I recently started studying the theory of schemes from "The geometry of schemes", by Eisenbud and Harris. At the very beginning of this volume I unfortunately found the first osbtacle: pick a commutative ring with unity $R$, and consider $f \in R$ as a function on Spec$R$. The authors define $f(\mathfrak{p})$ as the image of $f$ via the canonical maps $$R \rightarrow R/\mathfrak{p}\rightarrow k(\mathfrak{p}) $$ with $k(\mathfrak{p})$ the residue field of $R/\mathfrak{p}$.

If I'm not wrong, $k(\mathfrak{p})=((A/\mathfrak{p})\setminus \{[0]\})^{-1}A/\mathfrak{p}$, which should be equal to $k(\mathfrak{p}) \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ (but I can't figure out how, all this localisations confuse me).

The problem is that I'm getting confused with all this stuff, I mean, I always naively consider the function $15$ at $(7)$, then $15((7))=1$ mod $7$, or working over $k[x]$, $$x^3-1((x^2+1))= -x-1 (\text{ mod } (x^2+1)).$$

But I don't see all this machinery behind this computation, I can do the calculation but I feel I don't undertand what $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ is and why I'm working in this exact space.

Any help which clarifies the theoretic aspect of this calulations would be appreciate, I'm quite confused about this ring. Thanks in advance.


1 Answer 1


Maybe it helps to put in something more geometric. Let's say $k = \bar{k}$ is an algebraically closed field and you look at something like $\mathbb A^n := \mathrm{Spec} \ k[X_1,...,X_n]$.

Let's first consider a global section $f = \sum_{\underline i} a_{\underline i} X_1^{i_1}\cdot ... \cdot X_n^{i_n} \in \Gamma(\mathbb A^n, \mathcal O_{\mathbb A^n}) = k[X_1,...,X_n]$.

Take any closed point $\mathfrak{m} \in \mathbb A^n$. By Hilbert's Nullstellensatz we know that there exist $x_1,...,x_n \in k$ such that $\mathfrak{m} = (X_1 -x_1, ...., X_n-x_n)$.

Now observe that the residue field is given by $\kappa(\mathfrak m) = k[X_1,...,X_n]/\mathfrak m \cong k$ where the last isormophism is given by $X_i \mapsto x_i$.

Now we can compute:

$f(\mathfrak m) = \sum_{\underline i} a_{\underline i} X_1^{i_1}\cdot ... \cdot X_n^{i_n} \equiv \sum_{\underline i} a_{\underline i} x_1^{i_1}\cdot ... \cdot x_n^{i_n} \mod \mathfrak m = f(x_1,...,x_n) \in k$. Hence if we think of $\mathfrak m$ as the point of $\mathbb A^n$ with coordinates $(x_1,...,x_n)$, this definition of evaluation makes perfectly sense.

Also note that it makes sense that for example $\Gamma(D(g),\mathcal O_{\mathbb A^n}) = k[X_1,...,X_n]_g$:

A function on $D(g)$ is of the form $\frac{f}{g^k}$.

When is a closed point $(x_1,...,x_n)$ in $D(g)$? Well, $\mathrm{Spec} \ k[X_1,...,X_n]_g \cong \{\mathfrak p \in \mathrm{Spec} \ k[X_1,...,X_n] \mid g \not\in \mathfrak p\}$.

What does it mean of a maximal ideal $\mathfrak m := (X_1-x_1,...,X_n-x_n)$ to contain $g$?

$g \in \mathfrak m \iff g \equiv 0 \mod \mathfrak m \iff g(x_1,...,x_n) = 0$.

Hence: If $(x_1,...,x_n) \in D(g)$, then $g(x_1,...,x_n) \neq 0$ and hence $\frac{f(x_1,...,x_n)}{g(x_1,...,x_n)} \in k$ is well-defined (and you can check that this really is the evaluation you get using your definition above).

Note that we have non-closed points $\mathfrak p$ in $\mathbb A^n$ as well. For example the generic point $(0)$. The evaluation at these points won't give us actual numbers, but something different instead. For example $f((0)) = f \in k(X_1,...,X_n)$.

Working over non-algebraically closed fields we might end up getting different residue fields, even in the very "geometric" example of $\mathbb A^n_k$. For example if $n = 1$ and $k = \mathbb Q$ we have the closed point $(X^2 - 2)$ so that $\kappa( (X^2 - 2) ) = \mathbb Q(\sqrt 2)$.

