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This question already has an answer here:

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?

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

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  • $\begingroup$ 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? $\endgroup$ – Servaes Jun 27 at 22:44
  • $\begingroup$ The fraction field of the given ring must contain the ring, and the field $\Bbb Q$ (generated by $1$). $\endgroup$ – dan_fulea Jun 27 at 22:49
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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.

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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

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

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