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Let $R$ be any commutative ring, and consider the polynomial ring $R[x]$. Prove that $x$ is not a zero divisor in $R[x]$ and generalise the argument to prove that a monic polynomial $f=x^n + a_{n-1}x^{n-1}+ \dots +a_0$ is not a zero divisor in $R[x]$.

If $x$ was a zero divisor of $R[x]$, then there would exist a nonzero $g \in R[x]$ for which $xg = 0.$ This would mean that $$x(c_nx^n+\dots+c_0)=c_nx^{n+1}+ \dots + c_0x = 0$$ and since this is only zero when all of the coefficients are zero it would imply that $g$ is zero that's a contradiction.

Now if $f \in R[x]$ is a monic and a zero divisor then there would exist nonzero $p \in R[x]$ for which $fp=0$. That is $$(x^n+a_{n-1}x^{n-1}+ \dots+ a_0)(b_mx^m + \dots+ b_0)=0.$$ I'm having issues on grouping terms in $(x^n+a_{n-1}x^{n-1}+ \dots +a_0)(b_mx^m + \dots+ b_0)$, I think I should have no constant terms in this expression in order to conlcude a contradiction on $p$ being nonzero?

user26857
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Gregori
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1 Answers1

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Let $f$ be a nonzero polynomial in $R[x]$, such that $p(x)f(x) = 0$ and $f$ is of minimum degree, say $m$. Write $f = b_mx^m + \cdots + b_0$ where $b_m \neq 0$. Then,

$$(x^n + a_{n-1}x^{n-1} + \cdots + a_0)(b_mx^m + \cdots + b_0) = 0 $$ Expanding and looking at the highest degree term gives that $b_m = 0$, which means that there is a polynomial of lower degree which satisfies the condition, this is a contradiction. So $f$ must be the zero polynomial.

Rishabhs1711
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  • The highest degree term is $b_mx^{n+m}$ right? How will this imply that $b_m = 0$? I think we have a term $a_0b_0$ in the product without any powers of $x$ in it? – Gregori Apr 22 '22 at 16:06
  • The RHS is $0$, so all coefficients on the LHS must be zero. In particular, $b_m$, which is the coefficient of $x^{m+n}$ should be $0$ as well. – Rishabhs1711 Apr 22 '22 at 16:09
  • Oh right. Is minimum degree of $f$ a neccessary condition here? Rather is it something we can always assume? – Gregori Apr 22 '22 at 16:13
  • Yes, an $f$ with minimum degree will always exist, once we are given that there exists at least one suitable $f$. Since the possible degrees of suitable $f$'s are all natural numbers, and within a set of natural numbers, you can always find a minimum (This is the Well Ordering Principle). To be precise, the polynomial with minimum degree may not be unique, there might be many such polys, but there will be AT LEAST one with a minimum degree. – Rishabhs1711 Apr 22 '22 at 16:24
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