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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 $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S) = \ker \lambda.$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes for free.

To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in \ker \lambda$ 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 $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.

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

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 $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S) = \ker \lambda.$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes for free.

To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in \ker \lambda$ 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 $\overline{A}[T_1, \dots, T_{n - 1}]$ where $\overline{A} = A[T_n]/\left(s_n T_n - 1\right)$. By the induction hypothesis, we have $s \overline{a} = 0$ for some $s \in S$. This means that $sa \in (s_n T_n - 1)$ 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 $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.

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 $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S) = \ker \lambda.$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes for free.

To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in \ker \lambda$ 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 $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.

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 $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S) = \ker \lambda.$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes for free.

To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in \ker \lambda$ 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 $a$ by $s_1^m a$ 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 $\overline{A}[T_1, \dots, T_{n - 1}]$ where $\overline{A} = A[T_n]/\left(s_n T_n - 1\right)$. By the induction hypothesis, we have $s \overline{a} = 0$ for some $s \in S$. This means that $sa \in (s_n T_n - 1)$ 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 $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.

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 $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S) = \ker \lambda.$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes for free.

To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in \ker \lambda$ 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 $\overline{A}[T_1, \dots, T_{n - 1}]$ where $\overline{A} = A[T_n]/\left(s_n T_n - 1\right)$. By the induction hypothesis, we have $s \overline{a} = 0$ for some $s \in S$. This means that $sa \in (s_n T_n - 1)$ 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 $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.