Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.3
Let $R$ be a ring. Define the set $R[[x]]$ of formal power series in the indeterminate $x$ with coefficients from $R$ to be all formal sums $\sum_{n \geq 0} a_nx^n$. Define addition and multiplication on formal power series as follows.$$\left( \displaystyle\sum_{n \geq 0} a_nx^n \right) + \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) = \displaystyle\sum_{n \geq 0} (a_n + b_n)x^n$$ $$\left( \displaystyle\sum_{n \geq 0} a_nx^n \right) \cdot \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) = \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n}a_ib_j \right) x^n$$
The term “formal” is used here to indicate that convergence is not considered, so that formal power series need not represent functions on $R$.
(1) Prove that $R[[x]]$ is a ring. Prove that if $R$ is commutative, then so is $R[[x]]$. Prove that if $R$ has a 1, then so does $R[[x]]$.
(2) Show that $1-x$ is a unit in $R[[x]]$ with inverse $\sum_{n \geq 0} x^n$.
(3) Prove that $\sum_{n \geq 0} a_nx^n$ is a unit in $R[[x]]$ if and only if $a_0$ is a unit in $R$.
Solution: Let $\alpha = \sum_{n \geq 0} a_n x^n$, $\beta = \sum_{n \geq 0} b_n x^n$, and $\gamma = \sum_{n \geq 0} c_n x^n$.
We have\begin{align*}(\alpha + \beta) + \gamma =&\ \left( \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq 0} (a_n+b_n) x^n \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} ((a_n + b_n) + c_n) x^n\\
=&\ \displaystyle\sum_{n \geq 0} (a_n + (b_n + c_n)) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} (b_n + c_n) x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right) \right)\\
=&\ \alpha + (\beta + \gamma)\end{align*}so that addition is associative.
Consider $0 = \sum_{n \geq 0} 0 \cdot x^n$. Then
\begin{align*}\alpha + 0 =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} 0 \cdot x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} (a_n + 0) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) = \alpha\\
=&\ \displaystyle\sum_{n \geq 0} (0+a_n) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} 0 \cdot x^n \right) + \left( \displaystyle\sum_{n \geq 0} a_n x^n \right)\\
=&\ 0 + \alpha\end{align*}so that 0 is an additive identity.
Define $\overline{\alpha} \in R[[x]]$ as follows: $\overline{\alpha} = \sum_{n \geq 0} (-a_n) x^n$. Note then that\begin{align*}\alpha + \overline{\alpha} =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} (-a_n) x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} (a_n – a_n) x^n\\
=&\ \displaystyle\sum_{n \geq 0} 0 \cdot x^n\\
=&\ 0\end{align*}Similarly, $\overline{\alpha} + \alpha = 0$. Thus $\overline{\alpha}$ is an additive inverse for $\alpha$.
We have\begin{align*}\alpha + \beta =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} b_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} (a_n + b_n) x^n\\
=&\ \displaystyle\sum_{n \geq 0} (b_n + a_n) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} a_n x^n \right)\\
=&\ \beta + \alpha\end{align*}so that addition is commutative.
Note that\begin{align*}(\alpha\beta)\gamma =&\ \left( \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} a_ib_j \right) x^n \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{t+k = n} \left( \displaystyle\sum_{i+j = t} a_ib_j \right) c_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{t+k = n} \displaystyle\sum_{i+j = t} a_i b_j c_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j+k=n} a_ib_jc_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+s = n} \displaystyle\sum_{j+k=s} a_ib_jc_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+s=n} a_i \left( \displaystyle\sum_{j+k=s} b_jc_k \right) \right) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{j+k = n} b_jc_k \right) x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right) \right)\\
=&\ \alpha(\beta\gamma)\end{align*}so that multiplication is associative.
We have\begin{align*}
\alpha(\beta + \gamma) =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right) \right)\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} (b_n + c_n) x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} a_i(b_j + c_j) \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ib_j + a_i c_j \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \left( \displaystyle\sum_{i+j=n} a_ib_j \right) + \left( \displaystyle\sum_{i+j=n} a_ic_j \right) \right) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n \right) + \left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ic_j \right) x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)\\
=&\ \alpha\beta + \alpha\gamma\end{align*}Thus multiplication distributes over addition on the left. Similarly, multiplication distributes over addition on the right. Thus $R[[x]]$ is a ring.
