This question is from an assignment in algebraic geometry and I am struck on it.
Question: Let $A$ be a commutative ring and let $T =\{ t_i \mid i \in I\}$ be a family of elements in $A$ and let $S=\langle T,\cdot\rangle\subseteq A$ be the multiplicative submonoid of $(A,\cdot)$ generated by $T$, i.e. $S$ consists of all finite products of elements in $T$. Then there exists a canonical isomorphism of $A$-algebras $S^{-1} A \to A[X_i\mid i\in I]/\langle t_iX_i -1\mid i\in I\rangle$.
Unfortunately, I don't have much to show in work for this question. Can you please tell how exactly should I define the map? I can see that $\langle t_i X_i -1 \mid i \in I\rangle $ should be in the kernel of the map. But, except this I don't have any clues?
Kindly consider helping.
Just to write an answer I'll rewrite my comment here.
The fact that we are quotienting out by $\langle t_iX_i-1\mid i\in I\rangle$ implies that in the quotient we have $t_iX_i-1=0$ for all $i$, which is equivalent to $t_iX_i=1$. In other words, $X_i$ is the multiplicative inverse of $t_i$. This forces $t_i^{-1}\in S^{-1}$ to be mapped to $X_i$. And this is enough to define a map $S^{-1}A\to A[X_i\mid i\in I]$ if we leave $A$ fixed, i.e., $a\mapsto a$ for all $a\in A$. I'll let you finish the definition of the map from here.
With that definition it is more or less clear that the map is surjective and the kernel is precisely $\langle t_iX_i-1\mid i\in I\rangle$ since the only relation we are introducing is $t_iX_i=1$ from the fact that $t_it_i^{-1}=1$.