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I am trying to show an algebraic number field $F$ is the field of fractions of its ring of integers $D_F$.

I presume that what I have to show is that for any non-zero $b \in F$, where $b = c/d$ that $c, d \in D_F$

I have started by assuming $b\in F$ is an algebraic number. I can show that any element of $F$ (as an algebraic number field) satisfies a not-necessarily monic polynomial. And clearing denominators, the polynomial becomes:

$a_nb^n + a_{n - 1}b^{n - 1} + \dots + a_0 = 0$ where $a_i \in \mathbb{Z}$

Then multiplying through by $a_n^{n - 1}$ gives a monic polynomial with integral coefficients:

$(a_{n}b)^n + a_{n - 1}(a_{n}b)^{n - 1} + \dots + a_{n}^{n - 1}a_0 = 0$, so $a_{n}b \in D_F$

Hopefully I'm on the right track. I would appreciate any corrections to what I've done. And if this is correct, I would appreciate help finishing.

Thanks very much.

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2 Answers

up vote 3 down vote accepted

You are on the right track. However,

$(a_{n}b)^n + a_{n - 1}(a_{n}b)^{n - 1} + \dots + a_{n}a_0 = 0$, so $a_{n}b \in D_F$

This should be

$(a_{n}b)^n + a_{n - 1}(a_{n}b)^{n - 1} + \dots + a_{n}^{n-1}a_0 = 0$, so $a_{n}b \in D_F$

Since $b = \frac{a_nb}{a_n}$ and $a_n \in D_F$, $F$ is the field of fractions of $D_F$.

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Thanks. You hit on where I'm stuck, how do I show $a_n \in D_F$? Do I assume since $b$ is any element in $F$, then if $b = 1$ I have a monic integral polynomial in $a_n$? –  Andrew Nov 14 '12 at 0:32
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@Andrew $a_n$ is an algebraic integer because it is a root of $x - a_n$. –  Makoto Kato Nov 14 '12 at 0:35
    
Thanks (embarrassed). Maybe I could ask one more: maybe I should have asked how do I know $a_n \in F$ –  Andrew Nov 14 '12 at 0:43
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@Andrew Since $1 \in F$, $a_n \in F$. –  Makoto Kato Nov 14 '12 at 1:09
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I saw this proof on a set of notes by Katherine Strange (now at Colorado) which I thought I would post for anyone interested.

If $F$ is a number field and $D_F$ its ring of integers, then $F$ is the field of fractions of $D_F$

Let $L$ be the FOF of $D_F$. Then $L$ is the smallest field containing $D_F$. So $L \subseteq F$.

If $[F:L] \gt 1$, then there exists $\alpha \in F \setminus L$ which is algebraic over $\mathbb{Q}$ (since $F$ is a number field). Then there exists a $d \in \mathbb{Z}$ such that $d \alpha$ is an algebraic integer, and thus $\in D_F \subset L$. But $d \alpha \notin L$. A contradiction.

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Oddly enough, I had never seen the (very elementary) claim that some integer clears any algebraic number into an algebraic integer, and I had to think about it for a minute. Plus 1 for making me realize something obvious! –  Xander Flood Jun 21 '13 at 14:14
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