Determining the kernel of the localization map when defining the localization by generators and relations à la Serre

Question

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

Answer 1

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 $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.

Answer 2

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 $\{s_1, \dots, s_n\}$.