Polynomials of a ring form a ring

Dependencies:

Let $R$ be a commutative ring. Then $R[x]$ is a commutative ring.

Let $p, q, r \in R[x]$.

Closure of addition:

If $p = 0$, $p+q = q \in R[x]$. If $q = 0$, $p+q = p \in R[x]$.

$(p+q)_i = p_i + q_i \in R$. $\deg(p+q) \le \max(\deg(p), \deg(q))$. Therefore, $p+q \in R[x]$.
Associativity of addition:

\begin{align} & (p+(q+r))_i = p_i + (q+r)_i \\ &= p_i + (q_i + r_i) = (p_i + q_i) + r_i \\ &= (p+q)_i + r_i = ((p+q)+r)_i \end{align}
Commutativity of addition: $(p+q)_i = p_i + q_i = q_i + p_i = (q+p)_i$.
Additive identity: $(0+p)_i = 0_i + p_i = 0 + p_i = p_i$.
Additive inverse: $(-p)_i = -p_i$. Therefore, $(p+(-p))_i = p_i + (-p_i) = 0 = 0_i$.

Therefore, $R[x]$ is an abelian group under addition.

Closure of multiplication:

If $p = 0$ or $q = 0$, $pq = 0 \in R[x]$.

$(pq)_i = \sum_{j=0}^i p_jq_{i-j} \in R$ and $\deg(pq) \le \deg(p) + \deg(q)$. Therefore, $pq \in R[x]$.
Associativity of multiplication: $(pq)_i = \sum_{j=0}^i p_jq_{i-j} \in R$.

\begin{align} & ((pq)r)_i \\ &= \sum_{j=0}^i (pq)_jr_{i-j} \\ &= \sum_{j=0}^i (pq)_jr_{i-j} \\ &= \sum_{j=0}^i \left( \sum_{k=0}^j p_k q_{j-k} \right) r_{i-j} \\ &= \sum_{j=0}^i \sum_{k=0}^j p_k q_{j-k}r_{i-j} \tag{distributivity in $R$} \\ &= \sum_{k=0}^i \sum_{j=k}^i p_k q_{j-k}r_{i-j} \\ &= \sum_{k=0}^i \sum_{t=0}^{i-k} p_k q_t r_{i-k-t} \\ &= \sum_{k=0}^i p_k \left( \sum_{t=0}^{i-k} q_t r_{i-k-t} \right) \tag{distributivity in $R$} \\ &= \sum_{k=0}^i p_k (qr)_{i-k} \\ &= (p(qr))_i \end{align}
Commutativity of multiplication:

\begin{align} & (ab)_i \\ &= \sum_{j=0}^i a_jb_{i-j} \\ &= \sum_{j=0}^i b_{i-j}a_j \tag{$\because$ $R$ is commutative} \\ &= \sum_{k=0}^i b_ka_{i-k} \tag{$k = i - j$} \\ &= (ba)_i \end{align}
Distributivity:

\begin{align} & (p(q+r))_i \\ &= \sum_{j=0}^i p_j (q+r)_{i-j} \\ &= \sum_{j=0}^i p_j (q_{i-j} + r_{i-j}) \\ &= \sum_{j=0}^i (p_jq_{i-j} + p_jr_{i-j}) \tag{distributivity in $R$} \\ &= \sum_{j=0}^i p_jq_{i-j} + \sum_{j=0}^i p_jr_{i-j} \tag{additive commutativity in $R$} \\ &= (pq)_i + (pr)_i \\ &= (pq+pr)_i \end{align}

Distributivity in the other direction can be proved similarly.