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I would like to know if it is true that $\mathbb{Q}(\sqrt{2}-i, \sqrt{3}+i) = \mathbb{Q}(\sqrt{2}-i+2(\sqrt{3}+i))$.

I can prove, that $\mathbb{Q}(\sqrt{2}-i, \sqrt{3}+i) = \mathbb{Q}(\sqrt{2},\sqrt{3},i)$, so the degree of this extension is 8. Would it be enough to show that the minimal polynomial of $\sqrt{2}-i+2(\sqrt{3}+i)$ has also degree 8?

It follows from the proof of the primitive element theorem that only finitely many numbers $\mu$ have the property that $\mathbb{Q}(\sqrt{2}-i, \sqrt{3}+i)\neq \mathbb{Q}(\sqrt{2}-i+\mu(\sqrt{3}+i))$. Obviously $\mu=1$ is one of them, but how to check, whether 2 also has this property?

Thanks in advance,

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Yes; since $\mathbb{Q}(\alpha)$ (with $\alpha=\sqrt{2}-i+2(\sqrt{3}+i)$ is clearly contained in $\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i)$, the two fields are equal if and only if they have the same degree over $\mathbb{Q}$, if and only if the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $[\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i):\mathbb{Q}]$. –  Arturo Magidin Apr 1 '12 at 2:00
    
It's not the primitive element theorem itself that guarantees that only finitely many $\mu$ exist with that property, though the argument made in the proof is valid. –  Arturo Magidin Apr 1 '12 at 2:01
    
@Arturo: I don't see any reason why you shouldn't just call your comments an answer. Right? –  mixedmath Apr 1 '12 at 17:12
    
@mixedmath: Done. –  Arturo Magidin Apr 1 '12 at 20:08
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up vote 1 down vote accepted

Let $\alpha=\sqrt{2}-i+2(\sqrt{3}+i)$.

Since $\alpha\in\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i)$, it follows that $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i)$ if and only if their degrees over $\mathbb{Q}$ are equal. The degree $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is equal to the degree of the monic irreducible of $\alpha$ over $\mathbb{Q}$, so you are correct that if you can show that the monic irreducible of $\alpha$ is of degree $8$, then it follows that $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i)$.

I will note, however, that your interpretation of the Primitive Element Theorem is incorrect. The Theorem itself doesn't really tell you what you claim it tells you. The argument in the proof relies on the fact that there are only finitely many fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i)$, and so by the Pigeonhole Principle there are only finitely many rationals $\mu$ such that $\mathbb{Q}(\sqrt{2}-i,\sqrt{3}+i)\neq\mathbb{Q}(\sqrt{2}-i+\mu(\sqrt{3}+i))$. But this is not a consequence of the Primitive Element Theorem, but rather of the fact that there are only finitely many fields in between.

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