Exercise A0.III.1.13

Proposition

Let $R$ be a commutative ring, and let $a,b,c \in R[x]$ be arbitrary. Then $a(bc) = (ab)c.$

Proof

Recall that the product of two elements of $R[x]$ is given by

$$\begin{align*}\left(\sum_{n=0}^{m_1}a_nx^n\right)\left(\sum_{n=0}^{m_2}b_nx^n\right) &= \sum_{n=0}^{m_1+m_2} \left(\sum_{k=0}^n \underbrace{a_kb_{n-k}}_{\text{indices add up to } n}\right)x^n \\ &= \sum_{n=0}^{m_1+m_2} \left(\sum_{i+j=n}a_ib_j\right)x^n.\end{align*}$$

Note that it is commutative, as neither the coefficients nor the bounds depend on the order due to commutativity of addition and multiplication in $R$.

Using this, we can derive the following.

$$\begin{align*} \left(\left(\sum_{n=0}^{m_1}a_nx^n\right)\left(\sum_{n=0}^{m_2}b_nx^n\right)\right)\left(\sum_{n=0}^{m_3}c_nx^n\right) &= \left(\sum_{n=0}^{m_1+m_2} \left(\sum_{k=0}^na_kb_{n-k}\right)x^n\right)\left(\sum_{n=0}^{m_3}c_nx^n\right) \\ &= \sum_{n=0}^{m_1+m_2+m_3} \left(\sum_{l=0}^n\left( \sum_{k=0}^la_kb_{l-k} \right)c_{n-l}\right)x^n\\ &= \sum_{n=0}^{m_1+m_2+m_3} \left(\sum_{l=0}^n\left( \sum_{k=0}^l \underbrace{a_kb_{l-k}c_{n-l}}_{\text{indices add up to } n} \right)\right)x^n\\ &= \sum_{n=0}^{m_1+m_2+m_3} \left(\sum_{i+j+k=n} a_ib_jc_k \right)x^n\\ \end{align*}$$

Now, note that $a(bc) = (bc)a$. Applying the identity above to the product $(bc)a$ produces the same result as when we apply it to $(ab)c$ (modulo the order of multiplication of coefficients, but $R$ is commutative). Thus,

$$(bc)a=a(bc)=(ab)c.$$

$\blacksquare$

NOTE

Technically, we don’t need to assume commutativity of $R$. We could do the same manipulation the second time and produce the same result. I don’t think this would contribute to any further insight, however.