Can you give me an example of two number fields $L_1$ and $L_2$ such that $L_2/L_1\cap L_2$ is Galois but $L_1L_2/L_2$ is not Galois.

I think that if $L_2/L_1\cap L_2$ is Galois then $L_1L_2/L_2$ is also.. Am I right?

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    $\begingroup$ In the extension $L_2/L_1\cap L_2$ the adjoined elements come from $L_2$ whereas in the extension $L_1L_2/L_2$ the adjoined elements come from $L_1$. Therefore you should not expect any kind of relation between these two extension. One may have a property the other lacks. OTOH if you compare $L_1/L_2\cap L_1$ and $L_1L_2/L_2$, then you adjoining stuff from $L_1$ to some field, and can expect similarities. $\endgroup$ – Jyrki Lahtonen Oct 30 '15 at 11:41
  • $\begingroup$ thank you, my question is a little stupid yes :) $\endgroup$ – corcia candy Nov 1 '15 at 10:31

As $L_1L_2/L_2$ is not Galois, let's choose a number field $L_1$ such that $L_1/\mathbb{Q}$ is not Galois, eg $L_1=\mathbb{Q}(\sqrt[3]{2})$.

Now we want $L_2/L_1 \cap L_2$ to be Galois, so we ideally take a disjoint extension of $\mathbb{Q}$ to $L_2$ (by which I mean the intersection is just $\mathbb{Q}$. Now the obvious choice here is $L_2=\mathbb{Q}$ which does indeed work but for a more interesting example I will take $L_2=\mathbb{Q}(i)$.

Then $L_2/L_1 \cap L_2$ is just $\mathbb{Q}(i)/\mathbb{Q}$ so is Galois but $L_1L_2/L_2$ is $\mathbb{Q}(\sqrt[3]{2},i)/\mathbb{Q}(i)$ is not Galois since the polynomial $x^3-2 \in \mathbb{Q}(i)[x]$ obtains one root in $L_1L_2$ but not all so the extension cannot be normal and hence not Galois.

  • $\begingroup$ first thank you..:) but I have one more question, Does $L_1L_2$ mean directly $Q(\sqrt2, i)$? Is it because of defn? or something? $\endgroup$ – corcia candy Nov 1 '15 at 10:32
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    $\begingroup$ $L_1L_2$ is the compositum field which is defined as the smallest field that contains both $L_1$ and $L_2$ so in the example it does give $\mathbb{Q}(\sqrt[3]{2},i)$ by definition. $\endgroup$ – Matt B Nov 1 '15 at 10:58

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