Exercise 4.1.4

Proof. $F({\alpha }_1,\dots ,{\alpha }_r)\subset F({\alpha }_1,\dots ,{\alpha }_n),1\le r\le n$, since $F({\alpha }_1,\dots ,{\alpha }_n)$ contains $F$ and ${\alpha }_1,\dots ,{\alpha }_r$, and since $F({\alpha }_1,\dots ,{\alpha }_r)$ is the smallest subfield of $L$ containing $F$ and ${\alpha }_1,\dots ,{\alpha }_r$.

Moreover $F({\alpha }_1,\dots ,{\alpha }_n)$ contains ${\alpha }_{r+1},\dots ,{\alpha }_n$.

By Lemma 4.1.9, $F({\alpha }_1,\dots ,{\alpha }_r)({\alpha }_{r+1},\dots ,{\alpha }_n)$ is the smallest subfield of $L$ containing $F({\alpha }_1,\dots ,{\alpha }_r)$ and ${\alpha }_{r+1},\dots ,{\alpha }_n$, thus

From the reciprocal inclusion proved in section 4.1, we conclude that

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