# Why "every element is a divisor of zero in any ring of polynomials" is false?

This question originates from Pinter's Abstract Algebra Chapter 24 Exercise B7.

There are rings such as $$P_3$$ in which every element $$\ne 0,1$$ is a divisor of zero. Explain why this cannot happen in any ring of polynomials $$R[x]$$, even when $$R$$ is not an integral domain.

Note: $$P_3$$ is the power set of three elements with the group operation $$X*Y=(X\cup Y) - (X\cap Y)$$.

First (Incorrect) Attempt:

McCoy's theorem: Let $$F\in R[x]$$ be a polynomial over a commutative ring $$R$$. If $$F$$ is a zero-divisor then $$rF=0$$ for some nonzero $$r\in R$$.

Suppose $$F$$ is a zero-divisor such that $$rF=0$$ for some nonzero $$r\in R$$. Then $$F+1$$ has no divisor of zero, for $$r(F+1)=r\ne 0$$. Hence not every element $$\ne 0,1$$ in a commutative ring of polynomials is a divisor of zero.

Questions:

1. Is this correct ?
2. This proof relies on a theorem that specifically applies to polynomial over commutative ring. How do we prove when the ring is not commutative?

Second attempt:

Suppose the opposite, that every element in a ring of polynomial R[x] $$\ne 0,1$$ is a divisor of zero.

This implies $$x\in R[x]$$ is a zero-divisor such that $$x\cdot b(x)=0$$ for some nonzero $$b(x)\in R[x]$$. Suppose $$b(x) = b_0 + b_1x + \cdots b_nx^n$$ where $$b_0,\cdots,b_n \in R$$, $$n \ge 0$$, and $$b_n\ne 0$$.
Then $$x\cdot b(x) = b_0x + b_1x^2 + \cdots + b_nx^{n+1}=0$$. But this is absurd, as $$b_n\ne 0\implies b_n x^{n+1}\ne 0$$ and therefore $$x\cdot b(x)$$ cannot possibly be zero.

This proves the contradiction, as required. Correct?

• What is $P_3$? Also, $x\in R[x]$ is not a zero divisor. Commented Jan 17, 2020 at 16:18
• It's not correct. The $r\in R$ in McCoy's Theorem doesn't have to be the same for every $F$. Commented Jan 17, 2020 at 16:18
• The title should be edited, right now it's a trivial question that is unrelated to the body. Commented Jan 20, 2020 at 9:40
• @ArnaudMortier, how so? The body gives a proof (by contradiction) of exactly why the quoted statement is false. Commented Jan 21, 2020 at 15:58

The argument you've given seems to assume that if $$F+1$$ is a zero-divisor then $$r$$ itself witness this. But why should that be true? Maybe $$r(F+1)\not=0$$ (as you've shown correctly) but $$s(F+1)=0$$ for some $$s\not=r$$.
(For example, this happens in the ring $$\mathbb{Z}/6\mathbb{Z}$$: both $$[2]$$ and $$[2]+[1]=[3]$$ are zero-divisors.)
Instead, you should see if you can think up a specific element of $$R[x]$$ - regardless of what $$R$$ is! - which "obviously" isn't a zero (and then prove that). HINT: think about degrees of polynomials, and what multiplication does to them ...
Consider the polynomial "$$x$$" - what can we say about the degree of $$xp(x)$$ for any other nonzero polynomial $$p(x)$$, and why is that helpful to us? And what are some other polynomials this is guaranteed to work for (even if $$R$$ has zero divisors)?
• Namely, $x$ would work. Multiplying a polynomial by $x$ would always increase the degrees of all its terms by one, and clearly, could not result in the zero polynomial unless the original polynomial was itself the zero polynomial. Commented Jan 17, 2020 at 16:45