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#### cupofcoffee

##### New member

- Jun 5, 2013

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Let n belongs to N, let p be a prime number and let \(\displaystyle Z/p^n Z\)denote the ring of

integers modulo \(\displaystyle p^n\) under addition and multiplication modulo \(\displaystyle p^n\)

.Consider two polynomials \(\displaystyle f(x) = a_0 + a_1 x + a_2 x^2 +...a_n x^n\) and \(\displaystyle g(x)=b_0 + b_1 x + b_2 x^2 +...b_m x^m\),given the coefficients are in \(\displaystyle Z/p^nZ\).

\(\displaystyle f(x)*g(x)=0\). Show that \(\displaystyle a_i*b_j=0 $ for all $ 1<=i<=n, 1<=j<=m\)

I cant seem to use the fact that the coefficents belong to a ring of prime power order.I started by imposing an order on the coefficient and assuming that there is a smallest index in that order for which ai*bj is not 0, and proceeded from there, but that didn't lead anywhere.

I'll admit that my interest in algebra is of a casual nature, so maybe i'm not aware of a few important results that ones with professional interest would. In any case, i'd liek a pointer in the right direction.

integers modulo \(\displaystyle p^n\) under addition and multiplication modulo \(\displaystyle p^n\)

.Consider two polynomials \(\displaystyle f(x) = a_0 + a_1 x + a_2 x^2 +...a_n x^n\) and \(\displaystyle g(x)=b_0 + b_1 x + b_2 x^2 +...b_m x^m\),given the coefficients are in \(\displaystyle Z/p^nZ\).

\(\displaystyle f(x)*g(x)=0\). Show that \(\displaystyle a_i*b_j=0 $ for all $ 1<=i<=n, 1<=j<=m\)

I cant seem to use the fact that the coefficents belong to a ring of prime power order.I started by imposing an order on the coefficient and assuming that there is a smallest index in that order for which ai*bj is not 0, and proceeded from there, but that didn't lead anywhere.

I'll admit that my interest in algebra is of a casual nature, so maybe i'm not aware of a few important results that ones with professional interest would. In any case, i'd liek a pointer in the right direction.

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