a polynomial $p(x)$ is divisor of $0$ $\iff$ there is $c\in R\backslash \{0\}$ s.t. $cp(x)=0$.

Asked Feb 28 '17 at 16:25

Active Feb 28 '17 at 17:00

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Let $R$ a ring. I want to show that $p(x)\in R[x]$ is divisor of $0$ $\iff$ there is $c\in R\backslash \{0\}$ s.t. $cp(x)=0$.

The converse is obvious, but I have problems to show the implication. So let $q(x)\neq 0$ s.t. $$p(x)q(x)=0.$$ How can I show that $q(x)=c$ for a certain $c$ ? I tried as follow : Let $p(x)=\sum_{i=0}^n a_ix^i$ and $q(x)=\sum_{i=0}^n b_ix^i$ then $$p(x)q(x)=0\implies \sum_{i,j=0}^n a_ib_jx^{i+j}=0\implies \sum_{t=0}x^t\sum_{i+j=t}a_ib_j=0\implies \sum_{i+j=t}a_ib_j=0$$ for all $t=0,...,n$. How can I get $b_1=...=b_n=0$ ?

asked Feb 28 '17 at 16:25

user386627

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1

You can't. If $cp(x)=0$ then $cxp(x)=0$ , you are showing existence, not that only constants are zero divisors. – Feb 28 '17 at 16:35
@Noah: So how can I prove my statement ? – user386627 Feb 28 '17 at 16:41
@Andreas Caranti : The link you gave is not the same as my exercise (since $R$ is not commutative). – user386627 Feb 28 '17 at 16:59
OK, just reopened. – Andreas Caranti Feb 28 '17 at 17:00
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For the commutative case, see this post. – Andreas Caranti Feb 28 '17 at 17:00
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However, I do not think is true for an arbitrary ring. See an example in Section 3 of this paper. – Andreas Caranti Feb 28 '17 at 17:03

a polynomial $p(x)$ is divisor of $0$ $\iff$ there is $c\in R\backslash \{0\}$ s.t. $cp(x)=0$.

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