A fact about simple algebraic field extensions

Let $F$ be a field and let $E$ be an extension of $F$. Let $\alpha,\beta \in E$ be algebraic over $F$ with minimal polynomials $a(x)$ and $b(x)$, respectively. Show that $(F(\alpha))(\beta) = (F(\beta))(\alpha)$.

Recall that $F(\alpha)$ has a kind of universal property with respect to fields containing an element whose minimal polynomial over $F$ is $a(x)$. Since $(F(\beta))(\alpha)$ contains an element which is algebraic over $F$ with minimal polynomial $a(x)$, we have an injective ring homomorphism $F(\alpha) \rightarrow (F(\beta))(\alpha)$ fixing $F$ and $\alpha$. Similarly, we have an injective map $(F(\alpha))(\beta) \rightarrow (F(\beta))(\alpha)$ fixing $F$, $\alpha$, and $\beta$. Since every element of $(F(\beta))(\alpha)$ is a linear combination of $\alpha^i\beta^j$, this map is the identity. So $(F(\alpha))(\beta) = (F(\beta))(\alpha)$.