1

Note that by this I mean Z[[x]], a commutative ring with 1 and integer coefficients.

I've been racking my mind over this question. Considering the neutral multiplicative power here is 1 and thus is an inverse of itself, wouldn't all elements of the set have a multiplicative inverse of x^-i, i being the power of Z[[x]]? Or considering there are no negative powers in the original R[[x]], would there be no elements with inverses?

user126731
  • 81
  • Think about what happens to degree when you compute a product in your ring. What does this tell you about invertible elements? – Jonathan Mar 23 '14 at 17:28
  • The degree of the product x^m and x^n is x^m+n? Does this mean any invertible element would have to have negative exponents too? – user126731 Mar 23 '14 at 17:29
  • If you can do long division of polynomials, in the manner taught in high-school, then you can do long division of power series. Just write the series in •ascending* powers of your variable. The process is exactly the same as what you already know, but it doesn’t usually end. Once you do a few such divisions, you see what the answer to your question is. Of course you already know that $(1-x)^{-1}=1+x+x^2+x^3+x^4+\cdots$ – Lubin Mar 23 '14 at 17:31

1 Answers1

3

If you have an invertible power series $\sum a_ix^i$ in $R[[X]]$, with inverse $\sum b_jx^j$, then $a_0b_0 = 1$, so $a_0$ has to be invertible in $R$.

Conversely, if $a_0$ is invertible in $R$, you can divide by its inverse to reduce the problem to the case of a series starting with $1$, and then apply the power series $(1+z)^{-1} = 1 - z + z^2 - z^3 + z^4 \ldots$ with $z = \sum_{i \ge 1} a_ix^i$ to find its inverse.

mercio
  • 50,180
  • But considering the coefficients in this case can only be integers, wouldn't 1 be the only a and b that can work? – user126731 Mar 23 '14 at 17:42
  • in $\Bbb Z$, $1$ and $-1$ are the only invertible elements, if that's what you are asking. Then in $\Bbb R$, inevrtible elements are nonzero elements. – mercio Mar 23 '14 at 17:45
  • Right. But for this question, we're looking for elements of the set Z[[x]]*, where all coefficients are integers. Would this mean that a = 1 and a = -1 are the only solutions? – user126731 Mar 23 '14 at 17:49
  • If you think that $\Bbb Z[[X]]^*$ has only two elements, then no. – mercio Mar 23 '14 at 17:59
  • The invertible elements of $\mathbb{Z}[[x]]^*$ are precisely the elements with constant coefficient 1 or -1. – GuillaumeDufay Jan 25 '21 at 20:42