The problem

I've been trying to solve the above problem. There seems to already be some work regarding zero divisors in polynomial rings over here, but I'm not sure it is applicable to me, because it looks specifically for some $r \in R$ such that for a zero divisor $F \in R[x]$ holds $F \cdot r = 0$.

I, on the other hand, am looking to prove that if $R[x]$ has zero divisors, it implies $R$ has zero divisors.

My ideas so far

I've thought of the following: Let $p(x), q(x) \in R[x], \quad p(x) \neq 0 \neq q(x), \quad p(x) \cdot q(x) = 0$.

Since the product of $p(x)$ and $q(x)$ is zero, it shouldn't matter which $x \in R$ you choose. The product should still be zero regardless. So, freely choosing a single $x \in R$, we should be able to reduce both $p(x)$ and $q(x)$ to a value in $R$, with the equation still holding. As such, we'd have found our zero divisors in $R$.

Unfortunately, we have to watch out so that $x$ isn't a zero of either $p(x)$ or $q(x)$, or else we'd reduce one of them to $0$ and as such we wouldn't be finding a zero divisor. This leaves me stuck.

Obviously, for my approach I'd have to show that there is an $x \in R$, so that neither $p(x)$ nor $q(x)$ reduce to $0$.

  • 2
    $\begingroup$ Can you look at the leading coefficients ? $\endgroup$ – Max May 4 at 20:35

I typed this answer up before the notation was changed so that the base ring is referred to as $R$ instead of $\Bbb K$. So bear with me if you will . . .

It unnecessarily complicates things to look for $y \in \Bbb K$ with

$p(y) \ne 0 \ne q(y), \tag 1$


$p(x), q(x) \in \Bbb K[x]; \tag 2$

indeed, there is no a priori reason to suppose such a $y$ even exists; instead, it is easier, I'll warrant, to simply realize that the "$x$" in "$\Bbb K[x]$" does not represent an element of $\Bbb K$; rather, it is an indeterminate used to define, and allow the manipulation of, polynomials as abstract entities.

Bearing this in mind, we suppose that

$0 \ne p(x) \in \Bbb K[x] \tag 3$

is a zero divisor; then

$\exists 0 \ne q(x) \in \Bbb K[x] \tag 4$


$p(x)q(x) = 0; \tag 5$

now with

$p(x) = \displaystyle \sum_0^{\deg p} p_j x^j \tag 6$


$q(x) = \displaystyle \sum_0^{\deg q} q_j x^j, \tag 7$

we have

$p_{\deg p} \ne 0 \ne q_{\deg q}, \tag 8$

since they are the leading coefficients of $p(x)$ and $q(x)$, respectively; but

$0 = p(x)q(x) = \left (\displaystyle \sum_0^{\deg p} p_i x^i \right ) \left (\displaystyle \sum_0^{\deg q} q_j x^j \right )$ $= \displaystyle \sum_{k = 0}^{\deg p + \deg q} \left ( \sum_{i = 0}^k p_{k - i}q_i \right )x^k$ $= p_{\deg p}q_{\deg q}x^{\deg p + \deg q} +\displaystyle \sum_{k = 0}^{\deg p + \deg q - 1} \left ( \sum_{i = 0}^k p_{k - i}q_i \right )x^k,\tag 9$


$p_{\deg p}q_{\deg q} = 0; \tag{10}$

that is, $p_{\deg p}$ and $q_{\deg q}$ are zero-divisors in $\Bbb K$.

In closing, I think it is worthy of note that what we have directly proved here is but the contrapositive of the well-known but elementary result that $\Bbb K[x]$ is an integral domain if $\Bbb K$ is, the proof of which is also typically had via examining the leading coefficients of $p(x)$, $q(x)$, and $p(x)q(x)$.


You want to show that if $R$ is a domain, $R[x]$ is a domain.

So, suppose that $R$ is a domain, then take $P,Q\in R[x]$.

Let $aX^n$ and $bX^m$ be the monomials of maximal degree in $P$ resp. $Q$, then the maximal degree monomial of $PQ$ is $abX^{n+m}$, of course if $ab\neq 0$. But as $R$ has no zero divisor, this is true.

So $R[x]$ is a domain.


Be careful: when we write $R[x]$ for the ring of polynomials, $x$ is not an element of $R$. A polynomial $a_0+a_1x+\dots+a_nx^n$ is zero, by definition, if and only if $a_0=a_1=\dots=a_n=0$.

You're confusing with “polynomial functions”, but the concepts of polynomial and polynomial function are to be kept distinct.

If $f(x)=a_0+a_1x+\dots+a_nx^n\in R[x]$, then $f$ induces a polynomial function $\varphi_f\colon R\to R$ by $\varphi_f(r)=f(r)$. Note that it is very possible that for two distinct polynomials $f$ and $g$ it holds that $\varphi_f=\varphi_g$. It is the case when $R$ is a finite ring.

Suppose $f(x)=a_0+a_1x+\dots+a_nx^n$ is a zero divisor; in particular it is not the zero polynomial, so we can assume $a_n\ne0$. Then there exists a nonzero polynomial $g(x)=b_0+b_1x+\dots+b_mx^m$ and $b_m\ne0$ such that $f(x)g(x)=0$. Since the coefficient of $x^{m+n}$ in $f(x)g(x)$ is $a_nb_m$, we conclude that $R$ has zero divisors.


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