Discrete logarithm tables for the fields $\Bbb{F}_8$ and $\Bbb{F}_{16}$.

Solution 1:

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 $\Bbb{F}_{256}$