Determining the kernel of the localization map when defining the localization by generators and relations à la Serre 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 $A[T] = A[T_s|s \in S]$ 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
 A: Here is a proof that $$\ker \lambda = \{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \quad (LC_3)$$ holds true assuming that the following definition is in use: $$A[S^{-1}] = A[T_s \vert s \in S] /\left(sT_s -1 \vert s \in S\right).$$
If $ta = 0$ for some $a \in A$ and some $t \in S$, then we have $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S).$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes (almost) for free.
To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in A[T_1, \dots, T_n]$ and reason by induction on $n \ge 1$.
Let us suppose that $n = 1$, i.e., $a = p_1(T_1)(s_1 T_1 - 1)$. Replacing simultaneously $a$ by $s_1^m a$ and $p_1(T_1)$ by $s_1^m p_1(T_1)$ for some $m > 0$ if need be, we can assume that either $p_1(T_1) = 0 = a$, or $\deg(s_1 p_1(T_1)) = \deg(p_1(T_1))$. As the latter identity is clearly impossible, the induction base is settled.
Suppose now that $n > 1$ and let $\overline{a}$ be the image of $a$ in $\overline{A}[T_1, \dots, T_{n - 1}] \simeq A[T_1, \dots, T_n]/(s_nT_n - 1)$ where $\overline{A} = A[T_n]/\left(s_n T_n - 1\right)$ (see claim below).
Since $\overline{a} = \sum_{i = 1}^{n - 1} \overline{p_i}(T_1, \dots, T_{n - 1}) (s_i T_i -1)$ where $\overline{p_i} \in\overline{A}[T_1, \dots, T_{n - 1}]$ is obtained from $p_i$ by assigning the last indeterminate $T_n$ to its image in $\overline{A}$, the induction hypothesis yields $s \overline{a} = 0$ for some $s \in S$. This means that $sa \in (s_n T_n - 1) \subset A[T_n]$ so that we can conclude by resorting to the case $n = 1$.$\square$
Note that we have used the following:

Claim. Let $R$ be a commutative and unital ring. Let $R[T_1, \dots, T_n]$ be the ring of multivariate polynomials over $R$ with $n$ indeterminates $T_1, \dots, T_n$.  Let $P_1, \dots, P_k \in R[T_n]$ with $k \ge 0$. Then the natural isomorphism $R[T_1, \dots, ,T_n] \rightarrow (R[T_n])[T_1, \dots, T_{n - 1}]$ induces a ring isomorphism $R[T_1, \dots, T_n]/(P_1, \dots, P_k) \rightarrow \overline{R}[T_1, \dots, T_{n - 1}]$ where $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.

A: The proof of (LC3), in the given setting, is surprisingly difficult, or, at least, elaborate. Let $a \in A$ with $\lambda_S(a) = [a] = 0$ in $A[S^{-1}]$,
i.e. one has
\begin{equation}
\tag{1}
    a \in (sT_s-1\,|\,s \in S).
\end{equation}
To show is that
\begin{equation}
\tag{2}
    sa = 0
\end{equation}
for some $s \in S$. Because of (1), there are elements $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots,p_m(T) \in A[T]$  such that
\begin{equation}
    a = \sum_{i=1}^n p_i(T) (s_iT_i - 1) \quad
    \text{in $A[T]$},\quad,\quad T_i := T_{s_i}.
    \end{equation}
As a first reduction, we may assume $p_i(T) = p_i(T_1, \dots,T_n)$ for all $i$, so that
\begin{equation}
\tag{3}
    a = \sum_{i=1}^n p_i(T_1, \dots, T_n) (s_iT_i - 1)  
    \quad \text{in $A[T]$}.
\end{equation}
Namely, let $T' \subseteq T$ be those indeterminates which either equal some $T_i$, or which appear in at least one $p_i(T)$, $i = 1, \dots, n$, so that we may write $T' = \{T_1, \dots, T_n, T_{n+1}, \dots, T_q\}$.
By eventually introducing dummy terms with coefficient 0, we may assume $p_i(T) = p_i(T') = p_i(T_1, \dots, T_q)$, so that $$a = \sum_{i=1}^n p_i(T_1, \dots, T_q) (s_iT_i - 1)$$. Putting $p_i(T):=0$ for $i=n+1, \dots, q$ then gives
\begin{equation*}
    a = \sum_{i=1}^q p_i(T_1, \dots, T_q) (s_iT_i - 1) 
    \quad \text{in $A[T]$},
\end{equation*}
which upon renaming $q$ by $n$ gives (3).
