Does there exist a ring $R[x]$ without a zero divisor but the ring R is having a zero divisor.


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  • $\begingroup$ I know the the theorem that states if R is a ring without zero divisor, then $R[x]$ is a ring without zero divisor. Is the converse true? $\endgroup$ – Atul Anurag Sharma Apr 22 '18 at 16:34

This is a trivial result since any zero-divisor of $R$ is still a zero-divisor of $R[x]$. This is the more interesting result: McCoy's theorem. If $f(x)\in Z(R[x])$, then there is a non-zero $r\in R$ with $rf(x)=0$. See Zero divisor in $R[x]$

  • $\begingroup$ By $Z(R[x])$ do you mean the set of zero divisors in $R[x]$, or the center of $R[x]$? $\endgroup$ – Robert Lewis Apr 22 '18 at 17:22
  • 1
    $\begingroup$ Zero-divisors of $R[x]$. I think McCoy's theorem needs commutativity. Polynomials over non-commutative rings seem like they would be nasty. $\endgroup$ – CPM Apr 22 '18 at 17:24

No, because $R\subseteq R[x]$. Hence, if $a,b\in R$ with $a,b\neq 0$ and $ab=0$, then also $a,b\in R[x]$ with $ab=0$.

  • $\begingroup$ very good answer :-) $\endgroup$ – Surb Apr 22 '18 at 17:17

Indeed $R[x]$ has zero divisors of every degree; for if

$\exists a, b \in R, \; a, b \ne 0 \mid ab = 0, \tag 1$

then setting

$f(x) = \displaystyle \sum_0^n a x^i \in R[x], \tag 2$


$g(x) = \displaystyle \sum_0^m b x^j \in R[x], \tag 3$

we have

$f(x)g(x) = \displaystyle \left ( \sum_0^n a x^i \right) \left (\sum_0^m b x^j\right ) = \sum_{k = 0, i + j = k}^{m + n} ab x^{i + j} = 0. \tag 4$

As for the question our OP Atul Anurag Sharma posed in his comment to this question, it is easy to see that if $R[x]$ has no zero divisors, then neither does $R$, since $R \subset R[x]$ as the ring of polynomials of degree zero:

$(ax^0)(bx^0) = abx^0, \tag 5$

$ax^0 + bx^0 = (a + b)x^0, \tag 6$

etc. Thus the existence of zero divisors in $R$ forces their existence in $R[x]$; if there are none in $R[x]$, there are none in $R$.


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