Exercise 1.5

Question

Let $A$ be a ring and let $A[[x]]$ be the ring of formal power series $f = \sum_{n=0}^\infty a_nx^n$ with coefficients in $A$. Show that
  \begin{enumerate}[label=\roman*)]
    \item $f$ is a unit in $A[[x]] \Leftrightarrow a_0$ is a unit in $A$.
    \item If $f$ is nilpotent, then $a_n$ is nilpotent for all $n \ge 0$. Is the converse true? (See \href{./exercise_7-2}{Chapter $7$, Exercise $2$}.)
    \item $f$ belongs to the Jacobson radical of $A[[x]] \Leftrightarrow a_0$ belongs to the Jacobson radical of $A$.
    \item The contraction of a maximal ideal $\mathfrak{m}$ of $A[[x]]$ is a maximal ideal of $A$, and $\mathfrak{m}$ is generated by $\mathfrak{m}^c$ and $x$.
    \item Every prime ideal of $A$ is the contraction of a prime ideal of $A[[x]]$.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  Write $g = \sum_{n=0}^\infty b_nx^n$.
  \par $\Rightarrow$. Any unit $f$ must have $a_0 \in A^	imes$, for $fg = 1$ implies $a_0b_0 = 1$.
  \par $\Leftarrow$. Define $g$ recursively where $b_0 = a_0^{-1}$, and 
    $b_i = -a_0^{-1}\sum_{j=1}^i a_jb_{i-j}$ for $i>0$.
  Then, $a_0b_0 = 1$, and $fg = 1$ since all other coefficients are of the form
  \begin{equation*}
    \sum_{i=0}^k a_{k-i}b_{i} = a_0b_k + a_1b_{k-1} + a_2b_{k-2} + \cdots + a_kb_0 = 0.\qedhere
  \end{equation*}
\end{proof}
\begin{proof}[Proof of $ii)$]
  Suppose $f \in \mathfrak{N}$; we claim $a_n \in \mathfrak{N}$ by induction on $n$. $f^N = 0$ implies $a_0^N = 0$ by looking at the coefficient of least degree. Now suppose $a_n \in \mathfrak{N}$ for all $n < n_0$. Then, $f - a_0 - a_1x - \cdots - a_{n_0-1}x^{n_0} \in \mathfrak{N}$ by inductive hypothesis since $\mathfrak{N}$ is an ideal by Prop.~1.7. If $(f - a_0 - a_1x - \cdots - a_{n_0-1}x^{n_0})^N = 0$, then $a_{n_0}^N = 0$ by looking at the coefficient of least degree, hence by induction $a_n$ is nilpotent for $n \ge 0$.
  \par We claim the converse is false. Let $A = \prod_{i=0}^\infty \mathbb{Z}/(2^{i+2})$ and $a_n \in A$ such that $p_i(a_n) = 2\delta_{in}$, where $p_i$ are the projection maps $A 	o \mathbb{Z}/(2^{i+2})$. Then, $a_n \in \mathfrak{N}$ for $n \ge 0$, and since $a_na_m = 0$ if $n 
e m$, letting $f = \sum_{n=0}^\infty a_nx^n$, we see for any $N$,
  \begin{equation*}
    f^N = a_0^N + a_1^Nx^N + a_2^Nx^{2N} + \cdots 
e 0.\qedhere
  \end{equation*}
\end{proof}
\begin{proof}[Proof of $iii)$]
  $f \in \mathfrak{R}_{A[[x]]}$ if and only if $1 - fg \in A[[x]]^	imes$ for all $g \in A[[x]]$ by Prop.~1.9. By $i)$, this holds if and only if $1-a_0b_0 \in A^	imes$ for $b_0 \in A$ by considering $g = b_0 + xg'$. This is true if and only if $a_0 \in \mathfrak{R}_A$ by Prop.~1.9.
\end{proof}
\begin{proof}[Proof of $iv)$]
  It suffices to show $A/\mathfrak{m}^c$ is a field. By $iii)$, $x \in \mathfrak{R}_{A[[x]]} \subseteq \mathfrak{m}$ for any maximal ideal $\mathfrak{m} \subseteq A[[x]]$ since $0 \in \mathfrak{R}_A$. Now if $0 
e a \in A$, then since $A[[x]]/\mathfrak{m}$ is a field, there exists $a_0 + xg \in A[[x]]$ such that $a(a_0 + xg) \equiv 1 \bmod \mathfrak{m}$. Since $x \in \mathfrak{m}$, $aa_0 \equiv a(a_0 + xg) \bmod \mathfrak{m}$, hence $aa_0 \equiv 1 \bmod \mathfrak{m}^c$.
  \par Now $(\mathfrak{m}^c,x) \subseteq \mathfrak{m}$ since $a_0 \in \mathfrak{m}^c$ and $x \in \mathfrak{m}$ imply $a_0f + gx \in \mathfrak{m}$. Conversely, if $f = a_0 + gx \in \mathfrak{m}$, then $a_0 = f - gx \in \mathfrak{m}^c$ since $x \in \mathfrak{m}$, hence $\mathfrak{m} \subseteq (\mathfrak{m}^c,x)$.
\end{proof}
\begin{proof}[Proof of $v)$]
  Suppose $\mathfrak{p} \subseteq A$ is a prime ideal; then, the ideal $\mathfrak{q} \coloneqq (\mathfrak{p},x)$ of $f \in A[[x]]$ with $a_0 \in \mathfrak{p}$ is prime since $fg \in \mathfrak{q}$ implies $a_0b_0 \in \mathfrak{p}$, and so either $f$ or $g$ is in $\mathfrak{q}$. Since $\mathfrak{p} = \mathfrak{q}^c$, we are done.
\end{proof}

Exercise 1.5

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Exercise 1.5

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