Union of closed points of $\mathbb{A}^2_k = \operatorname{Spec}(k[x,y])$ The question is as follows: For finitely many closed points $x_1,\dots,x_n \in \mathbb{A}^2_k$, for $k$ a field (not assumed algebraically closed) show that their union can be written as $V(f,g)$ for $f,g\in k[x,y]$.
This is part 2 of a problem, where in the first problem we classify closed points of $\mathbb{A}^2_k$. These correspond to maximal ideals in $k[x,y]$, and I can show that these are of the form $V(f,g)$ where $f\in k[x]$ and $g\in k[x,y]$ are irreducible.
One can try to prove the above by induction, so for two closed points $x_1= V(f_1,g_1)$ and $x_2=V(f_2,g_2)$, we have that $\{x_1,x_2\}=V(f_1,g_1)\cup V(f_2,g_2)=V((f_1,g_1)\cdot (f_2,g_2))$ where
$(f_1,g_1)\cdot (f_2,g_2)=(f_1f_2,f_1g_2,g_1f_2,g_1g_2)$ denotes the product of the ideals (also equal to the intersection by the Chinese Remainder Theorem). The problem is that is will have $4$ generators, and we need to cut that down to two and this is where I am stuck. The immediate try of $V(f_1f_2,g_1g_2)$ consists of four points rather than two.
Any help or hints are appreciated (the linked duplicates seem to deal with the case of an algebraically closed field. The closed points don't have coordinates in $k^2$)!
 A: I claim that we can always reduce to the case where all our points are $k$-rational points, and then the answer here shows that any finite collection of $k$-rational points in $\Bbb A^n_k$ can be defined by $n$ polynomials with coefficients in $k$.
To do this, we want to find a Galois extension $k\subset F$ so that all our closed points become rational points over $F$. To do this, we use the fact that any maximal ideal in $k[x_1,\cdots,x_n]$ contains a unique monic irreducible polynomial in $x_i$: simply take the minimal polynomial of the image of $x_i$ in $k[x_1,\cdots,x_n]/m$, a finite extension of $k$. Now compile all of these minimal polynomials in to a finite list, take the splitting field of the first polynomial, remove all the polynomials in the list that split over this extension, and repeat: because the splitting field of any polynomial is Galois and the composition of Galois extensions is Galois, we get a Galois extension $k\subset F$ so that after base change, our finite collection of points is $F$-rational.
From here, we may apply the linked answer to find $n$ polynomials which vanish exactly at the base change of our finite collection of points. By construction of these polynomials, we see that they are $Gal(F/k)$ invariant, and so they're all actually in $k[x_1,\cdots,x_n]$ and therefore their vanishing locus in $\Bbb A^n_k$ is exactly the collection of closed points we started with.
