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Product of polynomials over non-integral domain is 0

cupofcoffee

New member
Jun 5, 2013
7
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.
 
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cupofcoffee

New member
Jun 5, 2013
7
Kindly ignore the degrees of the two polynomials, they have no relation to the index n of the ring \(\displaystyle Z/Z p^n\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 \(\displaystyle f(x) = a_0 + a_1 x + a_2 x^2 +...a_{\color{red}r} x^{\color{red}r}\) (not the same n, of course) 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.
Hi cupofcoffee and welcome to MHB!

Thinking about this problem, I started by trying to concoct a counterexample, in the hope that this would point towards a proof of the result. So suppose we take $p=2$ and $n=4$, so that all the coefficients are in the ring $Z/16$. Let $f(x) = 8+8x+4x^2$ and $g(x) = 2+4x+8x^2$. The constant term in $f(x)*g(x)$ is $8*2$, which is $0$ mod $16$. The coefficient of $x$ is $8*4 + 8*2$ (also $0$ mod $16$). But the coefficient of $x^2$ is $8*8 + 8*4 + 4*2$, which is congruent to $8$ mod $16$ and is therefore not $0$. Thus the product $f(x)*g(x)$ is not zero, and this is reflected by the fact that the product of coefficients $a_2*b_0$ (the coefficient of $x^2$ in $f(x)$ times the constant term in $g(x)$) is nonzero.

Can we use that example to form a proof? For $g\in Z/p^nZ$, let $d(g)$ be the power of $p$ that occurs in the prime factorisation of $g$. We want to show that, for the coefficients of $f(x)$ and $g(x)$, $d(a_i) + d(b_j) \geqslant n$ for all $i$ and all $j$. Choose $a_i$ with $d(a_i)$ minimal among all the coefficients of $f(x)$. If there is more than one such $a_i$, choose the first one (in other words, the one with the smallest $i$). Similarly, choose $b_j$ with $d(b_j)$ minimal among all the coefficients of $g(x)$, and again if there is more than one, choose the first one. Your job is now to show that if $d(a_i) + d(b_j) < n$ then the coefficient of $x^{i+j}$ in $f(x)*g(x)$ cannot be $0$ (mod $p^n$) and therefore $f(x)*g(x) \ne0$.
 

cupofcoffee

New member
Jun 5, 2013
7
Umm , lets say i and j are our chosen indices. I's not clear to me why the coefficient of
\(\displaystyle x^(i+j)\) should be non zero, (as i and j appear in the coefficients of other terms as well.)
While it might seem natural to check for \(\displaystyle x^(i+j)\) first, it would be rather pointless to go looking for a proof if we're no sure one exists.
I hope u undersand the issue
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Umm , lets say i and j are our chosen indices. I's not clear to me why the coefficient of
\(\displaystyle x^{i+j}\) should be non zero, (as i and j appear in the coefficients of other terms as well.)
While it might seem natural to check for \(\displaystyle x^{i+j}\) first, it would be rather pointless to go looking for a proof if we're no sure one exists.
I hope u undersand the issue
Let $k=i+j$, and let $d_0 = d(a_i) + d(b_j)$. The coefficient of $x^{i+j}$ in $f(x)*g(x)$ will be a sum of terms of the form $a_r*b_s$, where $r+s=k$. For each such term (other than $a_i*b_j$), either $r<i$ or $s<j$. In the first case, since $r<i$ and $i$ is the smallest index for which $d(a_i)$ is minimal, it follows that $d(a_r)>d(a_i).$ Also, since $d(b_j)$ is minimal, we must have $d(b_s)\geqslant d(b_j).$ Thus $d(a_r*b_s) = d(a_r) + d(b_s) > d_0.$ In the second case, where $s<j$, a similar argument shows that, again, $d(a_r*b_s) > d_0.$

Therefore, in the coefficient \(\displaystyle \sum_{r+s=k}a_r*b_s\) of $x^k$, the only term in which the power of $p$ is as low as $d_0$ is the single term $a_i*b_j.$ That is therefore the power of $p$ that occurs in the coefficient of $x^k$. But that coefficient has to be zero (in $Z/p^nZ$), because $f(x)*g(x) = 0.$ Therefore $d_0\geqslant n$, so that $a_i*b_j = 0.$
 
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TheBigBadBen

Active member
May 12, 2013
84
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.
The first thing that comes to my mind is that there is a correspondence between this question and this one:

http://www.mathhelpboards.com/f14/polynomial-rings-4768/

The idea would be as follows:
Let $f(x) = a_0 + a_1 x + a_2 x^2 +...a_n x^n$ and $g(x)=b_0 + b_1 x + b_2 x^2 +...b_m x^m$ be polynomials in $\mathbb{Z}_{p^k}$. Then, we can take the following corresponding polynomials in $\mathbb{Q}[x]$, namely consider the polynomials
$$p(x)=\frac{a_0}{p^k}+\frac{a_1}{p^k}x+\cdots+ \frac{a_n}{p^k} x^n\\
q(x)=\frac{b_0}{p^k}+\frac{b_1}{p^k}x+\cdots+\frac{b_n}{p^k}x^n
$$
$f(x)\,g(x)$ is equal to zero in $\mathbb{Z_{p^k}}[x]$ iff $p(x)\,q(x)\in \mathbb{Z}[x]$.

Just an initial idea, I'll try to put more thought into it.

EDIT: in fact, this is a valid way to prove the statement. However, it does presuppose Gauss's lemma. Also, my other proof wasn't very well stated.

DOUBLE EDIT: ends up Opalg's steps amount to the same under sufficient scrutiny
 
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