Can the Extended Field Generate More Polynomials than the Base Field? 
Question. Let $L:K$ be a field extension. Let $p, q, r\in K[x_1, \ldots, x_n]$, and suppose that there are polynomials $A, B\in L[x_1, \ldots, x_n]$ such that $pA+qB=r$. Do there necessarily exist polynomials $a, b\in K[x_1, \ldots, x_n]$ such that $pa+qb=r$?

In other words, if $I$ denotes the ideal generated by $p$ and $q$ in $K[x_1, \ldots, x_n]$, and $J$ denotes the ideal generated by $p$ and $q$ in $L[x_1, \ldots, x_n]$, then is it true that $I=J\cap K[x_1, \ldots, x_n]$?
I tried using induction on the sum of degrees of $p$ and $q$ but couldn't make it work.
 A: We prove a general statement in the form of Lemma 2 below which easily implies the above speculated fact.

Lemma 1. Let $K:F$ be a field extension, giving a map $F[x_1 , \ldots, x_n]\to K[x_1 , \ldots, x_n]$.
  If $\{v_\alpha\}_{\alpha\in J}$ is an $F$-basis for $K$, then we have
  $$K[x_1 , \ldots, x_n]= \bigoplus_{\alpha\in J} F[x_1 , \ldots, x_n]v_\alpha$$
  as $F[x_1 , \ldots, x_n]$-modules.

We write $\mathbf x$ to denote the list $x_1 , \ldots, x_n$.
It is clear that $\sum_{\alpha\in J}F[\mathbf x]v_\alpha= K[\mathbf x]$.
So let $p_1 , \ldots, p_k\in F[\mathbf x]$ be such that $p_1 v_{\alpha_1}+ \cdots + p_kv_{\alpha_k}=0$.
We want to show that each $p_i$ is $0$ to finish the proof.
Let $m:=x_1^{a_1} \cdots x_n^{a_n}$ be a monomial.
Let the coefficient of $m$ in $p_i$ be written as $c_i$, which is a member of $F$.
Then the coefficient of $m$ in $p_1v_{\alpha_1}+ \cdots + p_kv_{\alpha_k}$ is $c_1v_{\alpha_1}+ \cdots + c_kv_{\alpha_k}$.
But this is also equal to $0$.
Since $\{v_\alpha\}_{\alpha\in J}$ form an $F$-basis for $K$, we have each $c_i$ is $0$.
    Therefore, the coefficient of any monomial in each $p_i$ is $0$, meaning that each $p_i$ is $0$, and we are done.

Lemma 2.   Let $K:F$ be a field extension, giving a map $F[x_1 , \ldots, x_n]\to K[x_1 , \ldots, x_n]$.
  Let $I$ be an ideal in $F[x_1 , \ldots, x_n]$.
  Then $I^{ec}=I$.

We write $\mathbf x$ to denote the list $x_1 , \ldots, x_n$.
Since $F[\mathbf x]$ is Noetherian, there are finitely many polynomials $p_1 , \ldots, p_k\in F[\mathbf x]$ such that $I=\langle {p_1 , \ldots, p_k}\rangle$.
Let $p\in I^{ec}$.
Then there are polynomials $f_1 , \ldots, f_k\in K[\mathbf x]$ such that $p=f_1 p_1 + \cdots + f_kp_k$.
Let $\{v_\alpha\}_{\alpha\in J}$ be an $F$-basis for $K$, where, say $v_{\alpha_0}=1\in F$.
By Lemma 1, each $f_i$ can be written as
$$ f_i = \sum_{\alpha\in J} v_\alpha f_i^\alpha$$
where all but finitely many $f_i^\alpha$ are $0$ and each $f_i^\alpha\in F[\mathbf x]$.
Thus we have
$$p = \sum_{\alpha\in J} v_\alpha(f_1^\alpha p_1 + \cdots + f_k^\alpha p_k)$$
which gives
$$p = (f_1^{\alpha_0}p_1+ \cdots + f_k^{\alpha_0}p_k) + {\sum_{\alpha\in J\setminus\{\alpha_0\}} v_\alpha(f_1^\alpha p_1 + \cdots + f_k^\alpha p_k)
}$$
But since $p\in F[\mathbf x]$, by Lemma 1, we must have
$$p= f_1^{\alpha_0} p_1 + \cdots + f^{\alpha_0}_k p_k$$
and thus $p\in I$.
So we have shown that $I^{ec}\subseteq I$.
The reverse containment is clear, and thus we have $I^{ec}=I$.
