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I want to get all the intermediate fields of $L=\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$.

Please help me and give me advice.

3 Answers 3

1

Let $L=\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)$. Then $L/\mathbb Q$ has degree 8. By the way, element of Galois group has to through permute roots of $x^2-2$, permute roots of $x^2-3$ and roots of $x^2-5$. Then all element has order $2$, and thus $$\text{Gal}(L/\mathbb Q)=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z.$$

Find the subgroup of $\text{Gal}(L/\mathbb Q)$ and conclude using the Galois correspondance theorem.

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    Sorry but degree of L over $\mathbb{Q}$ is not equal to 8? – 2017-01-31
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    I think you're confusing degree of the polynomial with degree of extension. degree of L over Q is 8. @Surb – 2017-01-31
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    @SSepehr: yes of course, it's degree 8. Thanks – 2017-01-31
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    @JosephCurwen: I made a mistake as SSepehr mentionned. I indeed made a confusion. I corrected my post – 2017-01-31
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The Galois Group of Your Extenson is $\left(\mathbb{Z} / 2\mathbb{Z}\right)^3$.

Now use the Galois Correspondence to link every subgroup of $\left(\mathbb{Z} / 2\mathbb{Z}\right)^3$ to a subfields of $L$.

Note that these are all the subextension of $L$ (anf that are all Galois Extensions)

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    Please write all the subextension of $L$ specifically. – 2017-01-31
-1

The Galois group of the extension is $G=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, now:

$1)$ Show that besides trivial subgroups, $G$ has $6$ subgroups of order $2$, and $6$ subgroups of order $4$.

$2)$ Conclude that these are all the intermediate fields of $L/\mathbb{Q}$.

$$\mathbb{Q},$$$$\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{5}),\mathbb{Q}(\sqrt{6}),\mathbb{Q}(\sqrt{10}),\mathbb{Q}(\sqrt{15}),$$$$\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}(\sqrt{2},\sqrt{5}),\mathbb{Q}(\sqrt{3},\sqrt{5}),\mathbb{Q}(\sqrt{5},\sqrt{6}),\mathbb{Q}(\sqrt{10},\sqrt{3}),\mathbb{Q}(\sqrt{15},\sqrt{2}),$$$$\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}).$$

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    Because $G$ has seven subgroups of order two and, by duality, also seven subgroups of order four you should have seven intermediate fields of both extensions degrees four and two respectively. You seem to have left out $\Bbb{Q}(\sqrt{30})$ and $\Bbb{Q}(\sqrt6,\sqrt{10})$. – 2017-02-05
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    Each of the seven non-identity elements is of order two, and thus generates a subgroup of order 2. You get the seven subgroups of order four as the orthogonal complements of the orders two subgroups. Think of $G$ as a 3-dimensional vector space over the field of two elements. It has a natural non-degenerate bilinear form (the same recipe as the usual inner product of a 3D-space). – 2017-02-05
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    Sorry about having to downvote this. Just fix the answer, and I will turn it into an upvote and delete the comments. – 2017-02-27