In the case of $\mathrm{Spec} \ \mathbb Z$ the situation is even more weird:

The residue field at the generic point $(0)$ is given by $\mathbb Q$ and the residue field at a prime $(p)$ is given by $\mathbb F_p$. Hence a function $x \in \Gamma(\mathrm{Spec} \ \mathbb Z, \mathcal O_{\mathbb Z}) = \mathbb Z$ takes values in different fields at every point.

However, it still makes sense to evaluate at every point. And the same intution works for sets like $D(2) = \{(2)\}^{c}.$ As on a point $\mathfrak p \in D(2)$, one has that $2 \in \kappa(\mathfrak p)^\times$, it makes sense to compute $\frac{f}{2^k}(\mathfrak p) = \frac{f(\mathfrak p)}{2(\mathfrak p)} \kappa(\mathfrak p)$.

// Edit: Also, regarding your question about the residue fields: Let $A$ be a ring, $S$ a multiplicative subset and $I \subseteq A$ an ideal.

Take the exact sequence $0 \to I \to A \to A/I \to 0$ of $A$-modules.

Applying the exact functor $S^{-1}$ yields the exact sequence:

$0 \to S^{-1}I \to S^{-1}A \to S^{-1}(A/I) \to 0$ of $S^{-1}A$-modules, hence also of $A$-modules via the canonical map $A \to S^{-1}A$.

That is: $S^{-1}A/I \cong S^{-1}A / S^{-1}I$ as $A$-modules.

Both sides are rings, and it is not too difficult to check that the above map sends $1 \mapsto 1$ and that it is multiplicative, hence a ring isomorphism.

This gives you the desired isomorphism that you described in your question.

PS: Ravi Vakil's notes try to be very geometric and are filled with exercises that try to explain, why Algebraic Geometry deserves to be called Geometry.

  • $\begingroup$ thanks for your answer, I really appreciate that. Before accepting it, I have only a small doubt: you say that $f((0))=f\in k(X_1,\ldots, X_n)$, so one shold define the structure sheaf for a open set $U$ as $\mathcal{O}_X(U)=\{s: U \rightarrow \dot{\bigcup}_{\mathfrak{p} \in U} k(\mathfrak{p}) \}$, and not as $\{s: U \rightarrow \dot{\bigcup}_{\mathfrak{p} \in U} A_{\mathfrak{p}} \}$ as Hartshorne does. So, am I missing something? These disjoint unions are different as rings $\endgroup$
    – blob
    Apr 4, 2019 at 15:11
  • $\begingroup$ I don't know the definition of Hartshorne at all @christmas_light. But my claim that $f((0))$ is just putting in your given definition: As we're looking at the $(0)$ ideal, by the isomorphism you mentioned we get that $\kappa(\mathfrak p) = k[X_1,...,X_n]_{(0)} / (0) = \mathrm{Quot}(k[X_1,...,X_n]) = k(X_1,...,X_n)$ in this case. $\endgroup$
    – lush
    Apr 4, 2019 at 15:57
  • $\begingroup$ As I'm not really sure what Hartshorne is doing I'm blind guessing a little bit here, but if I'm wrong some other people might be able to figure out what's causing the confusion here: A function $f \in \Gamma(X,\mathcal O)$ of a scheme $X$ is not determined by all of its values $f(x), x\in X$ in general. So.. there are (even affine!) schemes that have functions $f,g$ such that $f(x) = g(x)$ on every point, but $f \neq g$. $\endgroup$
    – lush
    Apr 4, 2019 at 16:00
  • $\begingroup$ You can see this in the affine case as follows: Because $(f+g)(x) = f(x) + g(x)$ the problem immediately reduces to finding all functions $f$ such that $f(x) = 0$ on all points $x$. $f(x) = 0$ means that $f \in A \cap \mathfrak p_x A_{\mathfrak p_x} = \mathfrak p_x$. Hence a function $f \in A$ vanishes at every point iff. $f \in \cap_{x \in \mathrm{Spec} A} \mathfrak p_x = \mathrm{nil} \ A$. And hence functions on $A$ are determined by their values if and only if $A$ is reduced. $\endgroup$
    – lush
    Apr 4, 2019 at 16:03
  • $\begingroup$ Is that what you were worrying about @christmas_light? $\endgroup$
    – lush
    Apr 4, 2019 at 16:03

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