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I wish to show that $\mathbb{Q}(\sqrt2, \sqrt3)$ is not isomorphic to $\mathbb{Q}(\sqrt[4]2)$.

What I did so far was to show that $\mathbb{Q}(\sqrt2,\sqrt3) =\text{span}\{1,\sqrt2,\sqrt3,\sqrt6\}$.

Assuming they are isomorphic, denoting $f$ the isomorphism, then $f(1) = 1$, so $x^2-3$ has a root in $\mathbb{Q}(\sqrt2,\sqrt3)$ so $\sqrt 3\in\mathbb{Q}(\sqrt[4]2)$. ($\sqrt2 \in \mathbb{Q}(\sqrt[4]2)$, $\sqrt[4]2\cdot\sqrt[4]2=\sqrt2)$.

Thus under the assumption, $\mathbb{Q}(\sqrt2, \sqrt3)\subseteq \mathbb{Q}(\sqrt[4]2)$. As a vector spaces of the same dimension over $\mathbb{Q}$, it means that $\mathbb{Q}(\sqrt2, \sqrt3)= \mathbb{Q}(\sqrt[4]2)$.

Now all is left to do it to show a contradiction: $\sqrt[4]2\in\mathbb{Q}(\sqrt2,\sqrt3)$, then $\sqrt[4]2 = a+b\sqrt2+c\sqrt3+d\sqrt6$ so $2=(a+b\sqrt2+c\sqrt3+d\sqrt6)^4$, I believe that opening the expression, I would find out that it can't be an element of $\mathbb{Z}$. However this is a really inelegant way (if it really works), and I wish for a better way to do so.

An other approach: $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2 + \sqrt3)$ by the irreducibility of $x^4-10x^2+1$ over $\mathbb{Q}$ (an some dimension theorem usage). Also, $x^4-10x^2+1$ splits in $\mathbb{Q(\sqrt2, \sqrt3)}$.

Is there a way to show that this polynomial doesn't split over $\mathbb{Q}(\sqrt[2]4)$? with different method then assuming that $\sqrt3\in\mathbb{Q}(\sqrt[4]2)$ and getting a contradiction? Maybe the fact that $\sqrt[4]2$ has complex roots might help?

user26857
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user5721565
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4 Answers4

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$\mathbb{Q}(\sqrt[4]2)$ has exactly two automorphisms: those induced by $\sqrt[4]2 \mapsto \pm\sqrt[4]2$.

$\mathbb{Q}(\sqrt2, \sqrt3)$ has at least four automorphisms: those induced by $\sqrt2 \mapsto \pm\sqrt2$ and $\sqrt3 \mapsto \pm\sqrt3$.

lhf
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Another approach: $\Bbb Q(\sqrt2,\sqrt3)$ is normal over $\Bbb Q$, as it's the splitting field of $(X^2-2)(X^2-3)$.

But $\Bbb Q(\sqrt[4]2)$ is not normal over $\Bbb Q$. The number $\sqrt[4]2$ has minimal polynomial $X^4-2$ which also has the zero $i\sqrt[4]2\notin\Bbb Q(\sqrt[4]2)$.

Angina Seng
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Compute the discriminants of the fields $\Bbb Q(\sqrt{2},\sqrt{3})=\Bbb Q(\sqrt{2}+\sqrt{3})$ and $\Bbb Q(\sqrt[4]{2})$. They are $2304=2^{8}\cdot 3^2$ and $-2048=-2^{11}$ respectively. Since they are different, the fields are not isomorphic.

References:

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

Discriminant of Number Fields

Dietrich Burde
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  • could you double-check and/or explain the calculation of the discriminant of $\mathbb (\sqrt2,\sqrt3)$? – J. W. Tanner Apr 20 '20 at 18:39
  • @J.W.Tanner It is $2\cdot 3\cdot6\cdot8^2=2^8\cdot 3^2=2304$, see here. – Dietrich Burde Apr 20 '20 at 18:53
  • That link suggests $\det\begin{vmatrix}1&&1&&1&&1\1&&-1&&1&&-1\1&&1&&-1&&-1\1&&-1&&-1&&1\end{vmatrix}=8$, but I got $16$ – J. W. Tanner Apr 20 '20 at 19:16
  • Thank you, and I don't question that your approach works to answer OP's question (+$1$); I just questioned the particular value of that discriminant (I actually got $147456=2^{14}3^2$ not using the link you gave, but I could be wrong) – J. W. Tanner Apr 20 '20 at 21:11
  • I used a few other methods -- one of them was this formula for $x^4-10x^2+1$ – J. W. Tanner Apr 20 '20 at 21:20
  • @J.W.Tanner I used the definition of discriminant for number fields. There is also a discriminant of a (minimal) polynomial. Pari confirms that $\operatorname{poldisc}(x^4-10x^2+1)=147456$, but $\operatorname{nfdisc}(x^4-10x^2+1)=2304$ for $K=\Bbb Q(\sqrt{2}+\sqrt{3})$. – Dietrich Burde Apr 21 '20 at 08:27
  • Thanks again. I was wondering if it was something like that. I suppose it's like the example that $\mathbb Z(\sqrt5)$ is not the entire ring of integers in $\mathbb Q(\sqrt5)$. Apparently that link you provided calculated the correct number field discriminant with a flawed method ($8$ vs. $16$) – J. W. Tanner Apr 21 '20 at 14:06
  • Actually that link has $2^8\cdot3^2$ as the intermediate answer and $2^6\cdot3^2$ as the final answer, but based on our determinant calculations it should be $2^{10}\cdot3^2$ as the intermediate answer and $2^8\cdot3^2$ as the final answer – J. W. Tanner Apr 22 '20 at 00:35
  • Yes, it happens that there are documents in the web having typos or errors. I linked a file which I didn't verify myself, because the final result seemed correct. Next time I will be more careful, or just say " I computed it with pari". – Dietrich Burde Apr 22 '20 at 09:01
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It seems to me the easiest solution is likely to be to show that $x^2-3$ has no solutions in $\Bbb Q [\sqrt[4]{2}]$.

Robert Shore
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  • but this is just has showing $\sqrt(3) \notin \mathbb{Q}(\sqrt[4]2)$ which is described above. Am I wrong? – user5721565 Feb 28 '19 at 08:40
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    You're not wrong. I just suspect that focusing on that element is probably simplest. But I'll confess I've tried to expand the formula and it's not immediately obvious to me that it has no solution with coefficients in $\Bbb Q$. – Robert Shore Feb 28 '19 at 08:45