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Given a polynomial $f_1=a_1x+a_2x^2+\cdots+a_{p-1}x^{p-1}\in\mathbb{Z}[x]/(x^{p}-1)$, with prime $p$, we may generate the other $p-1$ polynomials: \begin{eqnarray*} f_2&=&a_1^2x^2+\cdots+a_{p-1}^2x^{2(p-1)}(\bmod x^{p}-1)\\ f_3&=&a_1^3x^3+\cdots+a_{p-1}^3x^{3(p-1)}(\bmod x^{p}-1)\\ &\cdots&\\ f_{p-1}&=&a_1^{p-1}x^{p-1}+\cdots+a_{p-1}^{p-1}x^{(p-1)(p-1)}(\bmod x^{p}-1) \end{eqnarray*}

Are $f_1,f_2,\cdots,f_{p-1}$ $\mathbb{Z}$-linearly dependent as elements of $\mathbb{Z}[x]/(x^p-1)$?

NOTE: I want to figure out whether $f_1,f_2,\cdots,f_{p-1}$ are $\mathbb{Z}$-linearly independent, i.e. whether there exists $\alpha_1,\alpha_2,\cdots,\alpha_{p-1}\in \mathbb{Z}$, with $\alpha_1\alpha_2\cdots \alpha_{p-1}\neq 0$, satisfying $$\alpha_1f_1+\alpha_2f_2+\cdots+\alpha_{p-1}f_{p-1}=0$$.

One may note that when putting $f_1,f_2,\cdots,f_{p-1}$ into the polynomial ring $\mathbb{Z}[x]/(x^{p}-1)$, they are of degree less than $p$ without constant term. Thus we may write these coefficients into a matrix (under the module operation, it is not a Vandermonde matrix). What are the properities of the matrix? Including its eigenvalues, its kernels and so on.

Has anybody ever studied this kind of matrices? What if I replace $\mathbb{Z}$ with a general ring $R$?

I think it is very interesting since it seems better than circulant matrix to design codes.

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  • $\begingroup$ They are certainly not l.i. over $\mathbb Z[x]/(x^p-1)$ because $f_2f_1-f_1f_2=0$, but probably you wanted to take the coefficients from $\mathbb Z$? $\endgroup$
    – Christian Remling
    Commented Oct 30, 2015 at 7:14
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    $\begingroup$ If $a_1=a_2=\dots=a_{p-1}$ they are linear dependent if $a_1\neq a_2=a_3=\dots=a_{p-1}$ they are linear independent. Are you looking for a criterion when they are dependent and when not? $\endgroup$
    – user35593
    Commented Oct 31, 2015 at 19:22
  • $\begingroup$ Could you tell me why they are linear independent when $a_1\neq a_2=a_3=\cdots=a_{p-1}$? Actually, I do care a lot about when they are dependent and when not, especially when considering a general ring $R$ instead of $\mathbb{Z}$.@user35593 $\endgroup$
    – Diane
    Commented Nov 2, 2015 at 1:44
  • $\begingroup$ Yes, I want to study that whether they are $\mathbb{Z}-$linearly independent. Or furthermore, $R$-independent if we consider a general ring $R$ instead of $\mathbb{Z}$. @ChristianRemling $\endgroup$
    – Diane
    Commented Nov 2, 2015 at 1:55

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