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Homework answers / question archive / Define the set R[[X]] of formal power series in the indeterminate X with coefficients from R to be all formal infinite sums sum(a_nX^n)=a_0 +a_1X+a_2X^2+

Define the set R[[X]] of formal power series in the indeterminate X with coefficients from R to be all formal infinite sums sum(a_nX^n)=a_0 +a_1X+a_2X^2+

Math

Define the set R[[X]] of formal power series in the indeterminate X with coefficients from R to be all formal infinite sums
sum(a_nX^n)=a_0 +a_1X+a_2X^2+...
Define addition and multiplication of power series in the same way as for power series with real or complex coeficients,i.e extend polynomial addition and multiplication to power series as though they were "polynomial of infinite degree":
sum(a_nX^n)+sum(b_nX^n)=sum(a_n+b_n)X^n and
(sum(a_nX^n))(sum(b_nX^n))=sum(sum(a_kb_n-k)X^n
(P.S: The term "formal" is used here to indicate that convergence is not considered, so that formal power series need not represent functions on R)
Assuming that R[[X]] is a commutative ring with 1 prove:

a)That 1-X is a unit in R[[X]] with inverse 1+X+X^2+X^3+.....
b)That sum(a_nX^n) is a unit in R[[X]] iff a_0 is a unit in R.

The sums here go from n=0 to infinity

pur-new-sol

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Proof:
a) Formally, we have
(1-x)(1+x+x^2+...)
=(1+x+x^2+...) - x(1+x+x^2+...)
=(1+x+x^2+...) - (x+x^2+x^3+...)
=1+(x+x^2+...) - (x+x^2+...) = 1
Thus (1-x) is a unit and its inverse is 1+x+x^2+...
b) "=>": If sum(anX^n) is a unit in R[[X]], then we can find
sum(bnX^n) in R[[X]], such that
sum(anX^n)*sum(bnX^n) = 1
This implies that
a0*b0 + c1X + c2X^2 + ... = 1
where c1=a0b1+a1b0, c2=a0b2+a1b1+a2b0, ...
Thus we must have a0*b0=1. This implies that a0 is a unit.
"<=": If a0 is a unit, then we can find b0 such that a0*b0=1.
So sum(anX^n)*b0=1+sum(an'X^n), where n from 1 to oo and
an'=an*b0. Let Y=-sum(an'X^n), then we have
sum(anX^n)*b0 = 1-Y, where Y is in R[[X]]. From a), we know
that 1-Y is a unit and its inverse is 1+Y+Y^2+...
Thus sum(anX^n) is a unit and its inverse is
b0*(1+Y+Y^2+... )

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