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