I cannot seem to find the answer to this question on the internet. It is a question about field extensions for an element $a,b \neq F$ but in some extension $K$.
I am wondering if $F(a,b)= \lbrace p(a,b) : p(x,y) \in \mathbb[x,y] \rbrace$. It seems like it should be the case since $F(a)= \lbrace p(a) : p(x) \in F[x] \rbrace$. Also, out of curiosity, is $F(a,b)=F(a) \cup F(b) $?
Also, if neither of these are the case, a helpful example or explanation would awesome. Any help would be appreciated.
You are correct in your first hunch. If we have fields $F \subset K$ and an element $a \in K$, then by definition, $F(a)$ is defined to be the smallest subfield of $K$ that contains $a$. However, if $a$ is algebraic over $F$, then $F(a) = \{p(a) \ | \ p(x) \in F[x] \}$. For a proof that these are equivalent statements when $a$ is algebraic, see my post here.
Following the above, provided $a$ and $b$ are algebraic over $F$, we have $F(a, b) = F(a)(b) = \{p(b) \ | \ p(x) \in F(a)[x] \}$, and when you think about it, this is equal to the set $\{p(a, b) \ | \ p(x, y) \in F[x, y] \}$.
However, it is not true that $F(a, b) = F(a) \cup F(b)$. To see this, let $F = \mathbb{Q}$, and let $a = \sqrt{2}$ and $b = \sqrt{3}$. Then $\sqrt{6} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$, but $\sqrt{6} \notin \mathbb{Q}(\sqrt{2})$, and likewise $\sqrt{6} \notin \mathbb{Q}(\sqrt{3})$. Therefore, it cannot be the case that $\sqrt{6} \in \mathbb{Q}(\sqrt{2}) \cup \mathbb{Q}(\sqrt{3})$.
The answer by Kaj is good. But I'll add one thing: This only works if $a,b$ are algebraic over $F$. Same goes for $F(a)=\{p(a):p(x)\in F[x]\}$
