I am given a commutative ring $R$, the polynomial ring $R[X_i, i \in I]$, and a family $(a_i)_{i \in I}$ in $R$. I want to prove that $(R[X_i, i \in I], (X_i - a_i)_{i \in I})$ is a polynomial ring as well. According to my lecture notes, I have to prove, that $(X_i - a_i)_{i \in I}$ is algebraically independent over $R$.
My idea so far: $(r_n)_{n \in \mathbb{N}}$ a family in $R$, $ r_n = 0 $ for almost all $n$ \begin{equation} \sum_{n \in \mathbb{N}^I}{r_n (X-a)^n} = \sum_{n \in \mathbb{N}^I}{r_n^* X^n} = 0 \end{equation} $(r_n^*)_{n \in \mathbb{N}}$ some family in $R$ and because $(X_i)_{i \in I}$ is algebraically independent, $ r_n^* = 0 , \forall n \in \mathbb{N}$. By using this fact I want to show that therefore $r_n = 0, \forall n \in \mathbb{N}$ but I don't know how to do that.
I would appreciate any kind of help.
Presumably you mean $(X-a)^n = \prod_{i\in I}(X_i-a_i)^{n_i}$ where $n=(n_i)_{i\in I}\in\mathbb{N}^I$. Then, since you're working in a polynomial ring only finitely many $n_i\neq0$. Let $S=\{n\in\mathbb{N}^I\mid r_n\neq 0\}$. We will show $S$ is empty.
To this end, fix an arbitrary total ordering, $\preceq$, on $I$, and extend to a total ordering, $\lhd$, on the finite set S by $n\lhd m$ if, for some $i$, $n_i<m_i$ and $n_j=m_j$ for all $j\prec i$.
If $S$ is nonempty, then $S$ has a maximal element $N$ relative to $\lhd$ and $$ 0=\sum_{n\in \mathbb{N}^I}(X-a)^n = r_NX^N + \sum_{n\lhd N} r_n^*X^n $$ for some $r_n^*\in \mathbb{N}^I$. But, $r_N\neq 0$ contradicting the algebraic independence of $\{X_i\mid i\in I\}$. Therefore, $S=\emptyset$ and $\{X_i-a_i\mid I\in I\}$ is algebraically independent.