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bits5.tex \section{Bytes}  The polynomials in $\Fb[x]$ are infinite. 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  \[  x^4+x^2=x(x^3)+x^2=x(x+1)+x^2=x^2+x+x^2=x.  \]  One can obtain the same result dividing the polynomial $x^4+x^2$ by the polynomial $x^3+x+1$. The remainder is $x$.  Hence the polynomials defined in the set  \[  \Fb[x], \mbox{ where }x^3+x+1=0  \]  are the polynomials  \[  0, 1, x, x+1, x^2, x^2+x, x^2+1, x^2+x+1.   \]  These polynomials are $8=2^3$. This number can be obtained directly if we consider that the generic polynomial of degree $2$ is  \[  a_2 x^2+ a_1 x+ a_0,  \]  where $a_2,a_1,a_0\in \Fb$.  \begin{Definition}\label{RemPol}  We define the set of polynomials $A$ as  \[  \Fb[x], \mbox{ where }p=0  \]  \end{Definition}  The finite set $A$ inherits the operations of sum and product defined in the polynomial ring $\Fb[x]$. Moreover the following fact holds  \begin{Theorem}  The finite set $A$ is a field if and only if $p$ is an irreducible polynomial.  \end{Theorem}  \begin{Example}  \textbf{Nota (M)}: non capisco cosa intendi comunicare con l'esempio...  Let $A$ defined by $\Fb[x]$ where we fix the relation $x^2+x+1=0$, that is $x^2=x+1$. The finite set $A$ is $\{0,1,x,x+1\}$. We observed that $x^2+x+1$ is irreducible. And in fact all the elements different from $0$ in $A$ have an inverse: the inverse of $1$ is $1$; the inverse of $x$ is $x+1$ (try it); the inverse of $x+1$ is $x$.  %In fact $x(x+1)=x^2+x=x+1+x=1$.  Now consider the reducible polynomial $x^2$. Then $A$ is defined by $\Fb[x]$ where $x^2=0$. Even in this case the set $A$ is $\{0,1,x,x+1\}$, but this time the inverse of $x$ does not exist.  In fact multiplying $x$ by each element in $\{0,1,x,x+1\}$ we have either $0$ or $x$. In fact  \[  x\cdot 0=0, x\cdot 1 =x, x\cdot x=x^2=0, x\cdot (x+1)=x^2+x=x.   \]  That is $x$ is not invertible.  %In algebra we call $x$ a $0$-divisor.  \end{Example}