deletions | additions       diff --git a/section_Multivariate_polynomials_on_bits__.tex b/section_Multivariate_polynomials_on_bits__.tex  index 50b13fa..6b9a367 100644  --- a/section_Multivariate_polynomials_on_bits__.tex  +++ b/section_Multivariate_polynomials_on_bits__.tex 
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Then, we can see that, in $\Fb[x,y]$, $x^3+0y^{12}=x^3=x^3+0xy^{32}$ and in $\Fb[x_1,x_2,x_3,x_4,x_5]$,   $x_5x_4-x_2=x_5x_4+0x_3-x_2$.  \\  As seen in section \ref{Sec:Polynomials} for univariate polynomials, the sum and sum,  the product and the powers  of polynomials in $\Fb[x_1,...x_n]$ are defined as for multivariate polynomials over $\RR$, only taking into account that their coefficients are bits.  Since we can multiply multivariate polynomials, we can raise them at a power $m$ as well, with follow  the usual meaning. rules.  \\  If, for example, we take the polynomial $f(x_1,x_2,x_3,x_4)=x_3x_1+x_4+1\in $f=x_3x_1+x_4+1\in  \Fb[x_1,x_2,x_3,x_4]$, we have that $$f(x_1,x_2,x_3,x_4)^2=(x_3x_1+x_4+1)^2=f(x_1,x_2,x_3,x_4)f(x_1,x_2,x_3,x_4)=(x_3x_1+x_4+1)(x_3x_1+x_4+1)$$ $$f^2=(x_3x_1+x_4+1)^2=f\cdot f=(x_3x_1+x_4+1)(x_3x_1+x_4+1)$$  $$x_3^2x_1^2+x_4^2+1 = (x_3x_1)^2+(x_4)^2+1^2.$$  As in the univariate case, raising a multivariate polynomial in $\Fb[x_1,...,x_n]$ to the power 2 is the same  as raising to the power 2 its monomials and summing them. \begin{Exercise}  In $\Fb[x_1,x_2,x_3,x_4,x_5]$ compute  \begin{itemize}  \item $(x_1^12x_5+x_3^3+x_4+x_2^6x_3)+(x_4^5+x_3^3+x_2^6x_3+x_5^6x_1)$  \item $(x_1^{10}x_2^6+x_3x_4x_5+x_2^5x_5)(x_1^3x_4+x_2x_5+x_3^5)$  \item $(x_1+x_5)^3$  \end{itemize}  \end{Exercise}