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\item[ii)] $A\setminus\{0\}$ is an abelian group w.r.t. the product of polynomials;  \item[iii)] $(f+g)\cdot h=fh+gh$, for any $f,g,h \in A$.  \end{itemize}  We have already observed that $g$ Since $A$  is irreducible (see Exercise \ref{FunzPoly}); now formed by polynomials, properties i) and iii) are obvious.  As regards ii),  we can need only to  proveit also showing  that each nonzero element in $A$ has an inverse: \begin{itemize}  \item $1\cdot 1=1$ so $1$ is 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 $x+1$ are mutually inverse. \underline{(x+1)}+x=x+1+x=1$, so in $A$,   $$\frac{1}{x+1}=x,\, \frac{1}{x}=x+1.$$  \end{itemize}  We have already observed that $g$ is irreducible (see Exercise \ref{FunzPoly}); now we can prove it also showing that  Now consider the polynomial $h=x^2+1$. We can easily note that it is reducible, since $h=(x+1)^2$.  The set of reminders modulo $h$ is the same $A =\{0,1,x,x+1\}$ as before, but the relation imposed on $A$ by $h$ does not make it a 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,\,