I'm struggling with this problem. Given $E/F$, $E_1/F$, and $E_2/F$ are field extensions such that $E_1, E_2 \leq E$ and $E_1/F$ and $E_2/F$ are finite, we want to show that $[E_1E_2:E_1] = [E_2 : E_1 \cap E_2]$. Any hints would be helpful.
Showing $[E_1E_2 : E_1] = [E_2 : E_1 \cap E_2]$.
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abstract-algebra
finite-fields
extension-field
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1This can't be right. Is there additional conditions? – 2017-01-29
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0These are the conditions given. It's a possibility there was a typo – 2017-01-29
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1Read my to answers to another post, generally it should be $[E_1E_2 : E_1] [E_1 : E_1\cap E_2]=[E_1E_2 : E_2] [E_2 : E_1\cap E_2]$. There is no reason that $[E_1E_2 : E_1]=[E_2 : E_1\cap E_2]$ must be true without other conditions – 2017-01-29
1 Answers
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Counter example:
$$E_1=\mathbb{Q}(\sqrt[3]{2})$$ $$E_2=\mathbb{Q}(\sqrt[3]{2}\zeta_3)$$
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0What does the $\zeta_3$ denote in this? – 2017-01-30
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0Primitive cube root of unity. – 2017-01-30