To prove that $sa = 0$ for some $s \in S$ we proceed by induction on $n$. For $n = 1$ we start with
\begin{equation*}
    a = p(T_s) (sT_s - 1) \quad \text{in $A[T]$}
\end{equation*}
for some indeterminate $T_s \in X$. We abbreviate notation by writing $u := T_s$, so that we have the equation
\begin{equation}
    a = p(u) (su - 1) \quad \text{in $A[T]$}.
\end{equation}
Let $p(u) = \sum_{k=0}^d a_k u^k$; then
\begin{equation*}
\begin{split}
    p(u) (su - 1) 
    &= \sum_{k=0}^d sa_k u^{k+1}-\sum_{k=0}^d a_ku^k\\
    &= \sum_{k=1}^{d+1} sa_{k-1} u^k - 
        \sum_{k=0}^d a_k u^k\\
    &= sa_du^d + \sum_{k=1}^d(sa_{k-1}-a_k) u^k-a_0\\
    &= a,
\end{split}
\end{equation*}
so that
\begin{equation*}
    a_0=-a \quad,\quad a_k=sa_{k-1}\,,\,k=1,\dots, d-1 
    \quad,\quad sa_d = 0,
\end{equation*}
hence
\begin{equation*}
    a_k = -s^ka \,,\, k=0, \dots, d \quad,\quad 
    sa_d = 0 ,
\end{equation*}
so that
\begin{equation*}
        s^{d+1}a = -sa_d = 0,
\end{equation*}
as was to be shown. This establishes the base clause of the induction.
We now assume that $n \ge 1$, and that, with $k < n$,
\begin{equation*}
    a = \sum_{i=1}^k p_i(T_1,\dots,T_n)(s_iT_i-1) 
    \quad \text{in $A[T]$}
\end{equation*}
implies that $sa = 0$ for some $s \in S$, and we want to show that the same is true for $k = n$. So we assume, with a given ring $A$, that $a \in \ker \lambda_S$ and (2) holds. We put $A' := A[T_n]/(s_nT_n - 1)$. The projection $A \longrightarrow A'$ then realizes(!) the  localization $$\lambda_{S'} : A \longrightarrow A[S'^{-1}]$$ with $S' := \{s_n\}$; in particular, $A'= A[S'^{-1}]$. The canonical map
\begin{equation*}
    A[T_n] \longrightarrow A[T] \longrightarrow 
    A[S^{-1}]
\end{equation*}
induces, by passing to the quotient, $$A'= A[S'^{-1}] \longrightarrow A[S^{-1}] = (A[S'^{-1}])[S^{-1}]$$, which realizes the localization
\begin{equation*}
    \lambda_S' : A[S'^{-1}] \longrightarrow (A[S'^{-1}])  
    [S^{-1}].
\end{equation*}
The localization map $\lambda_S : A \longrightarrow A[S^{-1}]$ then factors as the composite of localizations
\begin{equation*}
    A \longrightarrow A' \longrightarrow A[S^{-1}] = 
    A \longrightarrow A[S'^{-1}] \longrightarrow 
    (A[S'^{-1}])[S^{-1}].
\end{equation*}
Let $\overline{a} \in A' = A[S'^{-1}]$ be the image of $a \in A$ under $A \longrightarrow A'$. Then $\lambda_S(a) = \lambda_S'(\overline{a}) = 0$. and so, by (3),
\begin{equation*}
    \overline{a} = \sum_{i=1}^{n-1} 
    \overline{p_i}(T_1, \dots, T_{n-1}) (s_iT_i - 1) 
    \quad \text{in $A'[T]$}
\end{equation*}
with $\overline{p_i}(T_1, \dots, T_{n-1}) = p_i(T_1,\dots, T_{n-1},1/s_n)$, $i=1, \dots, n-1$, since $s_nT_n - 1 = 0$ in $A' = A[S'^{-1}]$. Therefore, by the induction  hypothesis, $s\overline{a} = \overline{sa} = 0$ for some $s \in S$. Thus $sa \in \ker \lambda_{S'}$, and so, by the base clause $n=1$ applied to $\lambda_{S'}$,
\begin{equation*}
    s_n^{d+1}(sa) = (s_n^{d+1}s)a = 0,
\end{equation*}
which finishes the proof. As a byproduct of the proof we obtain that $s$ in (2)
may be chosen as a product of the $s_i$'s (with repeated factors), i.e. as
an element of the multiplicative closure of $\{s_1, \dots, s_n\}$.
