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How do I show that $\sqrt{5}\notin \mathbb{Q}(\sqrt{2},\sqrt{3})$?

Since $X^2-5$ is the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ and its degree is not relatively prime to $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]$, it cannot be shown that $X^2-5$ is irreducible in $\mathbb{Q}(\sqrt{2},\sqrt{3})[X]$ in this way.

How do I show this?

Kaj Hansen
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Rubertos
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2 Answers2

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If you know a little Galois theory, then one way of going about this is to start by listing the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$, of which there are only three: $\mathbb{Q}[\sqrt{2}]$, $\mathbb{Q}[\sqrt{3}]$, and $\mathbb{Q}[\sqrt{6}]$. (Galois theory guarantees this list is exhaustive.)

If $\sqrt{5} \in \mathbb{Q}[\sqrt{2}, \sqrt{3}]$, then we necessarily have $\mathbb{Q}[\sqrt{5}] \subsetneq \mathbb{Q}[\sqrt{2}, \sqrt{3}] \implies \mathbb{Q}[\sqrt{5}] \cong \mathbb{Q}[\sqrt{a}]$ for $a = 2, 3, \text{ or } 6$, and a simple argument using the properties of isomorphisms shows that this is not possible.

Kaj Hansen
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  • This is a good approach, but you need also show that $\mathbb{Q}(\sqrt{5})$ is a proper subfield of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ in order to conclude that $\mathbb{Q}(\sqrt{5})\cong \mathbb{Q}(\sqrt{a})$ (since it obviously isn't $\mathbb{Q}$). Thus, instead of using $\mathbb{Q}(\sqrt{5})\cong \mathbb{Q}(\sqrt{a})$ to obtain a contradiction after having shown that it is proper, you can instead just use $\mathbb{Q}(\sqrt{5})\supset \mathbb{Q}(\sqrt{a})$ to obtain a contradiction. – Hayden Apr 19 '15 at 20:14
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    @Hayden, your suggestion would certainly work, but couldn't more elementary arguments can be applied to show that $\mathbb{Q}[\sqrt{5}]$ is a degree $2$ extension of $\mathbb{Q}$, whereas $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ is a degree $4$ extension, thereby guaranteeing a proper subfield? In particular, to show the last argument, one would need only show that $\sqrt{2} \notin \mathbb{Q}[\sqrt{3}]$. – Kaj Hansen Apr 19 '15 at 20:17
  • Yes, you're right that degree arguments immediately show that it can't be. I just wanted to point out that it's something that should be said, and my suggestion at the end is merely a suggestion, not necessarily something I would actually recommend doing. I suppose if one is unfamiliar with degrees, then it's a nice approach though. – Hayden Apr 19 '15 at 20:19
  • Though I suppose if you're unfamiliar with degrees, you probably wouldn't be familiar with the Fundamental Theorem of Galois Theory. XD – Hayden Apr 19 '15 at 20:21
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    Absolutely it should be said (in a student's write-up). I will often leave out chunks of detail on here though since (IMO) it is my task to get people pointed in the right direction -- not to do all their work for them. At any rate, I appreciate the feedback! – Kaj Hansen Apr 19 '15 at 20:22
  • That's a good point, it should be left to the student to fill in such details (unless that's what the question is about). – Hayden Apr 19 '15 at 20:23
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Every nonempty subset of $\,S = \{\sqrt2,\sqrt3,\sqrt 5\}\,$ has product $\not\in \Bbb Q\,$ thus $\,[\Bbb Q(\sqrt2,\sqrt3,\sqrt 5):\Bbb Q] = 8\,$ by this answer. Said informally, multiplicative independence implies linear independence (the reason for such becomes clearer when one studies Galois theory of Kummer extensions).

One can also give a direct elementary proof using the simple Lemma in the linked answer.

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Bill Dubuque
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