# Understanding valutation and residue field

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.

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.

• 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
– blob
Apr 4, 2019 at 15:11
• 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.
– lush
Apr 4, 2019 at 15:57
• 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 $f \neq g$.
– lush
Apr 4, 2019 at 16:00
• 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.
– lush
Apr 4, 2019 at 16:03
• Is that what you were worrying about @christmas_light?
– lush
Apr 4, 2019 at 16:03