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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 $k^2$)!

ThePuix
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  • I don't think the linked answers deal with the question at the level of schemes, but rather for varieties over an algebraically closed field, which is not assumed in this question. I therefore don't think this is a duplicate questions. – ThePuix Oct 30 '20 at 18:36
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    The answers at the linked duplicates suffice for your question: find a Galois extension $F$ of $k$ so that the base change of your finite set of closed points consists entirely of $F$-rational points, run the proof from the linked questions to get two polynomials in $F[x,y]$ which vanish on the base change, and note that the polynomials you get are Galois-invariant and thus are actually polynomials in $k[x,y]$. – KReiser Oct 30 '20 at 18:41
  • @KReiser Thanks for your comment! I don't know any Galois theory, and the course from which this question arises does not assume it either, so I am a little surprised you need it for this exercise. In particular, I have no idea how to construct the needed Galois extension $F$ that you are talking about. Would you mind elaborating, or directing me to somewhere I can better my understanding? – ThePuix Oct 30 '20 at 18:45
  • Surely if you're taking a course in schemes you've taken undergrad algebra? Where I am these usually cover Galois theory in some detail. Anyways, a maximal ideal in $k[x,y]$ contains a unique monic irreducible polynomial in $x$ and a unique monic irreducible polynomial in $y$. Take the list of all of these across all your maximal ideals, apply the fact that the splitting field of a polynomial is a Galois extension repeatedly plus the fact that a composition of Galois extensions is Galois. – KReiser Oct 30 '20 at 18:50
  • @KReiser I have indeed taken undergrad algebra, but at my university this does not cover Galois theory (this is a separate course entirely). I will try to see what I can understand from what you have linked. I don't understand though what you mean when you say every maximal ideal contains a unique irreducible polynomial only both $x$ and only $y$; one of them is a polynomial in $x$ and $y$? – ThePuix Oct 30 '20 at 18:58
  • $k[x,y]/m$ is an algebraic field extension of $k$, so the image of $x$ is algebraic over $k$. Thus the image of $x$ satisfies a monic irreducible polynomial with coefficients in $k$. So this polynomial when evaluated at $x$ must be in $m$. Similarly for $y$. – KReiser Oct 30 '20 at 19:03
  • @KReiser Okay, so if I understand your outline: I do this for every maximal ideal on my list, get some extension $F$ which is Galois and wherein my points become $F$-rational points. They then have coordinates (and roots) in $F$ which determine the maxinal ideals, and I can use the linked questions to construct two polynomials in $F[x,y]$ that only vanish at my maximal ideals (which are now $F$-rational points). Then $Gal(F/k)$ acts on these polynomial as the identity (how would I see this?), so they are actually polynomials in $k[x,y]$. And this does not alter their vanishing sets? – ThePuix Oct 30 '20 at 19:10
  • You can check that $Gal(F/k)$ acts as the identity on these polynomials by construction. As $F^{Gal(F/k)}=k$, we have that $(F[x,y])^{Gal(F/k)}=k[x,y]$. I don't understand what you're worried about with the "altering the vanishing sets" comment - if the polynomials vanished at some extra point in $\Bbb A^2_k$, then they would vanish at their preimage in $\Bbb A^2_F$, but that contradicts our construction. – KReiser Oct 30 '20 at 19:21
  • @KReiser I really appreciate your help. I will try to piece this together to something I am comfortable with. The idea of passing to a field extension is maybe obvious to trained algebraic geometers, but was not immediate for me, but it seems very useful for many things. I am still surprised this is marked as a duplicate, since there are some steps to reduce it to the linked questions, but I guess that could be a matter of interpretation. Thanks again! – ThePuix Oct 30 '20 at 19:29
  • After thinking about it a bit more, I guess I've come around to your point of view that your question is about reducing the general problem to the case of an algebraically closed field and that isn't available in the duplicates. Since I've basically answered that in the comments, I'll compile the comments in to an answer. – KReiser Oct 30 '20 at 19:52

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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.

KReiser
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