Question

Let R be a commutative ring. If $a_n\ne 0_R$and

$f\left(x\right)=a_0+a_{1}x+a_2{x}^2+.....+a_n{x}^n$(with $a_n\ne 0_R$) is a zero divisor in R[x] , prove that $a_n$ is a zero divisor in R.

Short answer

Answer:Hence it is proved that $a_n$is a zero divisor in R.

Step-by-step solution

Polynomial Arithmetic

If any given function R[x] is a ring, then the commutative, associative, and distributive laws hold such that the function f(x)+g(x) exists.

Polynomial Operations

The given function is a zero divisor:

$f\left(x\right)=\underset{i=0}{\overset{n}{\sum a_i}}{x}^i$

Now, according to the definition, we must have a function:

$g\left(x\right)=\sum _{j=0}^m{b}_j{x}^j\left(b_m\ne 0\right)$

Such that:

$f\left(x\right)+g\left(x\right)=0$,

Therefore, we have:

role="math" localid="1648446053254" $\sum _{i+j=k}\left(a_i{b}_j\right)=0......\left\{\forall 0\le k\le m+n\right\}$

Then, for k=m+n , we get:

$a_m{b}_n=0$.

This implies that $a_m$ is a zero divisor. Hence proved, $a_n$is a zero divisor in R.