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

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

• Prove by induction Commented Oct 13, 2020 at 17:03
• @Wuestenfux I tried same but didn't concluded Commented Oct 13, 2020 at 17:06
• @Wuestenfux,is the result true in general? Commented Oct 13, 2020 at 17:07

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