Primitive element theorem without building field extensions Is there are nice way to prove the primitive element theorem without using field extensions?
The primitive element theorem says that if $x$ and $y$ are algebraic over $F$ and $y$ is separable over $F$, then there exists a $z \in F(x,y)$ such that $F(x,y) = F(z)$.  In the case where $F$ is infinite, $z$ can be expressed in the form $x + {\lambda}y$ with $\lambda \in F$.  In fact, almost any $\lambda$ will do.  There are only a finite number of exceptions.  These exceptions are $\frac{\alpha_i - x}{y - \beta_j}$ where $\alpha_i$ and $\beta_j$ range over the (other) roots of the minimal polynomials of $x$ and $y$ respectively.
But in order to talk about $\alpha_i$ and $\beta_j$ we need to build a field extension where the minimal polynomials of $x$ and $y$ split.  This is the step I'm hoping to avoid.
Perhaps we can build a polynomial in $F[x]$ whose roots are $\frac{\alpha_{i_1} - \alpha_{i_2}}{\beta_{j_1} - \beta_{j_2}}$ and simply avoid picking $\lambda$s which are roots of this polynomial, and then use whatever properties this polynomials has to prove that this works.
Or maybe there is another completely different way of proving the primitive element theorem while avoiding building field extensions.
I found a nice proof that $x$ is separable over $F$ if and only if every $y \in F(x)$ is separable over $F$ that avoids building field extensions.  It uses derivations instead.  Now I'm hoping to do the same with the primitive element theorem.
Edit: I'll try to give some motivation. Adding roots of polynomials to fields in constructive mathematics is more difficult than in classical mathematics (because irreducibility is undecidable). It only works for countable fields, and then you have no guarantee that the original field will be a decidable subset of the new field. Yes, it can be done, but it seems like a pain. There is another way too using double negation translations, but it also seems like a pain. Instead I'd rather avoid the whole issue of building splitting fields, if I can, and it seems tantalizingly close to possible.  After all, the only reason splitting fields are used here is to build a finite set of elements in the base field to avoid.
 A: OK, I have a proof which meets your conditions. I relied on this write up of the standard proof as a reference.
Lemma 1: Let $K/F$ be an extension of fields, and let $f(x)$ and $g(x)$ be polynomials in $F[x]$. Let $d_F(x)$ be the GCD of $f$ and $g$ in $F[x]$ and let $d_K(x)$ be the GCD of $f$ and $g$ in $K[x]$. Then $d_F(x)$ and $d_K(x)$ coincide up to a scalar factor.
Proof: Since $d_F(x)|f(x)$ and $d_F(x)|g(x)$ in $K[x]$, we have $d_F(x)|d_K(x)$. Now, there are polynomials $p(x)$ and $q(x)$ in $F[x]$ such that $f(x) p(x) + g(x) q(x) = d_F(x)$. So $d_K(x) | d_F(X)$. Since $d_F(x)$ and $d_K(x)$ divide each other, they only differ by a scalar factor.
Lemma 2: Let $f(x)$ and $g(x)$ be polynomials with $g(0) \neq 0$. Then, for all but finitely many $t$, the polynomials $f(tx)$ and $g(x)$ have no common factor.
Proof: Let $f(x) = f_m x^m + \cdots + f_1 x + f_0$ and $g(x) = g_n x^n + \cdots + g_1 x + g_0$. If $f(tx)$ and $g(x)$ HAVE a nontrivial common factor, then there are polynomials $p(x)$ and $q(x)$, of degrees $\leq n-1$ and $\leq m-1$, such that 
$$f(tx) p(x)=g(tx) q(x).$$ 
This is $m+n$ linear equations on the $m+n$ coefficients of $p$ and $q$. 
