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If I consider the analogy of local ring at a point to the space of function germs at the point, then the residue field can be seen as the values that functions can take at the point.

But when I consider the residue field of generic point or the residue field of a point in a scheme over a non-algebraically closed field, the above analogy becomes unreasonable to me.

On Wikipedia entry "Residie field", it says "One can say a little loosely that the residue field of a point of an abstract algebraic variety is the 'natural domain' for the coordinates of the point." Can you elaborate also a bit on this?

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    $\begingroup$ Let me specialize to varieties to talk about the generic point. The residue field at the generic point is basically the field of possible "coherent collections of values at closed points," i.e. collections of values that extend to a collection of compatible germs (which, of course, then glue to a regular function). But we only actually look at some indeterminate dense open subset of the closed points, so in fact we only have enough germs to glue to a rational function. And that's what the residue field at the generic point is: the field of rational functions. $\endgroup$ Oct 9, 2021 at 20:29
  • $\begingroup$ the point is that we can do whatever people do in the language of varieties and functions with commutative algebra and replacing functions by element of a ring. but if you want a classical meaning for residue field of generic point there is one: sometimes you are not interested in the value of a function at a given point but you want to understand something about that function on a general point satisfying some condition, $\endgroup$
    – ali
    Oct 10, 2021 at 11:22
  • $\begingroup$ classically you write the function with parameter $x_1,,,.,x_n$ such that $x_1,...,x_n$ satisfy that condition $g(x_1,...,x_n)=0$. this is exactly the meaning of the residue field of the generic point of the $V(g)$. for example if you want to check that a variety is smooth at a given point you have to check the value of the determinant of some matrix on that given point. if this determinant does not vanish at a generic point of a subvariety you can say that it does not vanish at a dense open subset of that variety so the "generic point" on that subvaritey is smooth $\endgroup$
    – ali
    Oct 10, 2021 at 11:26
  • $\begingroup$ @TabesBridges see my edit, can you also help me clarify a bit about that sentence on wikipedia? $\endgroup$ Oct 10, 2021 at 14:39

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As you note, Wikipedia describes the residue field of an abstract variety $X$ at a point $p$ as the "natural domain" for coordinates at $p$; I'm not sure I like that statement, and I would probably say instead that the residue field is the natural codomain for functions defined at a given point.

Let me remain specialized to varieties to talk about the generic point. The first object one might consider in the search for a natural codomain would be the direct product of all residue fields at all closed points, $$ \displaystyle\prod_{p \in X \text{ closed}} k(p). $$

But this is WAY too big, and in particular its elements need have nothing to do with regular functions on $X$. For instance, by Lagrange interpolation a regular function on $\mathbb A^1$ over an algebraically closed field is determined by its values at finitely many points, but specifying an element of this big product requires specifying its values at every single closed point of $X$.

You should think of the residue field at the generic point as a certain subset of this giant product. It is basically the field of possible "coherent collections of values at closed points," i.e. collections of values that extend to a collection of compatible germs (which, of course, then glue to a regular function). But we only actually look at some indeterminate dense open subset of the closed points, so in fact we only have enough germs to glue to a rational function. And that's what the residue field at the generic point is: the field of rational functions.

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Question: "But when I consider the residue field of generic point or the residue field of a point in a scheme over a non-algebraically closed field, the above analogy becomes unreasonable to me."

Let $X$ be a locally Noetherian integral scheme and let $F$ be a coherent sheaf on $X$. We may construct the torsion subsheaf $T(F)$ of $F$ in the following way: for any affine open $U⊆X$, denote $A=Γ(U,O_X)$ and $M=Γ(U,F)$. Define

$$T(F)(U)=T(M)=\{m∈M|∃a∈A \text{ non-zero divisor such that $a⋅m=0$} \}.$$

Claim: $T(F)$ is a (quasi)coherent subsheaf of $F$.

Proof: If $X$ is a locally noetherian integral scheme, there is a canonical map

$$\phi: E \rightarrow E\otimes_{\mathcal{O}_X} K_X$$

where $K_X$ is the "sheaf of quotient fields" of $X$. For any open subscheme $U:=Spec(A) \subseteq X$ it follows $K:=K_X(U)\cong K(A) \cong \mathcal{O}_{X,\eta}$ where $\eta$ is the generic point. Let $\tilde{E}:=ker(\phi)$. It follows by an exercise in Atiyah-Macdonald that if $E(U) \cong M$ that $\tilde{E}(U)\cong T(M) \subseteq M$ is the torsion sub module of $M$. Hence since $\phi$ is a map of quasi coherent sheaves it follows $T(E):=ker(\phi)$ is a quasi coherentsheaf.

Hence one usage of the generic points is to use it to define the quotient field and the sheaf $K_X$ on an integral scheme $X$. This is used to define the torsion subsheaf $T(E)\subseteq E$ for any (quasi)coherent sheaf $E$.

