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Massimiliano Sala edited section_Bytes_The_polynomials_in__.tex
about 6 years ago
Commit id: 4113e577b71f8f1eaee13465470bdf156b6af5da
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...
x^5+x \,=\, x^2(\underline{x^3})+x \,=^{\mathrm{substitution}}\, x^2(\underline{x+1})+x\,=\,
\underline{x^3}+x^2+x=^{\mathrm{substitution}} \underline{x+1}+x^2+x \,=\, x^2+1 \,.
\]
One The crucial observation here is that we can obtain the same result
by dividing the polynomial
$x^4+x^2$ $x^5+x$ by the polynomial
$x^3+x+1$. The $g=x^3+x+1$: the remainder is
$x$.
Hence $x$ (see Exercise \label{QuoRemEx}).\\
Indeed, we can consider a set $T$, which collects the remainders of the
divisions of all polynomials
defined in
$F_2[x]$ by $g$. It is easy to see that $T$ contains $S$, because when
we divide a polynomial $f$ of degree less than $3$ by $g$, the
set
\[
\Fb[x], \mbox{ where }x^3+x+1=0
\]
are the polynomials
\[
0, remainder is $f$ itself
(see Example \ref{QuotRemStrange}), so
$$
\{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 x^2+x+1 \} subset T
$$
On the other hand, if we consider
that the generic any polynomial
$f \notin S$, this has $\deg(f)\geq 3$ and
when we divide it by $g$, we will get a remainder of degree
$2$ is
\[
a_2 x^2+ a_1 x+ a_0,
\]
where $a_2,a_1,a_0\in \Fb$. strictly less than $\deg(g)=3$,
and so $T$ can only contain polynomials of degree at most $2$, therefore
$$
T=S=\{0, 1, x, x+1, x^2, x^2+x, x^2+1, x^2+x+1 \}
$$
\begin{Definition}\label{RemPol} We
define can perform this construction in general. Let $g$ be in $F_2[x]$, we consider the set
$A$ of polynomials
$A$ as
\[
\Fb[x], \mbox{ where }p=0
\]
\end{Definition} that collects all remainders of the division by $g$
$$
A \,=\, \{ \mathrm{remainders by } g \} \,.
$$
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}