Show that $F((X))$ is a field and that $\mathbb Q((X))$ is the fraction field of $\mathbb Z[[X]]$.

Question

1) Let $F$ be a field and $$F((X))=\left\{\sum_{n=m}^\infty a_nX^n\mid m\in \mathbb Z, \ \ a_n\in F {\rm \ for \ all \ } n \geq m \right\}.$$

I have shown that $F((X))$ is a ring but how can I show that all elements are invertible ?

2) How can I show that $\mathbb Q((X))$ is the fraction field of $\mathbb Z[[X]]$ ? I tried to show that if $R$ is an integral domain and $K$ its fraction field then $K((X))$ is the fraction field of $R[[X]]$ but I didn't succeeded.

Answer 1

For (1), we can use explicit formula: if the degree of the leading term of $f$ is less than the degree of the leading term of $g$, then we have geometric series that converges in the field of Laurent series:

$$ \frac{1}{f - g} = \frac{1/f}{1 - g/f} = \frac{1}{f} + \frac{g}{f^2} + \frac{g^2}{f^3} + \ldots $$

For example, to invert $s$, we can take $f$ to be its leading term and $g = f-s$. The resulting formula only requires inverting scalars.

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I believe (2) to be false, but can't work out the details at this moment. Instead, I believe the fraction field to be the subfield of elements of the form $a X^n f(bX)$, where $a,b \in \mathbb{Q}$, $n \in \mathbb{Z}$, and $f \in \mathbb{Z}[[x]]$.

That is, I conjecture that the fraction field is the subfield of rational Laurent series for which the coefficients' denominators don't grow too fast. (and the above conjecture a specific meaning of 'too fast')

Answer 2

If you have a nonzero element in $F((X))$, then you can assume it is $$ f=\sum_{n\ge m}a_nX^n $$ with $a_m\ne0$. Then $$ f=X^m\sum_{n\ge m}a_nX^{n-m}= X^m\sum_{n\ge 0}a_{n+m}X^n $$ Now a power series $$ \sum_{n\ge0}b_nX^n $$ with $b_0\ne0$ is invertible in $F[[X]]$: you want to find $(c_n)$ so that $$ \biggl(\,\sum_{n\ge0}b_nX^n\biggr)\biggl(\,\sum_{n\ge0}c_nX^n\biggr)=1 $$ which means $b_0c_0=1$ and, for $n>0$, $$ \sum_{k=0}^n b_{n-k}c_k=0 $$ The equation for $n=1$ gives $c_1$, the one for $n=2$ gives $c_2$ and so on. More formally, $$ c_0=b_0^{-1}, \qquad c_n=-b_0^{-1}\sum_{k=1}^n b_{n-k}c_k\ \ (n>0) $$ so $c_n$ can be determined as soon as $c_0,\dots,c_{n-1}$ have been.

Since $f$ is the product of two invertible elements in $F((X))$, it is invertible.

Unfortunately, $\mathbb{Q}((X))$ is not the field of fractions of $\mathbb{Z}[[X]]$, so the statement you were given cannot be proved, see What is the fraction field of $R[[x]]$, the power series over some integral domain?