# Proving field of fractions

Basically, I'm trying to show that the field of formal Laurent series is the field of fractions for the ring of power series. My question here is, how exactly does one prove that a given field is a field of fractions? Not necessarily for this case, but what does one have to verify for in general?

I apologize for the somewhat elementary question, but I've looked elsewhere but can't find anything that helps. We didn't really go over this in class.

• What's your definition of field of fractions? Dec 13, 2013 at 21:42
• @GiorgioMossa basically the smallest field $Q=D^{-1}R$ for an integral domain $R$ and $D \subset R$ such that every element is either a zero divisor or an unit. As I only have my notes on me right now, this is the best definition I have, so I apologize for being unclear.
– Lost
Dec 13, 2013 at 21:59
• @Lost: This definition is not correct. It should be $Q = (R - \left\{0\right\})^{-1}R$ where $R$ is an integral domain. Dec 13, 2013 at 23:27
• @rghthndsd Yeah, that's right, thanks for correcting me.
– Lost
Dec 14, 2013 at 0:35

For a field $K$ to be a field of fractions of an integral domain $R$, it is necessary and sufficient that (i) $R \subset K$ and (ii) For all $k \in K$, there exists $s \in R$ with $s \neq 0$ such that $k \cdot s \in R$ (i.e. if we write $r = k \cdot s$ then $k = \frac{r}{s}$).

Let us verify the claim. As you mention in the comments, the definition of a field of fractions is being the smallest field $K$ such that $R \subset K$. So let's assume that $R \subset K$ and show: $$K \text{ is the field of fractions of } R \Leftrightarrow \text{Condition (ii) given above}.$$

($\Rightarrow$) Suppose that $K$ is the field of fractions of $R$. Define $$K'= \left\{k \in K : \text{ there exists } r \in R \text{ such that } r \neq 0 \text{ and } k \cdot r \in R\right\}.$$ Then one can check (you should do so!) that $K'$ is a field. Certainly $K'$ contains $R$. Therefore since $K$ is the field of fractions (the smallest field that contains $R$) and $K' \subset K$, we get that $K' = K$. Therefore $K$ satisfies condition (ii).

($\Leftarrow$) Now let $K$ be a field with $R \subset K$ and $K$ satisfying condition (ii). We will show $K$ is the field of fractions of $R$. Denote the field of fractions of $R$ by $F(R)$. To prove the claim, we will show that for any $k \in K$, $k \in F(R)$. Indeed, then since $F(R)$ is the smallest field containing $R$, and $K \subset F(R)$, it must be that $K = F(R)$.

So let $k \in K$, and by condition (ii) there exists $s \in R$ with $s \neq 0$ such that $k \cdot s \in R$. So write $k \cdot s = r$, and hence (in $K$) we have $k = rs^{-1}$. But $r \in F(R)$ and $s^{-1} \in F(R)$ so therefore $k = rs^{-1} \in F(R)$. This proves that $K \subset F(R)$, as claimed.

• Why is the denominator $s \in R$ not $K$?
– Lost
Dec 13, 2013 at 22:13
• Just to be clear, the denominator $s$ is in $R$, the fraction $\frac{1}{s}$ is in $K$. To see why replacing the statement with "$s \in K$" will invalidate it, you can check that it would imply that $\mathbb{Q}(\sqrt{2})$ is the field of fractions for $\mathbb{Z}$. Indeed, $\mathbb{Z} \subset \mathbb{Q}(\sqrt{2})$ and writing $k = a + b\sqrt{2}$ where $a,b \in \mathbb{Q}$, check that $s = (a-b\sqrt{2})d$ makes $k \cdot s \in \mathbb{Z}$ for $d$ a sufficiently large integer. Dec 13, 2013 at 22:17
• Alright, I see why. So, are you saying that given any $k$ that isn't a zero divisor, we need to show that $s$ is an unit?
– Lost
Dec 13, 2013 at 23:08
• I've given a proof of my claim, hopefully this clears somethings up. Since we are dealing with integral domains one should never worry about (nonzero) zero divisors (they don't exist!) and there is nothing in the statement about $s$ being a unit. Dec 13, 2013 at 23:28
• Oh yeah, of course. I worked through it, and I think I got it, thanks.
– Lost
Dec 14, 2013 at 0:34

If you use the Wikipedia definition. (field of fractions is the smallest field containing a given integral domain), you need to show that Laurent series form a field (easy), that they contain power series (easy), and finally they are the smallest. Since any such field has to contain $1/z$ and $z$ the rest follow by linearity.

• I guess what has gotten me confused is how to prove the "smallest" part. I've already proved the field and inclusion (trivial) parts.
– Lost
Dec 13, 2013 at 22:10
• @Lost I believe I address the "smallest" part. Dec 13, 2013 at 22:36
• Ah, so you're saying that it's immediate given linearity? I apologize for the confusion.
– Lost
Dec 13, 2013 at 23:04
• @lost no problem... Dec 13, 2013 at 23:51