## The set of formal power series is a ring

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 $R$.

Let $\alpha = \sum_{n \geq 0} a_n x^n$, $\beta = \sum_{n \geq 0} b_n x^n$, and $\gamma = \sum_{n \geq 0} c_n x^n$.

1. We have
 $(\alpha + \beta) + \gamma$ = $\left( \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)$ = $\left( \displaystyle\sum_{n \geq 0} (a_n+b_n) x^n \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} ((a_n + b_n) + c_n) x^n$ = $\displaystyle\sum_{n \geq 0} (a_n + (b_n + c_n)) x^n$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} (b_n + c_n) x^n \right)$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right) \right)$ = $\alpha + (\beta + \gamma)$

2. Consider $0 = \sum_{n \geq 0} 0 \cdot x^n$. Then
 $\alpha + 0$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} 0 \cdot x^n \right)$ = $\displaystyle\sum_{n \geq 0} (a_n + 0) x^n$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) = \alpha$ = $\displaystyle\sum_{n \geq 0} (0+a_n) x^n$ = $\left( \displaystyle\sum_{n \geq 0} 0 \cdot x^n \right) + \left( \displaystyle\sum_{n \geq 0} a_n x^n \right)$ = $0 + \alpha$

so that 0 is an additive identity.

3. Define $\overline{\alpha} \in R[[x]]$ as follows: $\overline{\alpha} = \sum_{n \geq 0} (-a_n) x^n$. Note then that
 $\alpha + \overline{\alpha}$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} (-a_n) x^n \right)$ = $\displaystyle\sum_{n \geq 0} (a_n - a_n) x^n$ = $\displaystyle\sum_{n \geq 0} 0 \cdot x^n$ = $0$

Similarly, $\overline{\alpha} + \alpha = 0$. Thus $\overline{\alpha}$ is an additive inverse for $\alpha$.

4. We have
 $\alpha + \beta$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} b_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} (a_n + b_n) x^n$ = $\displaystyle\sum_{n \geq 0} (b_n + a_n) x^n$ = $\left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} a_n x^n \right)$ = $\beta + \alpha$

5. Note that
 $(\alpha\beta)\gamma$ = $\left( \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)$ = $\left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} a_ib_j \right) x^n \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{t+k = n} \left( \displaystyle\sum_{i+j = t} a_ib_j \right) c_k \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{t+k = n} \displaystyle\sum_{i+j = t} a_i b_j c_k \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j+k=n} a_ib_jc_k \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+s = n} \displaystyle\sum_{j+k=s} a_ib_jc_k \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+s=n} a_i \left( \displaystyle\sum_{j+k=s} b_jc_k \right) \right) x^n$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{j+k = n} b_jc_k \right) x^n \right)$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right) \right)$ = $\alpha(\beta\gamma)$

so that multiplication is associative.

6. We have
 $\alpha(\beta + \gamma)$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} c_n x^n \right) \right)$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} (b_n + c_n) x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} a_i(b_j + c_j) \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ib_j + a_i c_j \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \left( \displaystyle\sum_{i+j=n} a_ib_j \right) + \left( \displaystyle\sum_{i+j=n} a_ic_j \right) \right) x^n$ = $\left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n \right) + \left( \displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ic_j \right) x^n \right)$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right) + \left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} c_n x^n \right)$ = $\alpha\beta + \alpha\gamma$

Thus multiplication distributes over addition on the left. Similarly, multiplication distributes over addition on the right. Thus $R[[x]]$ is a ring.

7. Suppose $R$ is commutative. Then we have
 $\alpha\beta$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{j+i=n} b_ja_i \right) x^n$ = $\left( \displaystyle\sum_{n \geq 0} b_n x^n \right) \left( \displaystyle\sum_{n \geq 0} a_n x^n \right)$ = $\beta\alpha$

Thus $R[[x]]$ is commutative.

8. 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
 $\alpha \cdot 1$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} e_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a_ie_j \right) x^n$ = $\displaystyle\sum_{n \geq 0} a_n e_0 x^n$ = $\displaystyle\sum_{n \geq 0} a_n x^n$ = $\alpha$

Similarly, $1 \cdot \alpha = \alpha$. Thus $R[[x]]$ has a 1.

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

 $(1-x)\delta$ = $\left( \displaystyle\sum_{n \geq 0} d_n x^n \right) \left( \displaystyle\sum_{n \geq 0} x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} d_i \right) x^n$

Note that if $n \geq 1$, then $\sum_{i+j=n} d_i = 0$, and $d_0 = 1$. Thus $(1-x)\delta = 1$.

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

 $\alpha\alpha^{-1}$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} \overline{a}_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j = n} a_i \overline{a}_j \right) x^n$

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

 $\alpha \overline{\alpha}$ = $\left( \displaystyle\sum_{n \geq 0} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} b_n x^n \right)$ = $\displaystyle\sum_{n \geq 0} \left( \displaystyle\sum_{i+j=n} a)ib_j \right) x^n$

Note that if $n = 0$, $\sum_{i+j=n} a_ib_j = a_0 \overline{a}_0 = 1$. If $n \geq 1$, then $\sum_{i+j=n} a_ib_j = a_0b_n + \sum_{i+j = n, j < n} a_ib_j$ $= a_0\left( -\overline{a}_0 \sum_{i+j=n, j < n} a_ib_j \right) + \left( \sum_{i+j=n, j < n} a_ib_j \right)$ $= 0$.

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 $R[[x]]$.