3
$\begingroup$

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 $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$.

$\endgroup$
  • $\begingroup$ $c+d\sqrt{2}$ is not needed since $1/(c+d\sqrt{2})=(c-d\sqrt{2})/(c^2-2d^2)$. $\endgroup$ – orole Jan 30 '18 at 19:09
  • $\begingroup$ 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 $a=b=0$. $\endgroup$ – Piquito Jan 30 '18 at 19:18
  • $\begingroup$ @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. $\endgroup$ – cactus314 Jan 30 '18 at 19:27
  • $\begingroup$ You can look here, here, here, here... $\endgroup$ – Jyrki Lahtonen Jan 31 '18 at 8:19
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ you're assuming $\sqrt{2} \notin \mathbb{Q}$ but that's cool $\endgroup$ – cactus314 Jan 30 '18 at 19:12
  • $\begingroup$ @cactus314: what ??? Did you think it was ? ;-) $\endgroup$ – idm Jan 30 '18 at 19:37
1
$\begingroup$

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 $\Bbb Q$.

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

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 $\sqrt6\not\in\mathbb{Q}$.

$\endgroup$
1
$\begingroup$

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, $\sqrt{p}\not\in\mathbb{Q}(\sqrt{q})$.

$\endgroup$
0
$\begingroup$

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 $\mathbf Z$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.