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I know there are elementary proofs of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ being irrational. Just want to apply field extension to solve it.

Attempt: It is sufficient to show $\mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{5})$ being a non-trivial extension. We know the minimal polynomial for $\sqrt2,\sqrt3,\sqrt5$ are \begin{align*} x^2-2,\quad x^2-3,\quad x^2-5 \end{align*} Galois group (fixing $\mathbb{Q}$) should permute the roots of within each minimal polynomials. $\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5)$ is Galois because it is the splitting field of a separable polynomial. (or the Galois group has the same cardinality as the extension dimension.)

If an automorphism in Galois group fixes $\sqrt2+\sqrt3+\sqrt5$, it should fix all the roots of the above minimal polynomials. Thus it is the identity. The Galois correspondence shows \begin{align*} Aut_\mathbb{Q}(\mathbb{Q}(\sqrt2+\sqrt3+\sqrt5)) \cong Aut_\mathbb{Q}(\mathbb{Q}(\sqrt2, \sqrt3, \sqrt5))/\{1\} = Aut_\mathbb{Q}(\mathbb{Q}(\sqrt2, \sqrt3, \sqrt5)) \end{align*} The RHS has cardinality 6, which shows $\mathbb{Q}(\sqrt2+\sqrt3+\sqrt5)$ is a non-trivial extension. (Indeed, I think this should suggest $\mathbb{Q}(\sqrt2+\sqrt3+\sqrt5) = \mathbb{Q}(\sqrt2, \sqrt3, \sqrt5)$).

The only thing I am not quite sure the the statement "Galois group (fixing $\mathbb{Q}$) should permute the roots of within each minimal polynomials'. I haven't find this explicitly in the textbook I have used (Algebra chapter 0), but somehow I feel this should be true.

My attempt for this statement is: Any automorphism fixing $\mathbb{Q}$ should permute within each minimal polynomial. On the other hand, I have to show all such permutation can be extended to an Automorphism. Extend sequentially: For example we know \begin{align*} \mathbb{Q}(\sqrt2, \sqrt3, \sqrt5) = \mathbb{Q}(\sqrt{2})(-\sqrt2)(\sqrt3)(-\sqrt3)(\sqrt5)(-\sqrt5) \end{align*} The textbook I used has the statement saying the extension of automorphism exists for simple field extension. For example, extend an isomorphism from $\mathbb{Q}(\sqrt2)$ to $\mathbb{Q}(-\sqrt2)$.

Then apply the extension sequentially, should be a global extension.

Zorualyh
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    I did something like this here. More or less in response to a more general variant. Do check out also Bill Dubuque's answer that does not use Galois theory at all. My answer uses Galois theory to reduce the question to a simpler one: equivalent to showing that $\sqrt3\notin\Bbb{Q}(\sqrt2)$. – Jyrki Lahtonen Aug 10 '23 at 03:39
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    Anyway, the field $\Bbb{Q}(\sqrt2, \sqrt3,\sqrt5)=\Bbb{Q}(\sqrt2+\sqrt3+\sqrt5)$ is a degree eight Galois extension of $\Bbb{Q}$. So you should recalibrate your intuition a bit. There are eight automorphisms, not six. – Jyrki Lahtonen Aug 10 '23 at 03:41
  • @JyrkiLahtonen Thanks for comment!. You are right, it should be 8. The proof will work if we apply the '$2^n$' result. – Zorualyh Aug 10 '23 at 04:23
  • @JyrkiLahtonen I revisited my proof. It seems we only need to know $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ is Galois (without knowing if $\sqrt5\not\in \mathbb{Q}(\sqrt2,\sqrt3))$. Then apply Galois correspondence to show the automorphism over $\mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{5})$ is non-trivial. – Zorualyh Aug 10 '23 at 04:27
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    This is generalised in https://math.stackexchange.com/a/30707/72031 – Paramanand Singh Aug 10 '23 at 04:34
  • $L=\Bbb{Q}(\sqrt2,\sqrt3,\sqrt5)$ is trivially Galois (over $\Bbb{Q}$) because it is the splitting field of $(x^2-2)(x^2-3)(x^2-5)$ and separability is given. You do need to determine by some other means whether $\sqrt5\in\Bbb{Q}(\sqrt3,\sqrt3)$ for otherwise you won't know the degree of the extension, nor the number of automorphisms. There are general results that can be applied, but you should explain how you are using them. – Jyrki Lahtonen Aug 10 '23 at 12:02

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