Are zero divisors in R[x] also zero divisors in R?

[chuy52506]

Let R be a commutative ring. If an doesn't equal 0 and
a0+a1x+a2x^2+...+anx^n is a zero divisor in R[x], prove that an is a zero divisor in R.


What I did was say if the polynomial is a zero divisor in R[x] then let that polynomial equal p(x) and any other polynomial be q(x) with coefficients b0,b1,...,bm, then p(x)*q(x)=0. And the leading coefficient and degree will be an*bm*x^(n+m) which will be a zero divisor in R. Therefore an will be a zero divisor. However I don't know what to say to show this? is it correct?

 

[ystael]

This is a mostly correct argument: you have the right basic idea, but a couple of points come across as nonsense or misstated. You say "let $$p$$ be a zero divisor and $$q$$ be any other polynomial, then $$pq = 0$$". This is false. What it means for $$p$$ to be a zero divisor is that there exists a polynomial $$q$$ such that $$pq = 0$$. Also, it doesn't make sense to say "$$a_n b_m x^{n+m}$$ is a zero divisor in $$R$$", as this is a monomial that lives in $$R[x]$$, not in $$R$$.

Other than that, your argument just needs text editing. Try correcting these points, and then one of us can show you how to clean up the wording.