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\section{Bytes}  The polynomials in $\Fb[x]$ are infinite. However, if we bound the degree, we find a finite set, as for example   $$S:=\{p \in \Fb[x] \vert \deg(p) < 3\}=\{0,1,x,x+1,x^2,x^2+x, x^2+1,x^2+x+1\}.$$  This set contains $8$ polynomials; indeed, $f \in S$ if and only if, $f$ is of the form  $$f=a_0+a_1x+a_2x^2,\, a_0,a_1,a_2 \in\Fb$$  Since there are two choices for each coefficient $a_0,a_1,a_2 $ the polynomials are $2^3=8$.  If we take two polynomials $f,g $ in $ S$, then $f+g$ belongs to $S$ as well, since when we sum two polynomials, the degree cannot grow (see Exercise \ref{degree}), according to the rule in $\Fb[x]$  $$\deg(f+g)\leq \deg(f), \deg(g); \qquad \deg(f+g)<\deg(f),\deg(g) \iff \deg(f)=\deg(g). $$  We describe a way to make it finite and bounded by a certain degree through a ''polynomial relation". For example fix the relations to be $x^3=x+1$, or equivalently $x^3+x+1=0$, as in the example above. We can limit the number of elements in $\Fb[x]$ in this way: each time we find a monomial of degree greater than or equal to $3$ we substitute $x^3$ by $x+1$. At the end of this process we obtain a new polynomial of degree strictly less than $3$. For example \[  x^4+x^2=x(x^3)+x^2=x(x+1)+x^2=x^2+x+x^2=x.  \]