Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove that a power series in more than one variable is invertible?

Is this still true if the variables are noncommutative?

share|improve this question
    
@Stultus: if by "invertible" you mean "units in the power series ring", then not every power series in more than one variable is invertible. For example, $(xy) + (xy)^2 + \cdots + (xy)^n + \cdots$ is not invertible in $\mathbb{R}[[x,y]]$. –  Arturo Magidin Nov 22 '10 at 20:03

3 Answers 3

(I am answering with respect to the commutative case.) This is not completely true. A power series with a nonzero constant term over a field is invertible. The power series ring in finitely many variables $k[[x_1,...,x_n]]$ over a field is a local ring and the ideal generated by all the variables is the unique maximal ideal. By an elementary result on local rings, every element outside the unique maximal ideal is invertible. So if a power series has a nonzero constant term, it is not in the unique maximal ideal and hence invertible. If the constant term is zero, it is in this maximal ideal and hence not invertible (this ideal is proper).

Edit: You only need the ring of coefficients to be local (and not necessarily a field) for the formal power series ring to be local. In this case the maximal ideal is $m+(x_1,...,x_n)$ where $m$ is the maximal ideal of the coefficient ring $R$. So in this case a power series is invertible iff the constant term is outside the maximal ideal of the coefficient ring.

I am not sure if this is true in the noncommutative case.

share|improve this answer
    
for arbitrary rings with commuting variables, an element of $R[[x]]$ is invertible if and only if the constant term is invertible (think about the lowest term of the product). Order the monomials lexicographically and by total degree and you should get the same answer in $R[[x_1,\ldots,x_n]]$. –  Arturo Magidin Nov 22 '10 at 20:15
    
@Arturo: Sorry I should have specified that the maximal ideal in this case is $m+(x_1,...,x_n)$ where $m$ is the maximal ideal of $R$. –  Timothy Wagner Nov 22 '10 at 21:07

Timothy Wagner's answer extends to the case of power series over a field with in noncommuting variables with nonzero constant terms.

Note that the power series with zero constant terms is both a maximal left ideal and a maximal right ideal. The same argument as in the commutative case gives left and right inverses, and by associativity both must be equal.

share|improve this answer

Since you are using $xy$ in all the terms and $x$ and $y$ are real numbers, we may represent it as a single variable i.e. $r=xy$ . Then the Sum becomes $S=r/(1-r)$ ;condition $-1 < r < +1$ $1/S =(1-r)/r$ ; condition $r$ not equal to zero.

share|improve this answer
1  
I think you are treating $x$, $y$ as numbers, when they are not. –  Martin Argerami Nov 28 '12 at 4:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.