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Consider $A$ and $B$ two intermediate fields of the field extension F/K.

I have already proved that $[AB:K]\leq {[A:K][B:K]\over [A\cap B: K]}$.

I would like to find a simple example (for example, taking K= $\mathbb Q)$ in which it doesn't hold the equality (a counterexaple to show that the equality only happens sometimes) , but all the examples that come to my mand follow the equality.

Do you have any idea?

Thank you so much!

Trian
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2 Answers2

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Let $K = \mathbb{Q}$, $A = \mathbb{Q}(2^{1/3})$ and $B = \mathbb{Q}(\zeta2^{1/3})$ (here $\zeta$ denotes a primitive cube root of one, so that $\zeta2^{1/3}$ is one of the other roots of the polynomial $x^3 - 2$).

The right hand side of your inequality is nine since these fields obviously intersect to $\mathbb{Q}$ (being unequal extensions of degree 3). The left hand side is six because the compositum of $A$ and $B$ is the splitting field of $x^3 - 2$, which splits in any field containing $2^{1/3}$ and $\zeta$, but the minimal polynomial of $\zeta$ over $\mathbb{Q}$ is just $x^2 + x + 1$.

hunter
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  • if you haven't encountered cube roots of unity before, it may be easier to picture the example where $A = \mathbb{Q}(2^{1/4})$ and $B = \mathbb{Q}(2^{1/4}i)$. – hunter Mar 04 '14 at 14:54
  • In this case, $AB=\mathbb Q(2^{1/4},2^{1/4}i)$ and which would be$[AB:\mathbb Q]$??? – Trian Mar 04 '14 at 15:05
  • @Triana in that case $[AB: \mathbb{Q}] = 8$ (since you can get it by adjoining $i$ to $\mathbb{Q}(2^{1/4})$.) – hunter Mar 04 '14 at 15:09
  • And in the other side we have 16, because the intersection is $\mathbb Q$? Thank you @hunter – Trian Mar 04 '14 at 15:13
  • @Triana uh-oh, sorry. the example in the comments is no good -- the intersection is $\mathbb{Q}(\sqrt{2})$ and we get $8 = 8$. The example in the answer is still correct. – hunter Mar 04 '14 at 15:40
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I think useful motivation would be to consider an extension of $AB$ (call $L$) that is Galois over $A\cap B$. Then if $A= Fix(G_1)$ and $B= Fix(G_2)$, then $AB = Fix(G_1\cap G_2)$ and $A\cap B = Fix(G_1G_2)$ (where $G_1G_2$ is the minimal group that contains $G_1$ and $G_2$), so now you get an equivalent problem on groups, which is more intuitive.

Kai Wang
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  • There is another idea using finite etale algebras. Note $[K \colon \mathbb{Q}]$ is the dimension of $L\otimes_{\mathbb{Q}} K$ as a finite etale $L$-algebra. $[KL\colon L ] = [K\colon \mathbb{Q}]$ is equivalent to $L\otimes_{\mathbb{Q}} K$ being a field. In this case, since $K/\mathbb{Q}$ is a simple extension, let $\alpha$ generate $K$ as a $\mathbb{Q}$-algebra and let $f$ be the minimal poly of $\alpha$ over $\mathbb{Q}$. This is equivalent to $f$ remains irreducible (inert) in $L$. – Kai Wang Jul 21 '24 at 15:34