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where $a_2,a_1,a_0\in \Fb$.  \begin{Definition}\label{RemPol}  We define the set of polynomials $A$ as  \[  \Fb[x]_{p(x)}=\Fb[x], \Fb[x],  \mbox{ where }p(x)=0 \]  \end{Definition}  The finite set \ref{RemPol} $A$  inherits the operations of sum and product defined in the polynomial ring $\Fb[x]$. Moreover the following fact holds \begin{Theorem}  The finite set \[  \Fb[x]_{p(x)}  \] $A$  is a field if and only if $p(x)$ 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 of polynomials are $A$ is  $\{0,1,x,x+1\}$. The 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 $1$, $1$;  the inverse of $x$ is $x+1$. In $x+1$ (try it); the inverse of $x+1$ is $x$.  %In  fact $x(x+1)=x^2+x=x+1+x=1$. Let Now consider the reducible polynomial $x^2$. Then $A$ is defined by  $\Fb[x]$ where $x^2=0$. Even in this case the set of polynomials are $\{0,1,x,x+1\}$, but in this case 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 $x$ that is $x$ is not invertible.  \[  x\cdot 0=0, x\cdot 1 =x, x\cdot x=x^2=0, x\cdot (x+1)=x^2+x=x.