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I have to prove that any non-zero element $b(z)\in\mathbb{C}((z))$ (the field of fractions of the ring $\mathbb{C}[[z]]$ of formal power series) has a unique presentation of the form $b(z)=z^ma(z)$ with $a(z)\in\mathbb{C}[[z]]$, $a_0\neq 0$ and $m\in\mathbb{Z}$.

I know that the ideals in the ring are of the form $(z^m)$ and that any formal power series with $a_0\neq 0$ is a unit. Can I somehow use this to prove the statement or how else should I prove it?

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$b(z)=f(z)/g(z)$ with $f,g\in\mathbb C[[z]]$, $g\ne0$. Write $f(z)=z^kf_1(z)$ with $f_1(0)\ne0$, and $g(z)=z^lg_1(z)$ with $g_1(0)\ne0$. Then $b(z)=z^{k-l}f_1(z)(g_1(z))^{-1}$. Now set $a(z)=f_1(z)(g_1(z))^{-1}$ and notice that $a(0)\ne0$.

The uniqueness is left as an exercise. (You can ask for help if need it.)

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  • $\begingroup$ That makes sense :) I guess that the uniqueness part is about assuming that there are two representations and then reaching a contradiction? $\endgroup$ – Alice Nov 23 '15 at 15:14
  • $\begingroup$ @Alice Yes. (Cancel a right power of $z$ and so on.) $\endgroup$ – user26857 Nov 23 '15 at 16:24
  • $\begingroup$ Assume for contradiction that $b(z)$ has to presentations such that $z^ma(z)=b(z)=z^nc(z)$. Then $a(z)=z^{n-m}c(z)$. Then $b(z)=z^mz^{n-m}c(z)=z^nc(z)$. I think I'm running in circles here... $\endgroup$ – Alice Nov 24 '15 at 10:53
  • $\begingroup$ @Alice $a(z)=z^{n-m}c(z)$ leads to two cases: 1. $n>m$; then set $z=0$ and find $a(z)=0$, impossible; 2. $n<m$; write $z^{m-n}a(z)=c(z)$, set $z=0$ and find $c(z)=0$, impossible. So $m=n$, and then $a(z)=c(z)$. $\endgroup$ – user26857 Nov 24 '15 at 10:58
  • $\begingroup$ Ah of course! Thank you :D $\endgroup$ – Alice Nov 24 '15 at 11:09

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