Show that if R has no zero divisors then R[x] also has no zero divisors.

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- Dec 25th 2012, 03:34 PMBernhardPolynomial rings - zero divisors in R and r[x]
Show that if R has no zero divisors then R[x] also has no zero divisors.

- Dec 25th 2012, 04:11 PMBernhardRe: Polynomial rings - zero divisors in R and r[x]
Thinking about this problem ...

Let $\displaystyle f(x) = a_n x^n + ... + a_1 x + a_0$

and $\displaystyle g(x) - b_n x^n + ... + b_1 x + b_0 $

Then $\displaystyle f(x)g(x) = c_{m+n} x^{m+n) + c_{m+n- 1} x^{m+n-1) + ... c_1 x + c_0$

where $\displaystyle c_k = a_kb_0 + a_{k-1}b_1 + ... + a_1b_{k-1} + a_0b_k$

But $\displaystyle c_k$ cannot equal zero since all of the $\displaystyle a_i b_j \ne 0$ since R has no zero divisors

Thus R[x] has no zero divisors

Can someone please confirm that this solution is OK?

Peter - Dec 25th 2012, 07:18 PMDevenoRe: Polynomial rings - zero divisors in R and r[x]
well, no.

c_{k}is a sum of elements, so there is no reason to suppose it can't be 0, even if all the terms in the sum are non-zero.

however, you're close...

c_{m+n}= a_{m}b_{n}≠ 0, since a_{m}≠ 0, and b_{n}≠ 0.

since this is the leading coefficient of f(x)g(x), f(x)g(x) cannot be the 0-polynomial.