# Show that $\mathbb{Q}[\sqrt{7}]$ is a field of fractions of $\mathbb{Z}[\sqrt{7}]$ [duplicate]

I've proved that $$\mathbb{Q}[\sqrt{7}]$$ is a field, but now I have to prove that $$\mathbb{Q}[\sqrt{7}]$$ is the smallest field containing $$\mathbb{Z}[\sqrt{7}]$$.

What is the best approach to solve this?

## marked as duplicate by Servaes, YuiTo Cheng, Lee David Chung Lin, Leucippus, postmortesJun 28 at 5:39

• Need a lot more context to be able to give a meaningful answer. What have you tried yourself? How do you define $\Bbb{Z}[\sqrt{7}]$ and $\Bbb{Q}[\sqrt{7}]$ and what do you know about these rings? – Servaes Jun 27 at 22:44
• The fraction field of the given ring must contain the ring, and the field $\Bbb Q$ (generated by $1$). – dan_fulea Jun 27 at 22:49

Suppose $$\;F\;$$ is a field containing $$\;\Bbb Z[\sqrt7]\;$$ . Then $$\;F\;$$ contains any number of the form $$\;\cfrac{f(\sqrt7)}{q(\sqrt7)}\;$$ , for $$\;f,q\in\Bbb Z[x]\;$$ , with $$\;q(\sqrt7)\neq0\;$$ , and this means $$\;\Bbb Q(\sqrt7)\subset F\;$$ . But $$\;\Bbb Q(\sqrt7)=\Bbb Q[\sqrt7]\;$$ as$$\;\sqrt7\;$$

is an algebraic number, so we're done.

Suppose $$F$$ is a field such that

$$\Bbb Z[\sqrt 7] \subset F; \tag 1$$

then

$$\Bbb Z \subset F, \tag 2$$

since

$$\Bbb Z \subset \Bbb Z[\sqrt 7]; \tag 3$$

also,

$$\sqrt 7 \in \Bbb Z[\sqrt 7] \subset F, \tag 4$$

and thus

$$\sqrt 7 \in F; \tag 5$$

since $$F$$ is a field, (2) implies

$$\Bbb Q \subset F, \tag 6$$

and combined with (5) this yields

$$\Bbb Q[\sqrt 7] \subset F; \tag 7$$

but $$\Bbb Q[\sqrt 7]$$ is itself a field, for it may be presented as

$$\Bbb Q[\sqrt 7] = \{a + b\sqrt 7; \; a, b \in \Bbb Q \}, \tag 8$$

and we may see that any

$$a + b\sqrt 7 \ne 0 \tag 9$$

has a multiplicative inverse in $$\Bbb Q[\sqrt 7]$$, for

$$(a + b\sqrt 7) \dfrac{a - b\sqrt 7}{a^2 - 7b^2} = \dfrac{(a + b\sqrt 7)(a - b\sqrt 7)}{a^2 - 7b^2} = \dfrac{a^2 - 7b^2}{a^2 - 7b^2} = 1; \tag{10}$$

these calculations are legitimized by the fact that the are no rationals $$a$$, $$b$$ such that

$$a^2 = 7b^2, \tag{11}$$

lest

$$\sqrt 7 = \dfrac{a}{b} \in \Bbb Q; \tag{12}$$

the proof of this assertion proceeds along the usual lines familiar from the classic case of $$\sqrt 2$$; we present a variant holding for any prime $$p$$; we assume

$$\sqrt p = \dfrac{r}{s}, \tag{13}$$

$$r, s \in \Bbb Z; \; \gcd(r, s) = 1; \tag{14}$$

then

$$r^2 = ps^2 \Longrightarrow p \mid r^2 \Longrightarrow p \mid r \Longrightarrow p^2 \mid ps^2 \Longrightarrow p \mid s^2 \Longrightarrow p \mid s \Rightarrow \Leftarrow \gcd(r, s) = 1. \tag{15}$$

We see from this argument that in fact the square root of any prime is irrational, and thus that $$\Bbb Q[\sqrt p]$$ is a field for any prime $$p$$.

Finally, since (7) binds for any $$F$$ satisfying (1), we see that $$\Bbb Q[7]$$ is the smallest field containing $$\Bbb Z[7]$$; furthermore, $$\Bbb Q[7]$$ is the field of quotients of $$\Bbb Z[7]$$, since

$$\dfrac{a + b\sqrt 7}{c + d\sqrt 7} = \dfrac{(a + b\sqrt 7)(c - d\sqrt 7)}{c^2 - 7d^2} \in \Bbb Q[\sqrt 7] \tag{16}$$

for any

$$a, b, c, d \in \Bbb Z, \tag{17}$$

provided of course that

$$c + d\sqrt 7 \ne 0. \tag{18}$$