order which we add elements to field extension doesn't matter Suppose I am given a field $F$. Suppose also we have $a, b$ which are algebraic over $F$. In every book I have read, it was written that it is trivial to show that $F(a)(b)=F(b)(a)$. I tried to prove this equality algebraically using ring homomorphism but could not. How is this equality is usually shown?
 A: The fact that $a$ and $b$ are algebraic is irrelevant.  We have a field $K$ with a subfield $F\subseteq K$ and two elements $a,b\in K$, and wish to show that $F(a)(b)=F(b)(a)$.  By definition, $F(a)$ is the smallest subfield of $K$ containing $F$ and $a$.  By definition again, $F(a)(b)$ is the smallest subfield of $K$ containing $F(a)$ and $b$.  Since a subfield of $K$ contains $F(a)$ iff it contains both $F$ and $a$, $F(a)(b)$ is the smallest subfield of $K$ containing $F$, $a$, and $b$.
Similarly, $F(b)(a)$ is the smallest subfield of $K$ containing $F$, $b$, and $a$.  Since "and" is commutative (that is, a subfield of $K$ contains $b$ and $a$ iff it contains $a$ and $b$), this is the same thing as $F(a)(b)$.
A: Hint: prove it for rings first, namely $F[a][b]=F[b][a]$. Then use the fact that $a$ and $b$ are algebraic over $F$.
A: we can use the theorem:

Theorem: Let be $K$ is extension field of $k$ and $\alpha_1,\alpha_2,...,\alpha_n \in K$ algebraic over $k$ and $\sigma\in S_n$ then:
$$k(\alpha_1)(\alpha_2)...(\alpha_n)=k(\alpha_1,\alpha_2,...,\alpha_n)= k(\alpha_{\sigma(1)},\alpha_{\sigma(2)},...,\alpha_{\sigma(n)})=k(\alpha_{\sigma(1)})(\alpha_{\sigma(2)})...(\alpha_{\sigma(n)})$$

