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Let $k$ be any field and $L/k$ be a field extension. Suppose $a, b \in L$ are algebraic over $k$. Is the formula $[k(a,b):k][k(a) \cap k(b):k] = [k(a):k][k(b):k]$ true?

This formula comes from page 228 (Remark 2.4) in "An invitation to Arithmetic Geometry" by Prof. Dino Lorenzini. But the author didn't explain why this formula is true. Thank you for four answer.

Peter Hu
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This is false. Consider $k=\Bbb{Q}$, $a=\root3\of2$, $b=e^{2\pi i/3}a$. We have $[k(a,b):k]=6$, $k(a)\cap k(b)=k$ and $[k(a):k]=[k(b):k]=3$.

Jyrki Lahtonen
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  • I will delete this, if we find a good duplicate. I have used this example myself a few times here, but the motivating context has been a bit different. – Jyrki Lahtonen Dec 07 '14 at 07:39
  • Oh! I see! Thank you very much! (so... I think I need to find another way to correct my note...) – Peter Hu Dec 07 '14 at 07:41
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    @Peter the formula does hold if you know that $k(a)$ and $k(b)$ are both Galois over $k$. – Jyrki Lahtonen Dec 07 '14 at 07:51
  • But the formula $[k(a,b):k][k(a) \cap k(b):k] \leq [k(a):k][k(b):k]$ is true. Right? (by my computation...) – Peter Hu Dec 07 '14 at 08:22
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    Correct, @PeterHu. That is a consequence of the inequality $$[k(a,b):k(a)]\le [k(b):k]$$ that is a simple onsequence of the fact that the minimal polynomial of $b$ over $k(a)$ is a factor of the minimal polynomial of $b$ over $k$. – Jyrki Lahtonen Dec 07 '14 at 08:34