All rings considered will be commutative and unitary. Let $A$ be a ring, $S \subseteq A$ a multiplicatively closed subset. The localization $\lambda_S : A \longrightarrow A[S^{-1}]$ can be characterized as a ring homomorphism $\lambda : A \longrightarrow B$ with the following three properties:
(LC1) $\lambda$ localizes $S$, i.e. $\lambda(s)$ is invertible in $B$ for all $s \in S$;
(LC2) for every $b \in B$ there is $s \in S$ such that $s b \in \text{im $\lambda$}$;
(LC3) $\ker \lambda = \{a \in A \,| \,\exists s \in S: sa = 0\}$.
One way to achieve this is to define the localization by means of generators and relations: take an indeterminate $T_s$ for each $s \in S$, form the polynomial ring over $A$ in these indeterminates and quotient out the ideal generated by the $sT_s - 1$ , $s \in S$, thus defining the localization $A[S^{-1}]$:
\begin{equation*}
A[S^{-1}] := A[T_s|s \in S]\,/\,(sT_s-1|s \in S).
\end{equation*}
The structure map $\lambda_S : A \longrightarrow A[S^{-1}]$ then comes along as the composite
\begin{equation*}
A \longrightarrow A[T_s|s \in S] \longrightarrow A[S^{-1}].
\end{equation*}
See [1], pp. I-7-8. The question is how to verify properties(LC1-3) for this construction. In fact, (LC1-2) are straightforward, but (LC3) seems hard. It is known to be true, since it holds in the other widespread model of localization, given by $\mu_S : A \longrightarrow S^{-1}A$ with
\begin{equation*}
S^{-1}A := A \times S / \sim,
\end{equation*}
where $\sim$ denotes the equivalence relation
\begin{equation*}
(a,s) \sim (b,t) :\iff \exists u \in S:\, u(ta-sb) = 0,
\end{equation*}
and
\begin{equation*}
\mu_S(a) := a/1,
\end{equation*}
where, for $(a,s) \in A \times S$, $a/s$ denotes its equivalence class in $S^{-1}A$. Here, $(LC3)$ is trivial for $\mu_S$, holding by construction. Since both $\lambda_S$ and $\mu_S$ are universal among the ring homomorphisms localizing $S$, it holds for $\lambda_S$, too. But to show this directly for $\lambda_S$ using its definition, is surprisingly difficult: if $\lambda_S(a) = 0$, this means there are $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots, p_n(T) \in A[T]$ such that ($T_i:=T_{s_i}$)
\begin{equation*}
a = \sum_{i=1}^n p_i(T)(s_iT_i - 1).
\end{equation*}
From this I can conclude
\begin{equation*}
a = -\sum_{i=1}^n a_i \quad,\quad a_i := p_i(0)
\end{equation*}
but this is, for the time being, the end of the flagpole. In the best
of all possible worlds, I would have $p_i(T) = a_i$; this would give
$a_is_i = 0$ for $i=1, \dots, n$, and so $sa = 0$ with $s := s_1 \cdots s_n$, but I see no reason for that.
So does somebody know what is needed to make progress towards (LC3)?
[1] Serre, J.-P., Algèbre locale - Multiplicités (Lecture Notes in Mathematics 11). Springer 1965