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)

    so that addition is associative.

  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

    so that addition is commutative.

  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]].

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