# Is $\mathbb{Q}(\sqrt2, \sqrt3)$ isomorphic to $\mathbb{Q}(\sqrt[4]2)$?

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?

$$\mathbb{Q}(\sqrt[4]2)$$ 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 $$\sqrt3 \mapsto \pm\sqrt3$$.

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 $$i\sqrt[4]2\notin\Bbb Q(\sqrt[4]2)$$.

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

It seems to me the easiest solution is likely to be to show that $$x^2-3$$ has no solutions in $$\Bbb Q [\sqrt[4]{2}]$$.

• but this is just has showing $\sqrt(3) \notin \mathbb{Q}(\sqrt[4]2)$ which is described above. Am I wrong? – dan Feb 28 at 8:40
• 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 at 8:45