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\end{Theorem}  We will denote $A=\FF_{2^n}$.\\  \begin{Example}  \textbf{Nota (M)}: non capisco cosa intendi comunicare con l'esempio... Let $A$ defined by $\Fb[x]$ where we fix us consider the polynomial $g=x^2+x+1 \in \Fb[x]$ and define  therelation $x^2+x+1=0$, that is $x^2=x+1$. The finite  set $A$ is $\{0,1,x,x+1\}$. $  A \,=\, \{ \textrm{remainders by } g \} \,.  $  For this particular $g$, we have $A =\{0,1,x,x+1\}$.  We have already  observed that $x^2+x+1$ $g$  is irreducible. And in fact all the elements different from $0$ irreducible (see Exercise \ref{FunzPoly}); now we can prove it also showing that each nonzero element  in $A$ have has  an inverse: the inverse of \begin{itemize}  \item $1\cdot 1=1$ so  $1$ is$1$; the inverse of $x$ is $x+1$ (try it);  the inverse of itself;  \item $x(x+1)=\underline{x^2}+x=^{\mathrm{substitution}}\, \underline{(x+1)}+x=x+1+x=1$ so,$x$ and  $x+1$ is $x$.  %In fact $x(x+1)=x^2+x=x+1+x=1$. are mutually inverse.  \end{itemize}  Now consider thereducible  polynomial $x^2$. Then $A$ $h=x^2+1$. We can easily note that it  is defined by $\Fb[x]$ where $x^2=0$. Even in this case the reducible, since $h=(x+1)^2$.  The  set $A$ of reminders modulo $h$  is $\{0,1,x,x+1\}$, the same $A =\{0,1,x,x+1\}$ as before,  butthis time  the inverse of $x$ does not exist.  In fact multiplying $x$ relation imposed on $A$  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 $h$ does  not invertible.  %In algebra we call $x$ make it  a $0$-divisor. field. Indeed, $x+1\in A$ has no inverse. Indeed   $$(x+1)\cdot 0=0,\, (x+1)\cdot 1=x+1,\, (x+1)\cdot x=\underline{x^2}+x=^{\mathrm{substitution}}\, \underline{1}+x,\,   (x+1)(x+1)=\underline{x^2}+1=^{\mathrm{substitution}}\, \underline{1}+1=0.$$  \end{Example}