# proof that $K((x)) = \operatorname{Frac}(K[[x]])$

I want to show $$K((x)) = \operatorname{Frac}(K[[x]])$$, i.e. K((x)) is the minimum field containing $$K[[x]]$$, where the latter is an integrity domain, since it is a commutative ring and has no zero divisors (the coefficients of each polynomial belong to a field).

We must then show that if $$f \in K((x))$$ then $$f=\frac{h}{g}$$ with $$f, g \in K[[x]]$$, first of all i consider

$$K((x))=\{f=\sum_{i \in \Bbb Z_{\geq \alpha}} a_i t^i: a_i \in K, \alpha \in \mathbb Z \}$$

And

$$K[[x]]=\{f=\sum_{i =0}^{\infty} a_i t^i: a_i \in K \}$$

If $$f \in K((x))$$ and $$\alpha \geq 0$$, clearly $$f \in K[[x]]$$ and I suppose that such an element can be expressed as a quotient, starting from the fact that the coefficients of each polynomial belong to a field; I really don't see clearly how to approach the proof, any suggestions?

As you said, if you know that $$K((x))$$ is a field (more on that below), you only have to show that $$K((x))$$ lies inside $$\operatorname{Frac}(K[[x]])$$. Let $$f \in K((x))$$ as in your description, if the lowest degree (often called valuation) of $$f$$ is non-negative, then $$f \in K[[x]] \subseteq \operatorname{Frac}(K[[x]])$$ (you can write it as $$\frac{f}{1}$$ if you insist on vewing it as a fraction). Now if $$f$$ has valuation $$-\alpha <0$$, notice that $$x^{\alpha} f \in K[[x]]$$, and if you denote the latter by $$g$$, this shows that $$f = \frac{g}{x^{\alpha}} \in \operatorname{Frac}(K[[x]])$$.
Now unless I understood the question wrong, I would say that what you asked is not really the trickiest part of showing $$K((x)) = \operatorname{Frac}(K[[x]])$$: so far, we assumed that $$K((x))$$ is a field, but that is not at all obvious that if $$f$$ is a power series in $$K((x))$$, then $$\frac{1}{f}$$ can also be written as such ! In other words, it is not obvious that every element of $$K((x))$$ has an inverse in that ring.
If you already know how to show this, great ! If not, here is a tip: by factoring cleverly, you can reduce to showing that $$\frac{1}{1-\sum_{i \ge 1} a_i t^i1}$$ lies again in $$K[[x]]$$. To do this, recall the power series expansion of $$x \mapsto \frac{1}{1-x}$$ for $$|x|<1$$.
• Thank you for your contribution, there is only one detail that I did not understand and that is because $x^{\alpha} f \in K[[x]]$ (i suppose that $f \in K((x))$) Commented Jul 12 at 12:21
• $f$ is a power series of the form $a_{-\alpha}x^{-\alpha} +a_{-\alpha+1}x^{-\alpha+1} + ...$, where I assume $-\alpha<0$. Multiplying by $x^{\alpha}$, you get $x^{\alpha}f = a_{-\alpha} + a_{-\alpha+1}x + ...$, which is an element of $K[[x]]$. Multiplying by $x^{\alpha}$ has the effect of "normalizing" the valuation of $f$ at 0. Commented Jul 12 at 12:40
• On a concrete example, if $f = \frac{1}{x} + 1 +x$, we are just saying that $xf = 1 + x + x^2$ is in $K[[x]]$. Commented Jul 12 at 12:40