# show that $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$

I'd like to prove that $\sqrt{3}$ is not in the field $\mathbb{Q}(\sqrt{2})$. Let's write out a system of equations:

$$x^2 = 3 y^2 \text{ with } x,y \in \mathbb{Z}[\sqrt{2}]$$ Then if we write $x = a + b \sqrt{2}$ and $y = c + d\sqrt{2}$ we obtain a system of diophantine equations: $$\left( \frac{a + b \sqrt{2}}{c + d \sqrt{2}} \right)^2 = 3$$ If we separate the variables we get two equations in 4 unknowns over $\mathbb{Z}$, which could have a solution: \begin{eqnarray*} a^2 + 2b^2 &=& 3(c^2 + 2d^2) \\ ab &=& 3\;cd \end{eqnarray*} One possibility is to bootstrap from unique factorization over $\mathbb{Z}$ and check that $3 | a \leftrightarrow 3 |b$ and obtain a descent this way.

Are there any other solutions?

One irrationality proof that $\sqrt{2} \notin \mathbb{Q}$ involves continued fractions:

$$\sqrt{2} = 1 + \frac{1}{2+ \frac{1}{2 + \dots }} = [1;\overline{2}]$$

such a continued fraction does not terminate, therefore it must not be an element of $\mathbb{Q}$. This line of reasoning doesn't seem so to help with $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$.

• $c+d\sqrt{2}$ is not needed since $1/(c+d\sqrt{2})=(c-d\sqrt{2})/(c^2-2d^2)$. – orole Jan 30 '18 at 19:09
• Another way is to prove that $\sqrt 2$ and $\sqrt 3$ are algebraically independant i. e. that $a\sqrt 2+b\sqrt 3=0$ with $a, b\in\mathbb Q$ implies $a=b=0$. – Piquito Jan 30 '18 at 19:18
• @Piquito algebraically independant or rationally independent (i.e. independent over $\mathbb{Q}$) ? So... you could prove that $\dim_\mathbb{Q}(\sqrt{2}, \sqrt{3}) = 2$ and then obtain the result. – cactus314 Jan 30 '18 at 19:27
• You can look here, here, here, here... – Jyrki Lahtonen Jan 31 '18 at 8:19

Suppose $\sqrt{3}=x+y\sqrt 2$ for $x,y\in \mathbb Q$. If $x,y\neq 0$, then $$3=x^2+2\sqrt 2xy+y^2\implies \sqrt{2}=\frac{3-x^2-y^2}{2xy},$$ no way that can happen. If $x$ or $y=0$ I let you adapt the proof.

• you're assuming $\sqrt{2} \notin \mathbb{Q}$ but that's cool – cactus314 Jan 30 '18 at 19:12
• @cactus314: what ??? Did you think it was ? ;-) – idm Jan 30 '18 at 19:37

Better: write $3=(u+v\sqrt2)^2$ with $u$, $v\in\Bbb Q$. Then $3=u^2+2v^2+2uv\sqrt2$. As $1$ and $\sqrt2$ are linearly independent over $\Bbb Q$ then $2uv=0$ so $u=0$ or $v=0$. This yields $3=2v^2$ or $3=u^2$ respectively; neither hard to show insoluble over $\Bbb Q$.

• OK. your proof reads $\dim_\mathbb{Q} (1, \sqrt{2} ) = 2$ then $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ which is fine. – cactus314 Jan 30 '18 at 19:14

Since $\frac1{a+b\sqrt2}=\frac{a-b\sqrt2}{a^2-2b^2}$, we have $\mathbb{Q}\!\left(\sqrt2\right)=\mathbb{Q}\!\left[\sqrt2\right]$ $$\left(\frac{a+b\sqrt2}c\right)^2=3\iff a^2+2b^2+2ab\sqrt2=3c^2$$ which means that $ab=0$ since $\sqrt2\not\in\mathbb{Q}$. Therefore, either $$a^2=3c^2$$ which is not possible since $\sqrt3\not\in\mathbb{Q}$, or $$4b^2=6c^2$$ which is not possible since $\sqrt6\not\in\mathbb{Q}$.

Assuming $\sqrt{3}\in\mathbb{Q}(\sqrt{2})$, every prime large enough such that $\left(\frac{2}{p}\right)=+1$ ($2$ is a quadratic reside $\pmod{p}$) would fulfill $\left(\frac{3}{p}\right)=+1$. On the other hand, by Dirichlet's theorem there are infinite primes of the form $24k+17$, and by quadratic reciprocity every prime of such a form fulfills $\left(\frac{2}{p}\right)=+1$ and $\left(\frac{3}{p}\right)=-1$. In general, the independence of Legendre symbols implies that if $p,q$ are distinct primes, $\sqrt{p}\not\in\mathbb{Q}(\sqrt{q})$.

There are "endlessly" many proofs, but they all boil down to the fact that $2$ and $3$ are coprime. Actually you can show the general criterion : Any quadratic field is of the form $\mathbf Q(\sqrt a)$, where $a\in \mathbf Q^*, \notin {\mathbf Q^*}^2$, and $\mathbf Q(\sqrt a)=\mathbf Q(\sqrt b)$ iff $ab \in {\mathbf Q^*}^2$. The first part is obvious ; the second can be shown "by hand", e.g. mimicking @idm's answer, or using a more elaborate tool such as Kummer's theory. In your particular case, obviously $2.3 \notin {\mathbf Q^*}^2$ because of the uniqueness of prime decomposition in $\mathbf Z$ .