Exercise 3.2

Question

Let $\mathfrak{a}$ be an ideal of a ring $A$, and let $S = 1 + \mathfrak{a}$. Show that $S^{-1}\mathfrak{a}$ is contained in the Jacobson radical of $S^{-1}A$.
  \par Use this result and Nakayama's lemma to give a proof of $(2.5)$ which does not depend on determinants.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of first claim]
  Suppose $x \in S^{-1}\mathfrak{a}$; recall that for $x \in
  \mathfrak{R}_{S^{-1}A}$ it suffices to show $1-xy$ is a unit in $S^{-1}A$ for
  all $y \in S^{-1}A$ by Prop.~$1.9$. Then, $x = m/(1+a)$ for $m,a \in \mathfrak{a}$. Let $y = n/(1+a^{\prime})$ for $a^{\prime} \in \mathfrak{a},n \in A$, then
  \begin{equation*}
    1 - xy = 1 - \frac{m}{1+a} \cdot \frac{n}{1+a^{\prime}} = 1 - \frac{mn}{1+a^{\prime\prime}} = \frac{1 + a^{\prime\prime}-mn}{1+a^{\prime\prime}} = \frac{1 + a^{\prime\prime\prime}}{1+a^{\prime\prime}},
  \end{equation*}
  for some $a^{\prime\prime},a^{\prime\prime\prime} \in \mathfrak{a}$, since $S$ is multiplicative and $a^{\prime\prime}-mn \in \mathfrak{a}$. Thus, we see that $(1+a^{\prime\prime})/(1+a^{\prime\prime\prime}) \in S^{-1}A$ is the inverse of $1-xy$, and so $x \in \mathfrak{R}_{S^{-1}A}$.
\end{proof}
\begin{corollary}[Cor.~$2.5$]
  Let $M$ be a finitely generated $A$-module and let $\mathfrak{a}$ be an ideal of $A$ such that $\mathfrak{a}M = M$. Then there exists $x \equiv 1\pmod{\mathfrak{a}}$ such that $xM = 0$.
\end{corollary}
\begin{proof}[Proof of Corollary]
  Since $M = \mathfrak{a}M$, we first have that, for any $m \in M$, $m = am^{\prime} \in
  \mathfrak{a}M$ for some $m^{\prime} \in M$. Then, we have $S^{-1}M =
  (S^{-1}\mathfrak{a})(S^{-1}M)$, by having $m/s = (a/1)(m^{\prime}/s)$ as above. Since
  $S^{-1}\mathfrak{a} \subset \mathfrak{R}_{S^{-1}A}$ by the first part of the
  problem, we apply Nakayama to get that $S^{-1}M = 0$. By Exercise
  3.1, we see that there is then $x \in S^{-1}$ such that $xM = 0$, and $x \equiv 1 \pmod{\mathfrak{a}}$ by definition of $S$.
\end{proof}

Exercise 3.2

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Exercise 3.2

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