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Let $F$ be a field and $F[t]$ be the ring of polynomials. Take $p(t)$, $q(t) \in F[t]$, and write $p(t) = \sum_{i=1}^{n} {a_i}{t^i}$, $q(t) = \sum_{k=1}^{n} {b_k}{t^k}$, where $n$ denotes the greater of the degree of $p(t), q(t)$. Then, we have:

$$p(t)q(t) = \big(\sum_{i=1}^{n} {a_i}{t^i}\big)\big(\sum_{k=1}^{n} {b_k}{t^k}\big)$$ $$ = a_0\sum_{k=1}^{n} {b_k}{t^k} + a_1\sum_{k=1}^{n} {b_k}t^{k+1} + a_2\sum_{k=1}^{n} {b_k}t^{k+2}... a_n\sum_{k=1}^{n} {b_k}t^{k+n}$$ $$ = \sum_{i=1}^{n} \sum_{k=1}^{n} {a_i}{b_k}t^{i+k}$$

If $p(t)$ is nonzero, then $a_i \neq 0$ for some $i \in \{0, 1, 2 ...n\}$. Let $M = \{i_m : a_{i_m} \neq 0\}$. So the double-sum above becomes:

$$ \sum_{i_m \in M} \sum_{k=1}^{n} {a_{i_m}}{b_k}t^{i_m+k} = \sum_{0 \le j \le 2n} \big(\sum_{i_m+k=j} {a_{i_m}}{b_k}\big)t^{j} $$

I see that the $i_m$'s are scattered, so how do I go from here to show that $\big(\sum_{i_m+k=j} {a_{i_m}}{b_k}\big) = 0 \rightarrow b_k = 0$ for all $k = 0$ to $n$?

Andy Tam
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1 Answers1

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Looking at the coefficients of $p(t)$ and $q(t)$ up to the maximum of the degrees of both, you lose track of the information what the highest degree coefficients / the degrees actually are. If you don't do that, the result follows very quickly, as hinted by Bill Dubuque in a comment.

Let $p(t), q(t) \in F[t]$ be non-zero polynomials. Say $$p(t) = \sum_{i=0}^m a_i t^i, \quad q(t) = \sum_{i=0}^n b_i t^i,$$ with $a_m, b_n \neq 0$. Then $$p(t) q(t) = \sum_{i=0}^m \sum_{j=0}^n a_i b_j t^{i+j} = a_m b_n t^{m+n} + (\text{terms of degree $< m + n$}).$$ Since $a_m, b_n \neq 0$, also $a_m b_n \neq 0$, so $p(t) q(t) \neq 0$.

Magdiragdag
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