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Let $R$ be a commutative ring. Given a polynomial $p=\sum_{k=0}^np_kX^k$ over $R$, let the value of $p$ at $x\in R$ be defined by $p(x)=\sum_{k=0}^np_kx^k$. This defines a function $\bar p\in R^R$. My book leaves the simple verification that $$\phi:R[X]\to R^R,\quad p\mapsto \bar p$$ to the reader. But when I tried to verify $\phi(p)\phi(q)=\phi(pq)$, it didn't seem so simple. I had to do a tedious proof by induction to show that $(\sum_{k=0}^m p_kx^k)(\sum_{j=0}^nq_jx^j)=\sum_{i=0}^{m+n}(\sum_{l=0}^i p_lq_{i-l})x^i$ for $x\in R$.

In this answer https://math.stackexchange.com/a/1743250/802314, the second bullet point says that it's enough to check that $\phi$ is multiplicative on monomials. Can someone explain why? How does the monomial case prove the general case?

tryst with freedom
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Lilypad
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    By bilinearity. – Qiaochu Yuan Sep 14 '20 at 03:27
  • Oh so basically it's because $(\sum_{k}p_kX^k)(\sum_j q_jX^j)=\sum_k\sum_j p_kq_jX^{k+j}$. Then $\phi(pq)=\phi(\sum_k\sum_j p_kq_jX^{k+j})=\sum_k\sum_j \phi(p_kq_jX^{k+j})=\sum_k\sum_j \phi(p_kX^k)\phi(q_jX^j)=(\sum_k\phi(p_kX^k))(\sum_j \phi(q_jX^j))=\phi(\sum_k p_kX^k)\phi(\sum_j q_jX^j)=\phi(p)\phi(q)$. – Lilypad Sep 14 '20 at 03:56
  • Yes. Polynomial multiplication is uniquely determined by the conditions that 1) $x^i x^j = x^{i+j}$ and that 2) it is bilinear. It is a conceptually simple operation and should be understood in a simple way! – Qiaochu Yuan Sep 14 '20 at 04:11
  • Please write the comment as an answer and answer your question to remove this question from the unanswered queue @Lilypad – tryst with freedom Apr 18 '23 at 23:59

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