Note: The residue field $\kappa(x)$ of any points $x\in X$ is defined using the local ring $\mathcal{O}_{X,x}$ and its unique maximal ideal $\mathfrak{m}_x$: By definition $\kappa(x):=\mathcal{O}_{X,x}/\mathfrak{m}_x$, hence for any local section $s\in \mathcal{O}_X(U)$ and any points $x\in U$ you may consider the "value of $s$ at $x$" $s_x:=\overline{s} \in \kappa(x)$. You "evaluate" a section $s$ at a point $x$. In algebra/geometry one seldom studies this "value" , one studies the zeros of $s$: The set of $x$ where $s_x=0$ and this is a subscheme of $U$.

Example: If $C:=\mathbb{P}^1_k$ and $s(x_0,x_1)\in H^0(D(x_0), \mathcal{O}(d))$ is a section at the open set $D(x_0)$ you may write

$$s(x_0,x_1)= \sum_i a_ix_1^{d-i}x_0^i=\sum_i a_it^ix_0^d=f(t)x_0^d$$

with $f(t)\in k[t]$ and $t:=x_1/x_0$ and it follows the zero scheme satisfies

$$Z(s):=V(f(t))) \cong Spec(k[t]/(f(t))\subseteq D(x_0).$$

If $k$ is algebraically closed and if $f(t)=\prod_j (t-b_j)^{l_j}$ it follows the $n$ $k$-points $b_1,..,b_n\in D(x_0)(k)$ have multiplicity $l_j$: The scheme $Z(s)$ is the set $b_j$ with multiplicities $l_j\geq 1$. One frequently studies the divisor $Z(s)$ of a section $s$ in terms of the points $b_j$ and the multiplicities $l_j$. More generally if $k$ is not algebraically closed you get a decomposition $f(t)=\prod_j p_j(t)^{l_j}$ with $p_j(t) \subseteq k[t]$ an irreducible polynomial and $l_j \geq 1$, and you may view the zero scheme $Z(s):=Spec(k[t]/(f(t))$ as the $n$ points $p_j:=(p_j(t)) \subseteq k[t]$ with multiplicities $l_j$. The residue field $\kappa(x_j):=k[t]/(p_j(t))$ will be a finite extension of $k$.

Note: You seldom study the "values of $s$ at closed points $x$". What you study is the "divisor of zeros" $Z(s)$ of $s$.

Question: "On Wikipedia entry "Residue field", it says "One can say a little loosely that the residue field of a point of an abstract algebraic variety is the 'natural domain' for the coordinates of the point." Can you elaborate also a bit on this?"

Answer: If $k$ is any algebraically closed field and $I:=(f_1,..,f_l) \subseteq A:=k[x_1,..,x_n]$ is a prime ideal, let $X:=Z(I)\subseteq \mathbb{A}^n_k$ be the corresponding irreducible algebraic variety. Let $g(x_1,..,x_n)\in A$ be any polynomial and view

$$g:\mathbb{A}^n_k(k) \rightarrow k$$

as a function where for any $n$-tuple $p:=(p_1,..,p_n)\in k^n$ you define $ g(p):=g(p_1,..,p_n)\in k$. You let $A(X):=A/I$ be the coordinate ring of $X$ and since $k$ is algebraically closed it follows a maximal ideal $\mathfrak{m}\subseteq A(X)$ is on the form $\mathfrak{m}:=(x_1-a_1,..,x_n-a_n)$ with $f_j(a_1,..,a_n)=0$ for all $j$. You may consider $g$ as an element in $A(X)$ and the "value" of $g$ at the "point " $\mathfrak{m}$ is the class

$$\overline{g}=g(a_1,..,a_n)\in A(X)/\mathfrak{m} \cong k.$$

This is how the polynomial $g$ is a function on the set of closed points of $X$. Hence the "natural domain" for the polynomial function $g$ is the set of closed (and non-closed) points of $X$ (and $\mathbb{A}^n_k$).

Example: If $A$ is a commutative ring and $M$ a left $A$-module and $s,t\in M$, let $I:=\{a(s-t):a\in A\}\subseteq M$ be the submodule generated by the element $s-t\in M$. By Atiyah-Macdonald Prop 3.8 it follows $I_{\mathfrak{m}}=0$ for all maximal ideals $\mathfrak{m}$ iff $I=0$. Hence there is an equality of sections $s=t$ iff $s_{\mathfrak{m}}=t_{\mathfrak{m}}$ for all maximal ideals $\mathfrak{m}$. Hence to check if the sections $s$ and $t$ give the same element in $M$, it is enough to check this at stalks at maximal ideals.

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  • $\begingroup$ I think what you wanted to say right at the end was "The set of $x$ where $s_x\in \mathfrak{m}_x$. $\endgroup$ Oct 10, 2021 at 10:42
  • $\begingroup$ @DouglasMolin: By the above notation $s_x\in \kappa(x)$. There is another element $s(x)\in \mathcal{O}_{X,x}$ and $s_x=0$ iff $s(x)\in \mathfrak{m}_x$. $\endgroup$
    – hm2020
    Oct 10, 2021 at 10:45
  • $\begingroup$ My apologies, I misread your notation. $\endgroup$ Oct 10, 2021 at 10:45
  • $\begingroup$ @DouglasMolin - the zero scheme $D:=Z(s)$ of a section $s$ is a divisor on $C$: In the above case it is a set of $n$ $k$-points $b_j$ with multiplicities $l_j \geq 1$. $\endgroup$
    – hm2020
    Oct 10, 2021 at 11:07

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