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Let $E/K$ be a field extension, and let $L_1$ and $L_2$ be intermediate fields of finite degree over $K$.

Prove that $[L_1L_2:K] = [L_1 : K][L_2 : K]$ implies $L_1\cap L_2 = K$.

My thinking process so far:

I've gotten that $K \subseteq L_1 \cap L_2$ because trivially both are intermediate fields over K.

I want to show that $L_1 \cap L_2 \subseteq K$, or equivalently that any element of $L_1 \cap L_2$ is also an element of $K$. So I suppose there exists some element $x\in L_1 \cap L_2\setminus K$. Well then I know that this element is algebraic over $K$, implying that $L_1:K=[L_1:K(x)][K(x):K]$, and similarly for $L_2$, implying that that these multiplied together equal $L_1L_2:K$ by hypothesis. And now I’m stuck in the mud... not knowing exactly where the contradiction is.

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You should prove that $[L_1 L_2 : L_1] = [L_2 : L_1 \cap L_2]$. – Zhen Lin Sep 12 '12 at 5:06
up vote 3 down vote accepted

The assumption implies $[L_1L_2:L_1]=[L_2:K]$. Hence $K$-linearly independent elements $b_1,\ldots ,b_m\in L_2$ are $L_1$-linearly independent, considered as elements of $L_1L_2$. In particular this holds for the powers $1,x,x^2,\ldots ,x^{m-1}$ of an element $x\in L_2$, where $m$ is the degree of the minimal polynomial of $x$ over $K$. Thus $[K(x):K]= [L_1(x):L_1]$. This shows that the minimal polynomial over $K$ of every $x\in L_1\cap L_2$ has degree $1$.

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