Marilyn Cameron

Answered

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 ${Z}_{p},\{0,1,...,p-1\}$. 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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Answer & Explanation

Tania Alvarado

Expert

2022-10-26Added 15 answers

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 $n-1$.

gasavasiv

Expert

2022-10-27Added 3 answers

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 $({a}_{0},{a}_{1})$, getting

$$\begin{array}{ccc}(0,0)\to {0}& (0,1)\to {x}& (0,2)\to {2x}\\ (1,0)\to {1}& (1,1)\to {1+x}& (1,2)\to {1+2x}\\ (2,0)\to {2}& (2,1)\to {2+x}& (2,2)\to {2+2x}\end{array}$$

Step 2

The second case goe along the same lines:

$$\begin{array}{cccc}& (0,0,1)& & (0,1,1)\\ & {{x}^{2}}& & {x+{x}^{2}}\\ (0,0,0)& & (0,1,0)\\ {0}& & {x}\\ & (1,0,1)& & (1,1,1)\\ & {1+x}& & {1+x+{x}^{2}}\\ (1,0,0)& & (1,1,0)\\ {1}& & {1+x}\end{array}$$

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