(1) Suppose $R$ is commutative. Then we have\begin{align*}\alpha\beta =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{j+i=n} b_ja_i \right) x^n\\
=&\ \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \left( \displaystyle\sum_{n \geq 0} a_n x^n \right)\\
=&\ \beta\alpha\end{align*}Thus $R[[x]]$ is commutative.
Suppose $R$ has a 1. Define $1 = \sum_{n \geq 0} e_n x^n$ by $e_0 = 1$ and $e_{i+1} = 0$. Then\begin{align*}\alpha \cdot 1 =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} e_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ie_j \right) x^n\\
=&\ \displaystyle\sum_{n \geq 0} a_n e_0 x^n\\
=&\ \displaystyle\sum_{n \geq 0} a_n x^n\\
=&\ \alpha\end{align*}Similarly, $1 \cdot \alpha = \alpha$. Thus $R[[x]]$ has a 1.
(2) Let $1-x = \sum_{n \geq 0} d_n x^n$; that is, $d_0 = 1$, $d_1 = -1$, and $d_{i+2} = 0$. Let $\delta = \sum_{n \geq 0} x^n$. Now\begin{align*}(1-x)\delta =&\ \left( \displaystyle\sum_{n \geq 0} d_n x^n \right) \left( \displaystyle\sum_{n \geq 0} x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} d_i \right) x^n\end{align*}Note that if $n \geq 1$, then $\sum_{i+j=n} d_i = 0$, and $d_0 = 1$. Thus $(1-x)\delta = 1$.
(3) We will now show that $\alpha$ is a unit in $R[[x]]$ if and only if $a_0$ is a unit in $R$. We assume that $R$ has a 1, but do not assume that $R$ is commutative.
($\Rightarrow$) Suppose $\alpha \in R[[x]]$ is a unit, with inverse $\alpha^{-1} = \sum_{n \geq 0} \overline{a}_n x^n$. Now\begin{align*}\alpha\alpha^{-1} =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} \overline{a}_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} a_i \overline{a}_j \right) x^n\end{align*}The coefficient of $x^0$ in this power series is 1 on one hand, and $a_0 \overline{a}_0$ on the other. Thus $a_0 \overline{a}_0 = 1$ in $R$. Similarly, from $\alpha^{-1}\alpha = 1$ we have $\overline{a}_0 a_0 = 1$. Thus $a_0$ is a unit in $R$.
($\Leftarrow$) Suppose $\alpha_0 \in R$ is a unit, with $\overline{a}_0 a_0 = a_0 \overline{a}_0 = 1$. Define $\overline{\alpha} = \sum_{n \geq 0} b_n x^n$ with $b_0 = \overline{a}_0$ and $$b_{k+1} = -\overline{a}_0 \sum_{i+j=k+1, j \leq k} a_i b_j.$$ Then\begin{align*}\alpha \overline{\alpha} =&\ \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a)ib_j \right) x^n\end{align*}Note that if $n = 0$, $$\sum_{i+j=n} a_ib_j = a_0 \overline{a}_0 = 1.$$ If $n \geq 1$, then \begin{align*}\sum_{i+j=n} a_ib_j =&\ a_0b_n + \sum_{i+j = n, j < n} a_ib_j \\=&\ a_0\left( -\overline{a}_0 \sum_{i+j=n, j < n} a_ib_j \right) + \left( \sum_{i+j=n, j < n} a_ib_j \right)\\ =&\ 0.\end{align*}Thus $\alpha \overline{\alpha} = 1$. Similarly, we can see that if $\underline{\alpha} = \sum_{n \geq 0} c_n x^n$ where $c_0 = \overline{a}_0$ and $$c_{k+1} = \left( \sum_{i+j=n, j \leq k} a_ic_j \right) (-\overline{a}_0)$$ then $\underline{\alpha}\alpha = 1$. Since $\alpha$ has both a left and a right inverse, they are in fact equal and $\alpha$ is a unit in $R[[x]]$.
