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Is $\mathbb{Q}(\sqrt 2,\sqrt 3,\sqrt 5)$ a simple extension?

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It is a general fact that a finite separable extension is a simple extension (the primitive element theorem, – Akhil Mathew Mar 30 '11 at 1:16

As Akhil mentions, the Primitive Element Theorem says that any finite separable extension of fields is a simple extension. Because the characteristic of $\mathbb{Q}$ is 0, any extension of $\mathbb{Q}$ is separable, and $$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{2},\sqrt{3})][\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2\cdot2\cdot 2=8$$ is finite. Thus the theorem guarantees that $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}$ is a simple extension.

Here is an explicit generator: $\alpha=\sqrt{2}+\sqrt{3}+\sqrt{5}$.

This is because the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is (as computed here) $$\text{Irr}(\alpha,\mathbb{Q})=x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$$ hence $8=\text{deg}(\text{Irr}(\alpha,\mathbb{Q}))=[\mathbb{Q}(\alpha):\mathbb{Q}]$. Thus $[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]$, and because $\mathbb{Q}(\alpha)\subseteq\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, this forces $$\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}).$$

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That's a good answer that could be better if you explicitly told why the degree is 8. – lhf Mar 30 '11 at 1:52

You can start by showing $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$ simply by showing you can get all the elements of one set using elements from the other. Clearly, the left hand side contains the right hand side so you just need to show that you can make $\sqrt{2}$ and $\sqrt{3}$ from polynomials in $\sqrt{2}+\sqrt{3}$.

In general, any finite extension of the rationals is a simple extension.

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