2

Starting with assuming $A[x]$ is a domain I only take the elements of $A$ and view them as polynomials of degree zero and use the hypothesis

My problem is going the other way:

Let $A[x]$ be a domain, we take two polynomials $\sum_{i=0}^n a_i x^i$ and $\sum_{j=0}^m b_j x^j$ with $a_i, b_j \in A$, $i \in \{1,...,n\}$ and $j \in \{1,...,m\}$ then:

$$\biggl(\sum_{i=0}^n a_i x^i\biggr)\biggl(\sum_{j=0}^mb_j x^j\biggr)=\sum_{k=0}^{m+n}\biggl(\sum_{i+j=k} a_i b_j \biggr)x^k =0$$

I believe that I have to prove the final sum is zero only if every $a_ib_j$ is zero but what if two elements cancel each other out without either being zero.

Any leads would be appreciated.

Bernard
  • 175,478
Rocketman
  • 95
  • 8

2 Answers2

0

If you can show that $\deg(fg) = \deg(f) + \deg(g)$ then this follows since if $f$ and $g$ are non-zero then $\deg(f), \deg(g) \ge 0$ and hence $\deg(fg) \ge 0$. By convention $\deg(0) = -\infty$ so this shows that $fg \ne 0$.

And if you don't like relying on a convention then you can try to synthesize the relevant part of the above proof sketch, namely looking at the leading terms.

Trevor Gunn
  • 27,041
0

Let $f=\sum_{i=0}^n a_i x^i$ and $g=\sum_{j=0}^m b_j x^j$ be two polynomials in $A[x]$ such that $fg=0$ and $f\neq0$. Then $a_n\neq0$ and since $fg=0$ we have $a_nb_m=0$. Being $A$ domain one has $b_m=0$, which is the leader coefficient of $g$. Hence $g=0$.

Rafael
  • 3,789