# Scheme over a not algebraically closed field

I'm trying to make a scheme out of an affine variety $V\subset k^n$, when the base field $k$ is not algebraically closed. I wonder if $V$ is the spectrum of its ring of regular functions $\mathcal{O}(V)$, ie functions $f:V\to k$ such that for any point $p\in V$, there exists a neighborhood $U$ of $p$ and two polynomials $Q,R\in k[X_1,\dots,X_n]$ such that $R$ has no zeros on $U$ and $f=\frac{Q}{R}$ on $U$. For example if $V=k=\mathbb{R}$ then $f(x)=\frac{(x−1)(x+2)}{x^2+1}$ is regular on $\mathbb{R}$, but not invertible because it has zeros in 1 and -2.

So unlike the coordinate ring $A(V)=k[X_1,\dots,X_n] / I(V)$, functions in $\mathcal{O}(V)$ have polynomial denominators (without zeros), because $k$ is not algebraically closed.

I don't manage to prove that

1. a maximal ideal $m$ of $\mathcal{O}(V)$ corresponds to a point of $V$. The denominators ensure that all elements of $m$ have zeros, but not necessarily a unique common zero.

2. a prime ideal $p$ of $\mathcal{O}(V)$ corresponds to an irreducible subvariety of $V$.

3. if there are more points or subvarieties in Spec $\mathcal{O}(V)$, such as the zero ideal, prove those extra points are generic, i.e. that their topological closure contains $V$.

I understand 2 and 3 are harder, I originally only asked 1 :)

Thanks, Vincent

To avoid any misunderstanding, the way I read your definition, if $k=\mathbb{R}$ and $V=\mathbb{R}$, you are considering the ring of rational functions that have no poles on real numbers, so it includes things like $\frac{1}{x^2+1}$ but not $\frac{1}{x}$. And you are asking (in that particular case) whether the maximal ideals of that ring are the real numbers.

I have to admit I didn't check too carefully, but I think it follows in general (or perhaps under mild assumptions like $k$ being infinite and perfect) from the results of EGA IV₃ §8 (esp. prop 8.2.13 and cor 8.2.10): if you take the open sets given by the complement of each closed point in $V$ whose residue field is not $k$, and the projective system of affine schemes given by finite intersections of such open sets (i.e., the complements of finitely many closed points whose residue fields are not $k$), the projective limit should (by 8.2.13) be the intersection of all the open sets in question, i.e., its closed points should be the points of $V$ defined over $k$, and it should also be given (by 8.2.10) as the spectrum of the inductive limit which is the ring you defined (because it is the localization which inverts every function that does not vanish on the closed points with residue field $k$).

• $(0)$ is an irreducible point of the set of "regular functions" – Tsemo Aristide Apr 30 '16 at 20:05
• Thanks Gro-Tsen, it is exactly the question. And thanks Tsemo Aristide for pointing out the generic point $(0)$ : I accept it because its topological closure contains $V$. I'm more concerned if there are extra points added by maximal ideals of $\mathcal{O}(V)$ and disconnected from $V$. Gro-Tsen's explanation looks like a proof there are none, I'll check it in details. – V. Semeria Apr 30 '16 at 20:21
• @TsemoAristide I think by now V. Semeria has made it amply clear that he is interested in the maximal spectrum (see the third paragraph of his question), or else that the non-closed points should be interpreted as some other kind of geometric objects on $V$ (e.g., $k$-subvarieties that contain a $k$-point, or with a dense set of $k$-points, I'm not exactly sure). Yes, the wording of the question could have been better: do you want to edit it? Because with the reasonable interpretation, it's an interesting question. – Gro-Tsen Apr 30 '16 at 20:54

Now that OP has precised is definition of regular functions, I believe that this result is not true since for $k^n$, $(0)$ is a prime ideal for the set of regular functions,

• What definition of ring of regular functions do you use? For me it is a polynomial ring in $n$ variables in this case. – Tobias Kildetoft Apr 30 '16 at 18:46
• I use the definition of OP for him a regular function is a function $P/Q$ where $P$ and $Q$ are polynomial functions, that is regular functions in the classical sense, – Tsemo Aristide Apr 30 '16 at 18:49
• If $n=1$, then $x$ is in $O(V)$ according to the OP 's definition. Is that inverttible there? – Mariano Suárez-Álvarez Apr 30 '16 at 18:56
• So for $n=1$ how is the function $\frac{1}{x}$ regular? There would for each point have to be some open neighborhood in which $x$ is not $0$. – Tobias Kildetoft Apr 30 '16 at 18:56
• it is not like this I understand OP, just a point where $x$ is not zero is sufficient for $1/x$ to be invertible. – Tsemo Aristide Apr 30 '16 at 19:00

I'm beginning to doubt about my claim. Firstly, consider the prime polynomial $Q=Y^2+(X^2-1)^2\in\mathbb{R}[X,Y]$. It generates a prime ideal in $\mathcal{O}(\mathbb{R}^2)$ that has 2 disconnected points : $(1,0)$ and $(-1,0)$. So its variety is reducible, even though $Q$ is prime.

Secondly, Bochnak, Coste and Roy, in their Real Algebraic Geometry, don't make schemes on real varieties. Instead of prime ideals they use prime cones (sort of half-ideals, positive elements of an ideal, a concept that uses the order of a real field). Then they define the real spectrum as the set of prime cones and build a topology and a sheaf on it. So instead of using the standard concepts of algebraic geometry, they rebuild parallel concepts that make use of the order of the base real field. In return they get semi-varieties defined by polynomial equalities and inequalities, leading to interesting shapes and problems.

Is there a general framework for algebraic geometry, that will handle both those real and algebraically closed cases ?

• There are two things you might mean by "real algebraic geometry." One is the study of subsets of $\mathbb{R}^n$ cut out by polynomial equations (this is the usual meaning), which turns out to be much more model-theoretic than algebro-geometric. Another is the study of schemes over $\text{Spec } \mathbb{R}$, which is covered by scheme theory. The two theories are very different, so you should think about which one you're interested in. – Qiaochu Yuan May 1 '16 at 17:47
• The difference is even more stark if you replace $\mathbb{R}$ with a finite field. "$\mathbb{F}_q$ algebraic geometry" in the first sense is just the study of arbitrary subsets of $\mathbb{F}_q^n$. But algebraic geometry over $\mathbb{F}_q$ properly means the study of schemes over $\text{Spec } \mathbb{F}_q$ which is much more interesting. – Qiaochu Yuan May 1 '16 at 17:50
• I would go for the subsets of $\mathbb{R}^n$, in which case you not only have polynomial equations but also inequations (thanks to the order on $\mathbb{R}$). Why model-theoretic ? Doesn't look linked to first-order logic. – V. Semeria May 1 '16 at 18:09
• I don't know much about this, but look up "o-minimal model." – Qiaochu Yuan May 1 '16 at 18:37