Let E be an extension of F and let a,b∈E. Prove that F(a,b)=F(a)(b)=F(b)(a).
This question was posed 11 months ago, and @GregoryGrant offered a brief explanation on the concept posed in the question: "You have to show rational functions in a and b with coefficients in F give the same space as rational functions in a with coefficients given by rational functions in b."
Running into this problem, I have utilized the Extension field definition to outline that F is a subfield of E. This shows that E is a vector space over F, with arbitrary elements a, b.
Arbitrary elements of E are vectors, with scalars elements of F. So from here, do you assume a and b are elements of F to use the splitting field?
Lost on this one!
$F(a,b)$ is the smallest field containing $a$,$b$ and $F$. Also, let $K=F(a)$. Then $K$ is the smallest field containing $F$ and $a$. Hence $K(b)$ is the smallest field containing $K$ and $b$. Now, clearly,
$F(a,b) \subseteq K(b)=F(a)(b)$, as $K(b)$ contains $F,a$ and $b$, hence must contain the smallest field containing $F,a$ and b.
Also, $F(a) \subseteq F(a,b)$, as $F(a,b)$ contains $F$ and $a$, hence contains the smallest field containing $F$ and $a$. Therefore $K \subseteq F(a,b)$. Hence, since $F(a,b)$ contains $K$ and $b$, it must contain the smallest field containing $K$ and $b$, i.e., $K(b)$. So we get that $K(b) \subseteq F(a,b)$.
Hence $F(a,b)=K(b)=F(a)(b)$, which, similarly, equals $F(b)(a)$.