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Here's what I've thought so far. Unless I'm very much mistaken, we have that $[\mathbb{Q}(\sqrt{3},\sqrt{5}):\mathbb{Q}] = 15$, so I'm looking for a $\theta$ such that $[\mathbb{Q}(\theta):\mathbb{Q}] = 15$. Because it's supposed to be a simple extension, I need the minimal polynomial of $\theta$ over $\mathbb{Q}$ to be of degree $15$. I'm quite sure $\theta$ cannot be $15$, although I have only briefly sketched out why I think that is.

I'm looking for some hints on how to approach this problem. I'll update this post with an edit once I have "solved" it and hopefully someone will tell me if it's correct or not.

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    $\begingroup$ $\Bbb Q(\sqrt 3, \sqrt 5)$ is the splitting field of the fourth (and not fifteenth) degree polynomial $(x^2 - 3)(x^2 - 5)$. You're thinking of $[\Bbb Q(\sqrt[3]a, \sqrt[5]b):\Bbb Q] = 15$. $\endgroup$
    – Arthur
    Apr 26, 2016 at 15:24
  • $\begingroup$ @Arthur Oh my god. I can't believe I did that...Now it all makes much more sense, thank you for pointing that out. $\endgroup$
    – Auclair
    Apr 26, 2016 at 15:27
  • $\begingroup$ As for the question itself, you do, of course, have the boring solution $\theta = -\sqrt3 - \sqrt5$. I assume you want something at least a bit less trivial. $\endgroup$
    – Arthur
    Apr 26, 2016 at 15:29
  • $\begingroup$ @Arthur Exactly. That was my initial thought, but it seemed too easy so I quickly cast it aside. $\endgroup$
    – Auclair
    Apr 26, 2016 at 15:30
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    $\begingroup$ I think that most numbers of the form $a\sqrt3 + b\sqrt 5$ works. You could also throw in ${}+c\sqrt{15}$ to spice it up. I don't think there are any other $\theta$ that works. $\endgroup$
    – Arthur
    Apr 26, 2016 at 15:32

1 Answer 1

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You should be thinking geometrically. The field $K=\Bbb Q(\sqrt3,\sqrt5\,)$ is of degree four over $\Bbb Q$, as the comments have pointed out. There are only four proper subfields, namely $\Bbb Q$, $\Bbb Q(\sqrt3\,)$, $\Bbb Q(\sqrt5\,)$, and $\Bbb Q(\sqrt{15}\,)$. Each of the fields involving a square root is two-dimensional. So here you are, in a four-dimensional space, with three two-dimensional subspaces, all intersecting in the one-dimensional subspace $\Bbb Q$.

You haven’t come anywhere near exhausting all the elements of $K$ — those fields are thin closed subsets of the whole of $K$. Any element $\alpha$ of $K$ that isn’t in the union of those three will have the property that $\Bbb Q(\alpha)=K$.

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  • $\begingroup$ Interesting. I'll have another go at it in the morning. Thanks for the reply. $\endgroup$
    – Auclair
    Apr 26, 2016 at 22:10
  • $\begingroup$ I don't understand the last sentence. The union of the three subfields is $K$. Shouldn't we be looking for an element $\alpha\in K$ such that $\alpha$ is not contained in one of the subfields? Because then $\mathbb{Q}(\alpha)$ is contained in $K$, with degree greater than two over $\mathbb{Q}$. Thus it must be $K$. $\endgroup$
    – Ryan
    Apr 27, 2016 at 8:56
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    $\begingroup$ The union of the three two-dimensional subspaces in the four-dimensional space is not $K$. You’re confusing the field-theoretic join (“compositum”) with set-theoretic join, which is the union. The rest of your argument is correct. $\endgroup$
    – Lubin
    Apr 27, 2016 at 13:13

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