# If $K$ is infinite and $u,v$ are separable and algebraic over $K$ then there is some $a \in K$ such that $K (u+av)=K (u,v)$ proof verification

Let $K$ be a field with infinite elements. Prove that if $u,v$ are separable and algebraic over $K$ then there is some $a \in K$ such that $K (u+av)=K (u,v)$. Is the result still true if $K$ is finite?

This question is from Kaplansky's fields and rings page 49.

Here is my proof for the first part:

As $K (u,v)$ is a simple extension of $K (u)$ which is in turn a simple extension of $K$, by the theorem on page 48 of the same book there is only a finite number of intermediate fields for $K (u,v)/K$. Considering the finite set $\{ K (u+av) : a \in K \}$, since $K$ is infinite there is some $K (u+av)=K (u+bv)$, and doing a similar process on page 48 proof we eventually get that $K (u,v) = K (u+av)$.

Here is my question: how does the fact that $u,v$ are separable are used? Is my proof correct or is it missing something?

Another update: it seems like, upon reflection, that the problem with my above 'proof' is that I am fixing a particular intermediate field $K(u)$. So, while for my particular field there may be only finite number of intermediate fields that 'fill the gap' between $K(u,v)$ and $K(u)$ and likewise 'fill the gap' between $K(u)$ and $K$ there may be infinite number of ways to choose an intermediate field. Is there a way to 'upgrade' my false proof in such a way that would solve the problem, and if not how should I approach this problem?