Is an Non-Zero Coefficient in a Polynomial a Zero Divisor in R[x]?

[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?

 

[micromass]

What you said sounds good! What more do you want to show??

 

[chuy52506]

Haha I'm not sure if it's legit?

 

[micromass]

Yes,, it's legit, but you'd have to watch your notations!

 

[blue_raver22]

Yes, your reasoning is correct. To show this formally, we can use the definition of a zero divisor in R[x].

A polynomial p(x) is a zero divisor in R[x] if there exists a non-zero polynomial q(x) such that p(x)*q(x)=0.

So, in this case, we have p(x)=a0+a1x+a2x^2+...+anx^n and q(x)=b0+b1x+...+bm*x^m.

Then, p(x)*q(x)=(a0+a1x+a2x^2+...+anx^n)(b0+b1x+...+bm*x^m)=0.

Now, we can see that the leading term of this product will be an*bm*x^(n+m), which is a non-zero polynomial since an and bm are both non-zero coefficients.

But, since p(x) is a zero divisor, this means that an*bm*x^(n+m) must also be a zero divisor in R[x]. And since R[x] is a commutative ring, we know that an and bm are both zero divisors in R.

Therefore, an is a zero divisor in R, as desired.