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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 $b_1=...=b_n=0$ ?

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    You can't. If $cp(x)=0$ then $cxp(x)=0$ , you are showing existence, not that only constants are zero divisors. – 2017-02-28
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    @Noah: So how can I prove my statement ? – 2017-02-28
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    @Andreas Caranti : The link you gave is not the same as my exercise (since $R$ is not commutative). – 2017-02-28
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    OK, just reopened. – 2017-02-28
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    For the commutative case, see [this post](http://math.stackexchange.com/questions/83121/zero-divisor-in-rx/83171#83171). – 2017-02-28
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    However, [I do not think is true for an arbitrary ring](https://arxiv.org/pdf/1204.5205.pdf). See [an example in Section 3 of this paper](http://ac.els-cdn.com/S0021869305005818/1-s2.0-S0021869305005818-main.pdf?_tid=1dc85236-fdd8-11e6-a5c7-00000aab0f6b&acdnat=1488301713_7f86df8118d6e599f709a3042cad0a6a). – 2017-02-28

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