ac.commutative algebra - Primitive element theorem without building field extensions

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.