This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise E2.
Recall the definition of $F(a)$. It is a field such that (i) $F \subseteq F(a)$; (ii)$a \in F(a)$; (iii) any field containing $F$ and $a$ contains $F(a)$.
Proves that $F(a^2) \subseteq F(a)$ and $F(a + b) \subseteq F(a,b)$. [$F(a, b)$ is the field containing $F, a$, and $b$, and contained in any other field containing $F, a$ and $b$.] Why are the reverse inclusions not necessarily true?
$a \in F(a)\implies a^2\in F(a)$ by closure of ring multiplication $\implies F(a^2)\subseteq F(a)$ by (iii). However, $a$ is not necessarily a member of $F(a^2)$. For example, $\sqrt{2}$ is not a member of $\Bbb{Q}(2)$. Hence, $F(a)\subseteq F(a^2)$ does not hold.
$a+b \in F(a,b)$ by closure of group addition $\implies F(a+b)\subseteq F(a,b)$ by (iii). $ab\in F(a, b)$ by closure of ring multiplication. However, $ab$ is not necessarily a member of $F(a+b)$. For example, $\sqrt{2}$ is not a member of $\Bbb{Q}(1+\sqrt{2})$. Hence, $F(a,b)\subseteq F(a+b)$ does not hold.
Correct?
I think you mean "$a, b$" (as in "$a$ and $b$") instead of "$ab$" a few places, but otherwise it's all correct.
However, your "is not necessarily a member" claims are completely without proofs. In this case the best proof would be an example.
Add a concrete example for each "is not necessarily a member" claim, and this is perfect. For instance, $\sqrt 2$ is well-known to not be a member of $\Bbb Q(\sqrt2^2)=\Bbb Q$.