Writing this out in coefficients, the matrix
$$\begin{pmatrix} 
f_m t^m & \cdots & f_1 t & f_0 & 0 & 0 & \cdots & 0 \\
0 & f_m t^m & \cdots & f_1 t & f_0 & 0 & \cdots & 0 \\
\ddots \\
0 & 0 & \cdots & 0 & f_m t^m & \cdots & f_1 t & f_0 \\
g_n & \cdots & g_1 & g_0 & 0 & 0 & \cdots & 0 \\
0 & g_n & \cdots & g_1 & g_0 & 0 & \cdots & 0 \\
\ddots \\
0 & 0 & \cdots & 0 & g_n & \cdots & g_1 & g_0
\end{pmatrix}$$
has nontrivial kernel. The determinant of this matrix is a polynomial in $t$, with leading term $(f_m)^n (g_0)^n t^{mn} + \cdots$. (Recall that $g_0 \neq 0$.) So, for all but finitely many $t$, this matrix has nonzero determinant and $f(tx)$ and $g(x)$ are relatively prime. QED.
Now, we prove the primitive element theorem (for infinite fields). Let $\alpha$ and $\beta$, in $K$, be algebraic and separable over $F$, with minimal polynomials $f$ and $g$. We will show that, for all but finitely many $t$ in $F$, we have $F(\alpha - t \beta) = F(\alpha, \beta)$.
Let $f(x) = (x -\alpha) f'(x -\alpha)$ and $g(x) = (x - \beta) g'(x - \beta)$. Since $\beta$ is separable, we know that $g'(0) \neq 0$. Note that $f'$ and $g'$ have coefficients in $K$. By Lemma 2, for all but infinitely many $t$ in $F$, the polynomials $f'(ty)$ and $g'(y)$ have no common factor. Choose a $t$ for which no common factor exists. Set $F' = F(\alpha - t \beta)$; our goal is to show that $F'=F(\alpha, \beta)$.
Set $h_t(x) = f(tx + \alpha-t \beta)$. Note that $h_t(x)$ has coefficients in $F'$. We consider the GCD of $h_t(x)$ and $g(x)$. 
Working in $K$, we can write $h_t(x) =  t (x - \beta) f'(t (x- \beta))$ and $g(x) = (x-\beta) g'(x-\beta)$. By the choice of $t$, the polynomials $f'(t (x- \beta))$ and $g'(x-\beta)$ have no common factor, so the GCD of $h_t(x)$ and $g(x)$, in the ring $K[x]$, is $x-\beta$.
By Lemma 1, this shows that $x - \beta$ is in the ring $F'[x]$. In particular, $\beta$ is in $F'$. Clearly, $\alpha$ is then also in $F'$, as $\alpha= (\alpha - t \beta) + t \beta$. We have never written down an element of any field larger than $K$. QED.
A: I assume you mean that you want to avoid constructing splitting fields; you can't avoid talking about field extensions when proving a theorem about field extensions! Here's a sort of a proof, but I think it is probably cheating. 
Lemma: Let $F$ be an infinite field. Let $V$ be a $K$ vector space, containing two linearly independent vectors $u$ and $v$. Let $W_1$, $W_2$, ..., $W_r$ be a finite list of subspaces of $V$, none of which contain the space spanned by $u$ and $v$. Then there are scalars $a$ and $b$ in $F$ such that $au+bv$ is not in any $W_i$.
Proof: Let $R$ be the plane spanned by $u$ and $v$. Since $W_i \not \supseteq R$, we know that $W_i \cap R$ has dimension at most $1$. So there is at most one ratio of $a/b$ for which $au+bv$ lies in $W_i$. Since $F$ is infinite, there are infinitely values of the ratio for which $au+bv$ is not in any $W_i$.
Now, take $V$ to be the field $F(\alpha, \beta)$, and take $W_i$ to be the list of proper subfields of $V$ containing $F$.
The problem is that this proof relies on the fact that a finite seperable extension has finitely many sub-extensions. I'm not sure it is possible to prove this without introducing splitting fields at some point. 
