Show that $[\mathbb{F}(a_1,a_2,....a_n):\mathbb{F}]=[\mathbb{F}(a_1):\mathbb{F}].[\mathbb{F}(a_2):\mathbb{F}].......[\mathbb{F}(a_n):\mathbb{F}]$

Question

There is an exercise in Gallian's contemporary abstract algebra to show that:Let $\mathbb{F}$ be a field and E be its extension,if $a_1,a_2\in E$ are algebraic over $\mathbb{F}$ and let degree of $a_1,a_2$ be relatively prime then $[\mathbb{F}(a_1,a_2):\mathbb{F}]=[\mathbb{F}(a_1):\mathbb{F}].[\mathbb{F}(a_2):\mathbb{F}]$

I want to know whether it is true in general i.e:

If $\mathbb{F}$ be a field and E be its extension,if $a_1,a_2...a_n\in E$ are algebraic over $\mathbb{F}$ and let degree of $a_i's$ be relatively prime for each i then $[\mathbb{F}(a_1,a_2,....a_n):\mathbb{F}]=[\mathbb{F}(a_1):\mathbb{F}].[\mathbb{F}(a_2):\mathbb{F}]...[\mathbb{F}(a_n):\mathbb{F}]$

Answer 1

Proof sketch.

For one thing, we can show

$$ [F(a_1,\cdots,a_k):F(a_1,\cdots,a_{k-1})] \le [F(a_k):F] $$

for each $k$. To do this, note the minimal polynomial of $a_k$ over $F$ (whose degree is the same as the dimension $[F(a_k):F]$) is divisible by the minimal polynomial of $a_k$ over $F(a_1,\cdots,a_{k-1})$. (Why?)

Thus, $[F(a_1,\cdots,a_n):F]\le [F(a_1):F]\cdots[F(a_n):F]$.

Then, you can show $[F(a_1,\cdots,a_n):F]$ is divisible by each of the $[F(a_i):F]$s to conclude.