$K(u,v)=K(u)(v)$ true or false

Question

Is it true that if $K$ is a field and $u,v$ are algebraic over $K$ then $K(u,v)=K(u)(v)$ for example $Q(\sqrt{2},\sqrt{3})=Q(\sqrt{2})(\sqrt3))$ if yes can someone provide a proof?

Answer 1

$K(X)=\langle K \cup X \rangle $ by definition, i.e. $K(X)$ is the field generated by $K \cup X$.

$K(u,v) \subset K(u)(v)$, proof:

$$K(u,v)=\langle K \cup \{u,v\}\rangle =\langle K \cup \{u\}\cup \{v\}\rangle\subset\langle \langle K \cup \{u\}\rangle \cup \{v\}\rangle=\langle K(u)\cup \{v\}\rangle=K(u)(v).$$

$K(u)(v) \subset K(u,v)$, proof:

$$K(u)(v)=\langle K(u)\cup \{v\}\rangle=\langle \langle K \cup \{u\}\rangle \cup \{v\}\rangle \subset \langle \langle K \cup \{u\}\cup \{v\}\rangle \cup \{v\}\rangle= \langle K(u,v)\cup \{v\}\rangle =\langle K(u,v)\rangle=K(u,v).$$

Hence, $K(u)(v) = K(u,v).$