Marilyn Cameron

2022-10-25

Modular Polynomial Arithmetic
For $p=3$ and $n=2$, the ${3}^{2}=9$ polynomials in the set are
$\begin{array}{|ccc|}\hline 0& x& 2x\\ 1& x+1& 2x+1\\ 2& x+2& 2x+2\\ \hline\end{array}$
For $p=2$ and $n=3$, the ${2}^{3}=8$ polynomials in the set are
$\begin{array}{|ccc|}\hline 0& x+1& {x}^{2}+x\\ 1& {x}^{2}& {x}^{2}+x+1\\ x& {x}^{2}+1\\ \hline\end{array}$
For the first example right, from what I understand it is the polynomial whose coefficient are from ${Z}_{p},\left\{0,1,...,p-1\right\}$. So given $p=3$, I can get 0,1,2. And if given $p=2$, I can get 0,1.
But I not sure how those x are derived. Any ideas?

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Expert

Step 1
If $n=2$, the polynomial has 2 terms, the x and constant term. If $x=3$, the polynomial has three terms, ${x}^{2}$, x, and the constant term.
Step 2
So the highest degree of the polynomial is $n-1$.

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gasavasiv

Expert

Step 1
It seems to be the list of all polynomials of degree less than 2, with coefficients in Z/3Z in the first case; of all polynomials of degree less than 3, with coefficients in Z/2Z in the second case.
Indeed, for the first case, a polynomial of degree less than 2 has the form ${a}_{0}+{a}_{1}x$. Plug in the different possible values for the pairs $\left({a}_{0},{a}_{1}\right)$, getting
$\begin{array}{ccc}\left(0,0\right)\to 0& \left(0,1\right)\to x& \left(0,2\right)\to 2x\\ \left(1,0\right)\to 1& \left(1,1\right)\to 1+x& \left(1,2\right)\to 1+2x\\ \left(2,0\right)\to 2& \left(2,1\right)\to 2+x& \left(2,2\right)\to 2+2x\end{array}$
Step 2
The second case goe along the same lines:
$\begin{array}{cccc}& \left(0,0,1\right)& & \left(0,1,1\right)\\ & {x}^{2}& & x+{x}^{2}\\ \left(0,0,0\right)& & \left(0,1,0\right)\\ 0& & x\\ & \left(1,0,1\right)& & \left(1,1,1\right)\\ & 1+x& & 1+x+{x}^{2}\\ \left(1,0,0\right)& & \left(1,1,0\right)\\ 1& & 1+x\end{array}$

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