$\mathbb{R}$, line segments and triangles as schemes

Question

The definition of a scheme is mind-blowing by its generality: since $\mathbb{Z}$ is a ring of Krull dimension 1, the set of prime numbers is a line. While this draws superb connections between geometry and arithmetic, simple geometric concepts are not so easy to define as schemes.

Consider the line $\mathbb{R}$, line segment $[0,1]$ and a triangle. Those are meant as intuitive geometric figures that we can draw on paper, and I try give them precise definitions as schemes. I expect them all to have Krull dimension 1, so I equip them with the topology of finite sets as closed sets. For the structural sheaf of rings $\mathcal{O}$, on the line and the segment I take the rational fractions valued in $\mathbb{R}$. It seems the stalk $\mathcal{O}_x$ at each point $x$ is a local ring, as required. So I have locally ringed spaces for the line and the segment, and wonder if they are affine schemes (isomorphic to the spectrum of some ring), or openly covered by affine schemes.

The triangle is a little harder because I failed to define the sheaf of rings of rational fractions with a single variable $x$ (embedded in the plane the triangle has 2 variables $x,y$ to index it). Can we recover the triangle as the gluing of 3 schemes isomorphic to the line segment? Whatever the definition of the triangle's scheme, I expect it to be nonsingular, except at its 3 tips, which should be singularities (tangent spaces of dimension 2 probably).

Answer 1

New answer:

Again, the topology you've picked shows that this cannot happen. The cofinite topology implies that single points are closed, so there are no nontrivial specializations or generalizations between points. Therefore for any affine open subscheme, we get no nontrivial inclusions between prime ideals. But that's condition 5 in 04MG, so again your space must have the discrete topology, contradiction.

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Old answer:

None of the geometric figures you mention (equipped with the usual topology) can ever be the underlying topological space of a scheme, because every Hausdorff scheme is totally disconnected (ref 04MG). You may find some joy in considering the maximal spectra of the ring $C(X)$ of continuous real-valued functions on a topological space $X$: if $X$ is compact Hausdorff, then $\operatorname{MaxSpec} C(X)$ has as its underlying topological space $X$ (ref MO). But this is not really "typical" algebraic geometry.

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Old advice, still valid on the new version:

To put it somewhat bluntly, what you're trying to do seems incredibly strange and like it's coming from a place of deep unfamiliarity with algebraic geometry. You may find it helpful to get a bit more experience with the subject so you can make more reasonable requests of this particular area of math.

Answer 2

The triangle does exist as a scheme, if we add to its set of points $T=([0,1]\times\{0\}) \cup (\{0\}\times [0,1]) \cup \{ (t,1-t) ; t\in [0,1] \}$ the ideals that represent the 3 infinite lines running through its sides.

To construct it, we first localize $\mathbb{R}[X,Y]$ by the multiplicative subset $S = \{P\in \mathbb{R}[X,Y] \;/\; \forall (x,y)\in \mathbb{R}^2, P(x,y) = 0 \Rightarrow x \notin T\}$. We denote $i:\mathbb{R}[X,Y] \to S^{-1}\mathbb{R}[X,Y]$ the localization morphism. We then quotient by the ideal generated by $i(XY(X+Y-1))$, to yield the ring $A_T$. Its prime spectrum represents a triangle, because it is made of the 3 ideals generated respectively by $\{X\}$, $\{Y\}$, $\{X+Y-1\}$, and also for each $(x,y)\in T$, of the ideal generated by the pair $\{X-x,Y-y\}$. Below is the proof of this spectrum.

By localization, $i^{-1}$ is a bijection from the prime ideals of $S^{-1}\mathbb{R}[X,Y]$, towards the prime ideals of $\mathbb{R}[X,Y]$ of which each polynomial has a zero in $T$. By quotient, the prime ideals of $A_T$ are in bijection with the prime ideals of $S^{-1}\mathbb{R}[X,Y]$ that contain $i(XY(X+Y-1))$. We therefore want to describe $$ \text{Spec }A_T \simeq \{ I\subset\mathbb{R}[X,Y] / I \text{ prime ideal and } \forall P\in I, \exists (x,y)\in V, P(x,y) =0 \text{ and } XY(X+Y-1)\in I \} $$ Let $I$ such an ideal. Since $XY(X+Y-1)\in I$ and $I$ is prime, we have $X\in I$ or $Y\in I$ or $X+Y-1\in I$.

$\blacktriangleright$ Assume $X\in I$, and also assume that $I$ is not generated by $\{X\}$. We prove there exists $y\in [0,1]$ such as $I$ is generated by the pair $\{X,Y-y\}$. Because there is a polynomial $P\in I$ not divisible by $X$, euclidean division yields $P=XQ+R$, with $R\in\mathbb{R}[Y]$ not zero. Then $R\in I$ and there is $(x,y)\in V$ such as $R(x,y)=0$. Since $R$ does not depend on $x$, we have $R(y)=0$. There exists $R_0\in\mathbb{R}[Y]$ such as $R=(Y-y)R_0$. Since $I$ is prime, we have $Y-y\in I$ or $R_0\in I$. The first case terminates the proof, otherwise we restart the root search on $R_0$, which will terminate because $\deg R_0 < \deg R$.

$\blacktriangleright$ Assume $Y\in I$. Then with a similar reasoning as in the previous paragraph, we get that $I$ is generated either by $\{Y\}$ or by the pair $\{X-x,Y\}$ with $x\in [0,1]$.

$\blacktriangleright$ Assume $X+Y-1\in I$, and $I$ is not generated by $\{X+Y-1\}$. Then there exists a polynomial $P\in I$ not divisible by $X+Y-1$, by euclidean division we have $P=(X+Y-1)Q+R$, avec $R\in\mathbb{R}[Y]$ not zero. By the first paragraph there exists $y\in [0,1]$ such as $Y-y\in I$. The combination with $X+Y-1\in I$ yields $X+y-1\in I$.