Fundamental Question on how to prove $a \not\in K(b)$ where $a,b$ algebraic over $K$ I have a very fundamental question on how to prove something like $\sqrt{3} \not\in \mathbb{Q}(\sqrt{2})$. In all of the proofs trying to show something similar eg. here, or here it is shown that (for the particular example above) the equation 
$$\sqrt{3} = a + b \sqrt{2}$$
for some $a,b \in \mathbb{Q}$ leads to a contradiction. How does this prove the proposition? 
My closest guess is that the approach comes from the fact that $\sqrt{2}$ is algebraic over $K$ and thus $\mathbb{Q}[\sqrt{2}]/(x^2-2) \simeq \mathbb{Q}(\sqrt{2})$. But then we needed to show that 
$$\sqrt{3} = a + b (x^2 - 2)$$
with $a,b \in \mathbb{Q}[x]$ leads to a contradiction, which is MUCH more freedom.
What am I missing?
 A: If $\alpha\in\mathbb C$, then $\mathbb{Q}(\alpha)$ is the smallest subfield of $\mathbb C$ containing $\alpha$. But$$\left\{a+b\sqrt2\,\middle|\,a,b\in\mathbb Q\right\}$$is a field which contains $\sqrt2$ and it is clearly the smallest such subfield. Therefore,$$\mathbb{Q}\left(\sqrt2\right)=\left\{a+b\sqrt2\,\middle|\,a,b\in\mathbb Q\right\}.$$
A: To begin, how does assuming
$\sqrt 3 = a + b\sqrt 2, \; a, b \in \Bbb Q, \tag 1$
help prove that
$\sqrt 3 \notin \Bbb Q(\sqrt 2)? \tag 2$
It may help to understand that
$\Bbb Q(\sqrt 2) = \{ a + b\sqrt 2; \; a, b \in \Bbb Q \}; \tag 3$
that is, $\Bbb Q(\sqrt 2)$, the smallest field contaning $\Bbb Q$ and $\sqrt 2$, is precisely the set on the right of (3); we can in fact easily demonstrate a more general fact, that is
$\Bbb Q(\sqrt p) = \{ a + b\sqrt p, \; a, b \in \Bbb Q \}, \tag 4$
where again $\Bbb Q(\sqrt p)$ is the smallest field containing $\Bbb Q$ and $\sqrt p$; here I assume, for the time being at least, that $p \in P$ is prime.  We note that the set on the right is clearly closed under addition, and also under multiplication, since 
$(a + b\sqrt p)(c + d\sqrt p) = (ac + bdp) + (ad + bc)\sqrt p; \tag 5$
as for multiplicative inverses, we observe that
$a^2 - b^2p \ne 0, \forall a, b \in \Bbb Q, \tag 6$
lest
$\sqrt p = \dfrac{\vert a \vert}{\vert b \vert } \in \Bbb Q. \tag 7$
(Here we assume the fact that no prime has a rational square root, the proof of which is easy, mimicing as it does the classic Euclidean proof that $\sqrt 2$ is irrational.)
By virtue of (6) we may write
$(a + b\sqrt p)\left (\dfrac{a - b\sqrt p}{a^2 - b^2 p} \right ) = \dfrac{a^2 - b^2 p}{a^2 - b^2p} = 1, \tag 8$
whence
$(a + b\sqrt p)^{-1} = \dfrac{a - b\sqrt p }{a^2 - b^2 p}; \tag 9$
in the light of (5), (8) and (9) we see that $\{ a + b \sqrt p; \; a, b \in \Bbb Q \}$ is closed under multiplication and reciprocation, and is hence a field; evidently
$\{ a + b \sqrt p; \; a, b \in \Bbb Q \} \subset \Bbb Q(\sqrt p), \tag{10}$
while certainly
$\Bbb Q(\sqrt p) \subset  \{ a + b \sqrt p; \; a, b \in \Bbb Q \}, \tag{11}$
since $\Bbb Q(\sqrt p)$ is the smallest field containing $\Bbb Q$ and $\sqrt p$; thus (4) binds.
Now let
$q \in \Bbb P, \; q \ne p; \tag{12}$
then if
$\sqrt q \in \Bbb Q(\sqrt p), \tag{13}$
we have via (4) that
$\sqrt q = a + b \sqrt p, \; a, b \in \Bbb Q; \tag{14}$
now we cannot have $b = 0$ lest
$\sqrt q = a \in \Bbb Q, \tag{15}$
and if $a = 0$,
$q = b^2 p; \tag{16}$
then we may write
$b = \dfrac{r}{s}; \; r, s \in \Bbb Z, \; \gcd(r, s) = 1; \tag{17}$
$r^2p = s^2q, \tag{18}$
whence
$q \mid r^2p \Longrightarrow q \mid r^2, \tag{19}$
since $p \ne q$ are primes; then
$q \mid r^2 \Longrightarrow q \mid r \Longrightarrow r = cq, \; c \in Z; \tag{20}$
$r = cq \Longrightarrow c^2q^2 p = s^2q \Longrightarrow q \mid s^2 \Longrightarrow q \mid s; \tag{21}$
but (20) and (21) together contradict $\gcd(r, s) = 1$; thus $a \ne 0$ and hence
$ab \ne 0; \tag{22}$
now it follows from (14) that
$q = a^2 + pb^2 + 2ab \sqrt p, \tag{23}$
which in the light of (22) implies
$\sqrt p = \dfrac{q - a^2 - pb^2}{2ab} \in \Bbb Q, \tag{24}$
which is impossible for a prime such as $p$; we conclude that
$\sqrt q \notin \Bbb Q(\sqrt p). \tag{25}$
