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On page 49 of Liu's Algebraic Geometry and Arithmetic Curves, we find Example 3.32. In it, he shows that if $k$ is a field and $X=k[T_1,\dots,T_n]/I$ is an affine scheme over $k$, the sections $X(k)$ (the $k$-rational points of $X$) are in bijection with the algebraic set $Z(I)$ cut out by $I$. This is the set of all $\alpha\in k^n$ with $P(\alpha)=0$ for every $P(T)\in I$.

At the end of the section, he gives this exercise:

$\textbf{3.6}$ Generalize Example 3.32 to the case where $k$ is an arbitrary ring.

How is this possible? The proof in the example makes essential use of the fact that $k$ is a field. Clearly we want to consider the set cut out by $I$ over $R^n$, where $R$ is some ring, and relate this to the sections $X(R)$, but I do not see how to proceed. In particular, the proof identified the $k$-rational points with the points where the residue field was $k$. Since the residue field is always a field, but $R$ may not be, we can no longer use the residue field to identify the $R$-rational points.

Potato
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    You spread your hands about yea far apart and then you start waving them about in circles... – rschwieb Jun 11 '13 at 20:29
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    I've known a few authors who have ridiculous things as exercises. My favorites so far are (paraphrased): "invent the Dorroh extension" and "Prove the Cartan-Brauer-Hua theorem." – rschwieb Jun 11 '13 at 21:07
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    @rschwieb Where can I find those? You have piqued my curiosity... – Potato Jun 11 '13 at 21:09
  • The results or the exercises?! Martin Isaacs assigned the first one, and Nathan Jacobson assigned the latter one. – rschwieb Jun 11 '13 at 21:10
  • @rschwieb The exercises of course! – Potato Jun 11 '13 at 21:13
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    Isaacs Graduate Algebra problem 12.4 page 172, and Jacobson's Basic algebra I problem 10 page 100. I don't know why you're interested: I think these problems are perfect examples of bad problems precisely because they are prohibitively open ended. – rschwieb Jun 11 '13 at 21:17
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    @rschwieb: I agree with your assessment of them - but that's exactly why I'd want concrete references to point to :) – Zev Chonoles Jun 12 '13 at 01:38

2 Answers2

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Let $X = \text{Spec}\,A$ be an affine scheme of finite type over $R$. Then $X$ comes with a map $X\rightarrow \text{Spec}\,R$ so that the corresponding map of rings $R\rightarrow A$ makes $A$ a finitely generated $R$-algebra. So we can write $A \cong R[T_1,\dots,T_n]/I$ for some ideal $I$.

Now $X(R)$, the $R$-points of $X$, are maps $\text{Spec}\,R \rightarrow X$ which are sections of the map $X \rightarrow \text{Spec}\,R$. Since we're talking about maps of affine schemes, these are in canonical bijection with ring maps $A \rightarrow R$ which are sections of the structure map $R\rightarrow A$, that is, $R$-algebra homomorphisms $R[T_1,\dots,T_n]/I\rightarrow R$.

Now since $R[T_1,\dots,T_n]$ is the free $R$-algebra on $n$ generators,

\begin{align} \text{Hom}_R(R[T_1,\dots,T_n]/I,R) &\cong \{f\in\text{Hom}_R(R[T_1,\dots,T_n],R)\,|\,I\subseteq \text{Ker}(f)\}\\ &\cong \{(r_1,\dots,r_n)\in R^n\,|\,p(r_1,\dots,r_n) = 0\,\text{for all}\,p\in I\}. \end{align}

Alex Kruckman
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  • We seem to have posted at the same time, but I like your answer better =D – Jeff Jun 11 '13 at 20:52
  • Thanks! The answers are more or less the same, mine is just a bit more verbose. Of course, in your answer $R$ should be a $k$-algebra... – Alex Kruckman Jun 11 '13 at 20:54
  • Yes, but that edit button is so far away... – Jeff Jun 11 '13 at 20:55
  • Why is $X(R)$ in canonical bijection with $R$-algebra homomorphisms $R[T_1,\ldots,T_n]/I\to R$? I know that the map $\rho: \operatorname{Mor}(\operatorname{Spec}R,X)\to\operatorname{Hom}\text{rings}(R[T_1,\ldots,T_n]/I,R), (\sigma, \sigma^#)\mapsto \sigma^#(X)$ is bijective. I also proved that $\rho(X(R))\subseteq\operatorname{Hom}\text{algebra}(R[T_1,\ldots,T_n]/I,R)$. But I can't prove the opposite inclusion. – zxcv Apr 25 '19 at 07:52
  • In my above comment, $\operatorname{Mor}(\operatorname{Spec} R, X)$ is the set of all morphisms of scheme from $\operatorname{Spec} R$ to $X$, $\operatorname{Hom}\text{rings}(R[T_1,\ldots,T_n]/I,R)$ is the set of all ring homomorphisms from $R[T_1,\ldots,T_n]/I$ to $R$, and $\operatorname{Hom}\text{algebra}(R[T_1,\ldots,T_n]/I,R)$ is the set of all algebra homomorphisms from $R[T_1,\ldots,T_n]/I$ to $R$. – zxcv Apr 25 '19 at 08:01
  • @zxcv Suppose $A$ is an $R$-algebra with structure map $\rho\colon R\to A$. Then an $R$-algebra morphism $A\to R$ is a ring homomorphism $f\colon A\to R$ such that $f\circ \rho = \text{id}_R$ (do you agree?). – Alex Kruckman Apr 25 '19 at 14:55
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Let $Y=$ Spec $R$. The $R$-valued points of $X$, $\hspace{1mm} X(R)$, are the elements of Hom$(Y,X)$, which are in bijection with Hom$(k[T_1,\ldots,T_n]/I,R)$. Thus, $X(R)$ is in bijection with the tuples $\alpha=(a_1,\ldots,a_n) \in R^n$ for which $P(\alpha)=0$ for all $P \in I$.

Jeff
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  • However, when $R$ is a field, this is not to be confused with the notion of $R$-rational points... – Zhen Lin Jun 11 '13 at 21:01
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    @ZhenLin: Dear Zhen, I don't understand your comment. These are precisely the $R$-rational points. Regards, – Matt E Jun 18 '13 at 04:09