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[parent] example of functor of points of a scheme (Example)

Let $ X$ be an affine scheme of finite type over a field $ k$. Then we must have

$\displaystyle X = {\mathrm{Spec}}k[X_1,\ldots,X_n]/\left<f_1,\ldots,f_m\right>, $
with the structure morphism $ X\to{\mathrm{Spec}}k$ induced from the natural embedding $ k\to k[X_1,\ldots,X_n]$.

Let $ k'$ be some field extension of $ k$. What are the $ k'$-points of $ X$? Recall that a $ k'$-point of $ X$ is by definition a morphism $ {\mathrm{Spec}}k' \to X$ (observe that since we have an embedding $ k\to k'$ we have a morphism $ {\mathrm{Spec}}k' \to {\mathrm{Spec}}k$, so $ {\mathrm{Spec}}k'$ is natuarlly a $ k$-scheme). Since $ X$ is affine, this must come from a ring homomorphism

$\displaystyle k[X_1,\ldots,X_n]/\left<f_1,\ldots,f_m\right> \to k' $
which takes elements of $ k$ to themselves inside $ k'$. Such a homomorphism is completely specified by specifying the images of $ X_1,\ldots,X_n$; for it to be a homomorphism, these images must satisfy $ f_1,\ldots,f_m$. In other words, a $ k'$-point on $ X$ is identified with an element of $ (k')^n$ satisfying all the polynomials $ f_i$.

If $ k'$ is an algebraically closed field, a point on $ X$ corresponds uniquely to a point on an affine variety defined by the same equations as $ X$. If $ k'$ is just any extension of $ k$, then we have simply found which new points belong on $ X$ when we extend the base field. T

For an example of why schemes contain much more information than the list of points over their base field, take $ X={\mathrm{Spec}}\mathbb{R}[X]/\left<X^2+1\right>$. Then $ X$ has no points over $ \mathbb{R}$, its natural base field. Over $ \mathbb{C}$, it has two points, corresponding to $ i$ and $ -i$.

This suggests that schemes may be the appropriate adaptation of varieties to deal with non-algebraically closed fields.

Observe that we never used the fact that $ k'$ (or in fact $ k$) was a field. One often chooses $ k'$ as something other than a field in order to solve a problem. For example, one can take $ k' = k[\epsilon]/\left<\epsilon^2\right>$. Then specifying a $ k'$-point on $ X$ amounts to choosing an image $ \kappa_i + \lambda_i\epsilon$ for each $ X_i$. It is clear that the $ \kappa_i$ must satisfy the $ f_j$. But upon reflection, we see that the $ \lambda_i$ must specify a tangent vector to $ X$ at the point specified by the $ \kappa_i$. So the $ k[\epsilon]/\left<\epsilon^2\right>$-points tell us about the tangent bundle to $ X$. Observe that we made no assumption about the field $ k$ -- we can extract these “tangent vectors” in positive characteristic or over a non-complete field.

The ring $ k[\epsilon]/\left<\epsilon^2\right>$ and rings like it (often any Artinian ring) can be used to define and study infinitesimal deformations of schemes, as a simple case of the study of families of schemes.



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Cross-references: simple, deformations, infinitesimal, artinian, ring, characteristic, positive, tangent bundle, tangent vector, reflection, clear, order, chooses, closed, varieties, information, contain, schemes, base field, extension, equations, affine variety, point, algebraically closed, polynomials, images, homomorphism, ring homomorphism, embedding, morphism, field extension, natural embedding, induced, structure morphism, field, finite type, affine scheme
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This is version 1 of example of functor of points of a scheme, born on 2004-02-22.
Object id is 5612, canonical name is ExampleOfFunctorOfPointsOfAScheme.
Accessed 1945 times total.

Classification:
AMS MSC: 14A15 (Algebraic geometry :: Foundations :: Schemes and morphisms)
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