This is an exercise in my professors notes. The notation $A[t_i \ | \ t_i \in I]$ means all finite linear combinations of monomials in $t_i$ with coefficients in $A$. I can see why it is true, but I had difficulty making my proof rigorous, and so I am asking for feedback to improve or simplify my solution, or identify any errors in it. This is essentially a generalization of: Prove that the Polynomial Ring $F[t]$ Is an Integral Domain.
Proof: Let $p, q \neq 0$ be in $B = A[t_i \ | \ i \in I]$. Suppose $p = p_0 + \sum_{i}^{n_p} \sum_j^{m_i} p_{ij} t_{s_i}^j$ and $q = q_0 + \sum_{k}^{n_q} \sum_j^{m_k} p_{kl} t_{s_k}^l$. Then suppose $t_{s_i} = t_{s_k}$ for some $(i,k)$. Choose $(i,j)$ and $(k,l)$ such that $t_{s_i} = t_{s_k}$ and $m_i + m_k$ is maximal. Then the coefficient corresponding to $t_{s_i}^{m_i} t_{s_k}^{m_k}$ is simply $p_{im_i}q_{km_k}$. Since $p_{im_i}, q_{km_k} \neq 0$, it follows that their product is nonzero since $A$ is a domain. Thus $pq \neq 0$. On the other hand, suppose that for no $(i,k)$ do we have $t_{s_i} = t_{s_k}$. Then the product all coefficients of the product $pq$ have the form $p_{ij}q_{kl}$. But since neither $p_{ij}$ nor $q_{kl}$ are zero, their product cannot be. Hence, $pq \neq 0$. Thus $B$ is a domain.
The point is uncertainty lies in examining the "maximal exponent." Is this necessary? Does it add problems I haven't addressed in the proof, which could be avoided by taking a simpler approach? While this is dealing with finite sets, so any problems involving Zorn's lemma or axiom of choice are avoided, my concern is mostly that perhaps there's a way to combine the two steps into one.
