This has a very easy proof if one generalizes it to (infinite-dimensional) linear algebra and forgets about commutative algebra. Let $K/F$ be an extension of fields (e.g., could take $F$ to be a prime field), $V$ an $F$-vector space (such as a polynomial ring over a prime field), and $W$ a $K$-subspace of $V_K := K \otimes_F V$.
Among all subfields $K_0$ of $K$ over $F$ such that $W = K \otimes_{K_0} W_0$ for a (visibly unique) $K_0$-subspace $W_0$ of $V_{K_0}$, we claim that the intersection of these fields works too.
(In case $V$ is an $F$-algebra and $W$ is an ideal of $V_K$, obviously $W_0$ is an ideal of $V_{K_0}$, so this really does imply Weil's result. In fact, it gives a more general result: no need to assume the algebras are finitely generated.)
Proof: Choose an $F$-basis $\{v_i\}_{i \in I}$ of $V$ and expand all elements V$, so there is a subset$w$J$ of $W$ I$such that $\{v_j \bmod W\}_{j \in J}$ is a$K$-linear combinations K$-basis of $$w = V/W. For i' \sum_i a_i(w) v_i$$ in I - J$, expand$v_{i'} \bmod W \in V/W$in this basis: $$v_{i'} \equiv \sum_{j \in J} a_{i'j} v_j \bmod W$$ with$a_i(w) a_{i'j} \in K$. The necessary and sufficient condition on$K_0$for$W_0$to exist is that$K_0$contains every$a_i(w)$a_{i'j}$ (for $w j \in W$ J$and$i i' \in I$)I - J$). So the subfield $F(a_i(w))_{i, w}$ F(a_{i'j})_{i', j}$is the desired minimal subextension of$K$over$F$. QED There is a very elegant modern discussion of the theme of field of definition (for closed subschemes, morphisms, etc.) without any finiteness hypotheses in EGA IV$_2$, 4.8. 1 [made Community Wiki] This has a very easy proof if one generalizes it to (infinite-dimensional) linear algebra and forgets about commutative algebra. Let$K/F$be an extension of fields (e.g., could take$F$to be a prime field),$V$an$F$-vector space (such as a polynomial ring over a prime field), and$W$a$K$-subspace of$V_K := K \otimes_F V$. Among all subfields$K_0$of$K$over$F$such that$W = K \otimes_{K_0} W_0$for a (visibly unique)$K_0$-subspace$W_0$of$V_{K_0}$, we claim that the intersection of these fields works too. (In case$V$is an$F$-algebra and$W$is an ideal of$V_K$, obviously$W_0$is an ideal of$V_{K_0}$, so this really does imply Weil's result. In fact, it gives a more general result: no need to assume the algebras are finitely generated.) Proof: Choose an$F$-basis $\{v_i\}_{i \in I}$ of$V$and expand all elements$w$of$W$in$K$-linear combinations $$w = \sum_i a_i(w) v_i$$ with$a_i(w) \in K$. The necessary and sufficient condition on$K_0$for$W_0$to exist is that$K_0$contains every$a_i(w)$(for$w \in W$and$i \in I$). So the subfield$F(a_i(w))_{i, w}$is the desired minimal subextension of$K$over$F$. QED There is a very elegant modern discussion of the theme of field of definition (for closed subschemes, morphisms, etc.) without any finiteness hypotheses in EGA IV$_2\$, 4.8.