6
$\begingroup$

I'm having trouble understanding the following argument (which I believe to be somewhat incomplete or flawed). Let $A=C(X)$ be the set of continuous functions from the topological space $X$ to the complex plane $\mathbb{C}$. We define $m_{x} = \{f \in C(X): f(x) = 0 \}$ and $A_x$ the ring of germs at point $x$. The statement is the following $A_x \simeq A_{m_x}$.

(1) I don't see how one defines $A_{m_x}$ since the set contains global functions that might not be well-defined as we quotient by functions $f$ such that $f(x) \neq 0$, but it doesn't necessarily mean that $f \neq 0$. Though it's indeed well defined in a neighborhood of $x$.

(2) Now using the universal property of localization, we sure want to define $\phi : A_{m_x} \rightarrow A_x$ s.t we have $\phi(a/s) = \iota(a)\iota(s)^{-1}$ where $\iota$ is the inclusion map $\iota: A \rightarrow A_x$. We want $\phi$ to be an isomorphism. It is surjective; now we want it to be one-on-one. Now I don't see how this is possible as $\phi(a/s) = 0$ iff $a = 0$ in a neighborhood of $x$, which doesn't imply that $a=0$ globally.

I guess there's something I don't really fathom, or my textbook might just be flawed. Anyway, thanks for your help.

$\endgroup$
0

2 Answers 2

4
$\begingroup$

You need to assume something about the space $X$ for the claimed statement to be true. A natural assumption to make is that $X$ is completely regular; I will use this assumption in addressing (2) below.

(1) Well, $A_{m_x}$ is not (a priori) a set of functions, it's just a ring of formal "fractions" (which may or may not make sense when evaluated as pointwise fractions of functions). You can still think of it perfectly well as a ring without having to think of its elements as functions on $X$ (which, as you observe, you can't exactly do).

(2) If $a\in A$ is such that $a=0$ in a neighborhood $U$ of $x$, then in fact the image of $a$ in the localization $A_{m_x}$ vanishes: the canonical "inclusion" $A\to A_{m_x}$ is not injective! To prove this, note that by complete regularity, there is a function $f:X\to[0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $y\not\in U$. We then have $f\not\in m_x$ and $fa=0$, so it follows that $a$ maps to $0$ in the localization $A_{m_x}$.

Note that you also need to use complete regularity to show that your map $\phi:A_{m_x}\to A_x$ is surjective: given a germ of a continuous function at $x$, it is not at all obvious a priori that you can write it as a quotient of two continuous functions that are defined on all of $X$. In detail, if you have a function $f:U\to\mathbb{C}$ where $U$ is an open neighborhood of $x$, let $V$ be an open neighborhood of $x$ whose closure is contained in $U$ (by regularity) and let $g:X\to[0,1]$ be a function such that $g(x)=1$ and $g(y)=0$ for all $y\not\in V$ (by complete regularity). Define $h(y)=\min(1,2g(y))$. Then $h=1$ on a neighborhood of $x$ (namely, the set where $g>1/2$), so $hf$ (which is defined on $U$) has the same germ at $x$ as $f$. But $hf$ vanishes on $U\setminus\overline{V}$, so we can continuously extend it to all of $X$ by setting it equal to $0$ outside of $U$. This continuous extension is then an element of $A$ whose germ at $x$ coincides with the germ of $f$. Thus the map $A\to A_x$ is surjective, and hence so is the map $A_{m_x}\to A_x$.

$\endgroup$
9
  • $\begingroup$ Ah, complete regularity being necessary to construct an inverse for a germ, in my scheme... This is an important point. $\endgroup$
    – rschwieb
    Commented Dec 29, 2015 at 11:52
  • $\begingroup$ Er, yes, my comment was directed at the earlier version of your answer. As for the current version, it is trivial that the germ of a function not vanishing at $x$ is invertible (just restrict it to the set where it is nonzero and take its inverse there). No complete regularity is needed for that. $\endgroup$ Commented Dec 29, 2015 at 12:07
  • $\begingroup$ I thought that was wher you comment applied, but yes, that is what I initially thought too. Time to wit for coffee to reevaluate everything... $\endgroup$
    – rschwieb
    Commented Dec 29, 2015 at 12:20
  • $\begingroup$ Yes indeed, when thinking about it, I had to assume for instance than X was a metric space, so I could put something like $f(x) = d(x, V^{c})$ where V is the open subset where our given function doesn't get to zero...So is it true, nonetheless, that $A_{x} \simeq A_{m_x}$? (It doesn't seem like it anymore) (and yes, $A=C(X)$ ;)) $\endgroup$
    – Hermès
    Commented Dec 29, 2015 at 12:34
  • 1
    $\begingroup$ No, it's not true for arbitrary $X$ that $A_x\simeq A_{m_x}$. For a counterexample, you can let $X=\mathbb{R}\cup\{x\}$, where a set $U\subseteq X$ is open iff it is an open subset of $\mathbb{R}$ not containing $0$ or it is of the form $V\cup\{x\}$ for an open set $V\subseteq\mathbb{R}$. Then $A_x\simeq\mathbb{C}$ (since $\{x\}$ is open) but you can show $A_{m_x}$ is not a field. $\endgroup$ Commented Dec 29, 2015 at 18:10
3
$\begingroup$
  1. The collection of functions that are nonzero at $x$ are certainly a multiplicatively closed subset $S$ of the ring, so there is no problem localizing at this set.

  2. We can always map a function $f$ to its germ $[f]$ at $x$. If $f$ is nonzero at $x$, it is nonzero on a neighborhood of $x$, and the function $\frac 1f$ is defined and nonzero on the same neighborhood. Then the germ of the inverse of $f$ on the interval is the inverse of the germ of $f$. So, things in $M_x$ get mapped to units of the ring of germs. By the universal property of localization, there is now a map of the localization ontointo the ring of germs at $x$. (Addendum: I never did catch that the map is not obviously onto and I won't pretend here by duplicating the work Eric Wofsey does in his solution to show that complete regularity of $X$ is sufficient to prove the map is onto.)

    Your last comment is on the right track. But your concern that the preimage is not globally zero is irrelevant. Starting with $f/s$ in your ring of fractions, if $[\frac fs]=[0]$ is the zero germ, then by definition of the extension $[f]=[0][s]=[0]$. As you noted, $f$ is zero on a neighborhood $U$ of $x$. To close the case we need to show that there exists $t\in S$ such that $tf=0$, demonstrating that $f\sim 0$ in the localization. For this, you need complete regularity (as pointed out by Eric Wofsey). By finding a continuous function which is zero on $U^c$ and $1$ at $x$, you will have such an element $t$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .