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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?

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

1 Answer 1

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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)?

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  • $\begingroup$ 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. $\endgroup$ Commented Jan 17, 2020 at 16:45
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    $\begingroup$ @GeoffreyTrang Yes, but I wanted the OP to come up with that themselves. (I've folded it in as a "hidden" answer, though.) $\endgroup$ Commented Jan 17, 2020 at 16:45
  • $\begingroup$ Thanks, Noah. Does the 2nd attempt look more believable now? $\endgroup$
    – hchar
    Commented Jan 18, 2020 at 6:05

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