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Suppose $f$ is a zero divisor. Choose polynomial $g = b_0 + b_1x + \ldots + b_mx^m$ of least degree $m$ such that $$fg = 0 \implies \Sigma_{k \geq 0} \Sigma_{i + j = k} a_i b_j x^k = 0.$$ From the formula of multiplication of two polynomials we must have $a_nb_m = 0 \implies a_n g = 0$ (Because $a_n g$ annihilates $f$ and has degree $< m$). We show by induction that $a_{n - r}g = 0$ for $ 0 \leq r \leq n$. If $r = 0$, then from the above reasoning we have $a_{n - 0} = a_n$, so we have $a_ng = 0$.

Suppose $a_{n - i}g = 0$ for $0 \leq i \leq n - 1$. We will show it is true for $r = n$. If $r = n$, then $a_{n - r} = a_0$. From the formula of polynomial multiplication we must have $a_0b_0 = 0$. We will show by induction on $j$ that $a_0b_j = 0$ for $0 \leq j \leq m$. If $j = 0$, then from the above reasoning we have $a_0b_0 = 0$. Suppose the result is true for $0 \leq j \leq m - 1$, we will show it is true for $j = m$.

Then from the formula of polynomial multiplication $$\Sigma_{i + j = m}a_ib_j x^m = 0 \implies a_0b_mx^m + \Sigma{i + j = m:\ i \neq 0}\ \ a_ib_jx^m = 0.$$ By induction hypothesis we have $a_ib_j = 0$ for $i \neq 0$. So from the formula above we must have $a_0b_m = 0$. Therefore $a_0g = a_0b_0 + a_0b_1x + \ldots a_0b_mx^m = 0$. Now we can finally finish the proof. From the above reasoning we have $$ a_ib_0 + a_ib_1x + \ldots a_ib_mx^m = 0.$$

Thus in particular for $a = a_n$ we must have $af = 0$. Thus result is proved.

Conversely if there exists $a \neq 0$ in A such that $af = 0$, then by definition $f$ is a zero divisor. $\square$

Xam
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  • Something strange is happening in the second paragraph - swapping between $i,r,n$ maybe? There's also a typo in the third paragraph with that $\sum_{i+j=m : i \neq 0}$. Finally, it seems you show $ag = 0$, but don't finish off why this implies $af = 0$ - did you swap $f,g$ somewhere? – Artimis Fowl May 04 '17 at 17:52
  • Yeah I did indeed swap it at the end. Thanks. I will edit it. –  May 04 '17 at 17:53
  • @ArtimisFowl second paragraph is correct since we need that in order to show that $\Sigma_{i + j = m : i \neq 0} = 0$ –  May 04 '17 at 17:55
  • @ArtimisFowl we want to show that $$a_ib_0 + a_ib_1x + \ldots a_ib_mx^m = 0$$ for all i. –  May 04 '17 at 18:00
  • Related: https://math.stackexchange.com/questions/83121/zero-divisor-in-rx – Xam May 04 '17 at 19:22
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    @Xam Here I am presenting my argument for verification. I know about that post. I just want to check if there is any gap in my argument. –  May 04 '17 at 20:31

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