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I am asked to show that the splitting field of $X^4-6X^2+4$ is $\mathbb{Q}(\sqrt{3}+\sqrt{5})$. By solving the quartic polynomial, I showed that the splitting field of this polynomial is in fact $\mathbb{Q}(\sqrt{3+\sqrt{5}})$. So my question is if these two fields are equal to each other and if so how.

confused
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    One thing to notice is that this polynomial does not vanish at $\sqrt{3}+\sqrt{5}$... – walkar Jul 03 '23 at 15:06
  • No, the fields are not equal, see here. There might be a typo, so that $X^4-16X^2+4$ was meant. – Dietrich Burde Jul 03 '23 at 15:11
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    @DietrichBurde OK, so that polynomial isn't the minimal polynomial of $\sqrt{3} + \sqrt{5}$. That doesn't immediately imply that $\sqrt{3 + \sqrt{5}}$ generates a different field than $\sqrt{3} + \sqrt{5}$ does. – Daniel Schepler Jul 03 '23 at 15:14
  • @anonymous Both have degree $4$ – jjagmath Jul 03 '23 at 15:14
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    If my calculations are correct, $\sqrt{3+\sqrt{5}} = \frac{1}{2}(\sqrt{2} + \sqrt{10})$. So, you can show $\mathbb{Q}(\sqrt{3+\sqrt{5}}) = \mathbb{Q}(\sqrt{2}, \sqrt{5})$, whose intermediate quadratic fields are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(\sqrt{10})$. But $\sqrt{15} \in \mathbb{Q}(\sqrt{3} + \sqrt{5})$, so... – Daniel Schepler Jul 03 '23 at 15:16
  • Exactly! I was also sensing that there might be a typo in the question. Either the polynomial was supposed to be $X^2-16X+4$ or they wrote the square root wrongly. – confused Jul 03 '23 at 15:17
  • @DanielSchepler yes you are right. Actually the following question is to find intermediate fields by using Galois theory and I also found the same fields. – confused Jul 03 '23 at 15:18
  • @DanielSchepler Great Observations! – Tim Jul 03 '23 at 15:19
  • We don't need the intermediate quadratic fields in order to conclude that $\sqrt3\notin\mathbb Q(\sqrt2,\sqrt5)$. It's just a matter of calculation with radicals. – user26857 Jul 03 '23 at 17:33
  • The fields are not equal and one can prove that $\sqrt{3}\notin\mathbb{Q}(\sqrt{3+\sqrt{5}})$ using the technique applied in this answer : https://math.stackexchange.com/a/4304716/72031 – Paramanand Singh Jul 19 '23 at 08:42

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To summarize the comments, let me mention that the usual exercise is to find the splitting field of $X^4-16X^2+4$. It is given by $\Bbb Q(\sqrt{3},\sqrt{5})=\Bbb Q(\sqrt{3}+\sqrt{5})$, see here:

Determine the minimal polynomial of $\sqrt 3+\sqrt 5$.

However, if really $X^4-6X^2+4$ was meant, then the splitting field is $\Bbb Q(\sqrt{3+\sqrt{5}})$. It is easy to see that $\sqrt{2}\in \Bbb Q(\sqrt{3+\sqrt{5}})$, but $\sqrt{2}$ is not contained in $\Bbb Q(\sqrt{3},\sqrt{5})$ - see here:

How to show that $\sqrt{2}$ is not in $\mathbb Q(\sqrt{3},\sqrt{5})$?

So the fields given in the title are not equal.

Dietrich Burde
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