^3\cdot\alpha^5=\alpha^8=\alpha^7\cdot\alpha=\alpha. $$ Note that both the base-$\alpha$ discrete logarithms and its inverse mapping are needed. I generate such a table as a part of the program initialization, whenever I carry out extensive computer-aided calculations involving a finite field. The above table gives the discrete logarithm when read from right to left, and the inverse mapping (that we actually produced above) when read from left to right.


Similarly with $q=16$ we use $\gamma$, a zero of $x^4+x+1$. This time the table looks like $$ \begin{aligned} \gamma^0&=&1\\ \gamma^1&=&\gamma\\ \gamma^2&=&\gamma^2\\ \gamma^3&=&\gamma^3\\ \gamma^4&=&\gamma+1\\ \gamma^5&=\gamma(\gamma+1)=&\gamma^2+\gamma\\ \gamma^6&=\gamma(\gamma^2+\gamma)=&\gamma^3+\gamma^2\\ \gamma^7&=\gamma^4+\gamma^3=&\gamma^3+\gamma+1\\ \gamma^8&=(\gamma^4)^2=&\gamma^2+1\\ \gamma^9&=\gamma(\gamma^2+1)=&\gamma^3+\gamma\\ \gamma^{10}&=\gamma^4+\gamma^2=&\gamma^2+\gamma+1\\ \gamma^{11}&=&\gamma^3+\gamma^2+\gamma\\ \gamma^{12}&=\gamma^4+\gamma^3+\gamma^2=&\gamma^3+\gamma^2+\gamma+1\\ \gamma^{13}&=\gamma^4+\gamma^3+\gamma^2+\gamma=&\gamma^3+\gamma^2+1\\ \gamma^{14}&=\gamma^4+\gamma^3+\gamma=&\gamma^3+1\\ (\gamma^{15}&=\gamma^4+\gamma=&1) \end{aligned} $$

Thus for example $$ (\gamma^3+1)(\gamma^2+1)=\gamma^{14}\cdot\gamma^8=\gamma^{22}=\gamma^7=\gamma^3+\gamma+1. $$


As another example of the use of this table I want to discuss the problem of factorizing $x^4+x+1$ over $\Bbb{F}_4$. To that end we need to first identify a copy of $\Bbb{F}_4$ as a subfield of $\Bbb{F}_{16}$. We just saw that $\gamma$ is of order fifteen. Therefore $\gamma^5=\gamma^2+\gamma$ and $\gamma^{10}=\gamma^2+\gamma+1$ are third roots of unity. It is then trivial to check that we have a homomorphism of fields $\sigma:\Bbb{F}_4\to\Bbb{F}_{16}$ given by $\sigma(\beta)=\gamma^5$. Note that composing this (from either end) by the Frobenius automorphism gives an alternative embedding $\beta\mapsto \gamma^{10}$.

Basic Galois theory tells us that $$ x^4+x+1=(x-\gamma)(x-\gamma^2)(x-\gamma^4)(x-\gamma^8) $$ as we get the other roots by repeatedly applying the Frobenius automorphism $F:x\mapsto x^2$. Here we see that the factor $$ (x-\gamma)(x-\gamma^4)=x^2+x(\gamma+\gamma^4)+\gamma^5=x^2+x+\gamma^5 $$ is stable under the automorphism $F^2$, and thus (as we also see directly!) has its coefficients in the subfield $\sigma(\Bbb{F}_4)$. The same holds for the remaining factor $$ (x-\gamma^2)(x-\gamma^8)=x^2+x(\gamma^2+\gamma^8)+\gamma^{10}=x^2+x+\gamma^{10}. $$ Pulling back the effect of $\sigma$ we get the desired factorization $$ x^4+x+1=(x^2+x+\beta)(x^2+x+\beta+1) $$ in $\Bbb{F}_4[x]$.


Here is a local version of similar tables for $\Bbb{F}_{256}$

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