# 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?

• @DietrichBurde Does that then not contradict the isomorphism $\mathbb{Q}[\sqrt{2}]/(x^2-2) \simeq \mathbb{Q}(\sqrt{2})$? – G. Chiusole Jan 7 at 19:55
• The constant $0$ function, isn't it? – G. Chiusole Jan 7 at 19:59
• Well isn't $\mathbb{Q}(\sqrt{2})$ defined to be the fraction field over $\mathbb{Q}[\sqrt{2}]$? Given the other answers I feel like I have faulty definitions – G. Chiusole Jan 7 at 20:02
• It clearly means $\Bbb Q[x]/(x^2-2)\cong \Bbb Q(\sqrt{2})$ and not what you have written. So, yes, there is some "faulty" concept - never mind. Now you have at least a clear answer to your question in the last line. – Dietrich Burde Jan 7 at 20:04
• Are you saying that the function is an isomorphism or not? Also, given the iso, how does $x^2 - 2$ being a polynomial in $\mathbb{Q}[x]$ contradict it being the constant $0$ function in $\mathbb{Q}/(x^2 - 2)$ and thus in $\mathbb{Q}(x)$? – G. Chiusole Jan 7 at 20:21

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

$$\sqrt q \notin \Bbb Q(\sqrt p). \tag{25}$$

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,$$\mathbb{Q}\left(\sqrt2\right)=\left\{a+b\sqrt2\,\middle|\,a,b\in\mathbb Q\right\}.$$