6
$\begingroup$

This is an example from Algebraic Topology, by Hatcher.

As far as I understand, I have to take the direct sum of all the $G_i$s (in this case, $\mathbb{Z}\oplus\mathbb{Z}\oplus...$) and quotient out all elements of the form $(g_1,g_2-\alpha_1(g_1),...)$ with $\alpha_i$ as given in the definition on the same page.

In this case, what are the elements I need to quotient out? Won't they be elements of the form $(z_1,z_2-pz_1,...)$? Because given that $z_2$ was obtained by taking an integer and multiplying it by $p$, $z_2-pz_1$ will be a multiple of $p$.

I just don't see how setting these to $0$ is the same as $\mathbb{Z}[1/p]$.

$\endgroup$
4
  • 1
    $\begingroup$ "free abelian group generated by [1/p]" Huh? $\Bbb Z[1/p]$ is the smallest subring of $\Bbb Q$ containing $\Bbb Z$ and $1/p$. All its elements are expressible as polynomials in $1/p$ with integer coefficients, all of which can be simplified to rationals of the form $a/p^n$ with $a$ an integer. This group is not free abelian, and is not finitely generated. $\endgroup$
    – whacka
    Dec 6, 2015 at 22:07
  • $\begingroup$ Your description of the direct limit looks to me like a mixture of the definitions of direct and inverse limits. $\endgroup$ Dec 6, 2015 at 22:11
  • 2
    $\begingroup$ @QiaochuYuan I disagree. The equations $z_2=\frac{z_1}p$, etc. in your comment won't be satisfiable by any integers other than all zeros. $\endgroup$ Dec 6, 2015 at 22:13
  • $\begingroup$ Ah, sorry, I had it backwards. $\endgroup$ Dec 7, 2015 at 0:25

1 Answer 1

8
$\begingroup$

There's a better way to see why $\Bbb Z[1/p]$ is the direct limit (in my opinion): set up an isomorphism from the original direct limit to another one that's easier to understand:

$$\require{AMScd} \begin{CD} \mathbb{Z} @>{p}>> \mathbb{Z} @>{p}>> \mathbb{Z} @>{p}>> \mathbb{Z} @>{p}>>\cdots \\ @VV{1}V @VV{1/p}V @VV{1/p^2}V @VV{1/p^3}V \\ \mathbb{Z} @>{}>> \frac{1}{p}\mathbb{Z} @>{}>> \frac{1}{p^2}\mathbb{Z} @>{}>> \frac{1}{p^3}\mathbb{Z} @>{}>> \cdots \end{CD}$$

The unlabelled arrows in the bottom row are simple inclusions. The other arrows are multiplication maps and are labelled by the scaling factor. In other words, the map $\mathbb{Z}\xrightarrow{p}\mathbb{Z}$ looks the exact same as the inclusion $\mathbb{Z}\hookrightarrow\frac{1}{p}\mathbb{Z}$, and similarly for all the other parts of the commutative diagram. The direct limit in the bottom is then $\bigcup \frac{1}{p^n}\mathbb{Z}=\mathbb{Z}[1/p]$.

I haven't seen the construction you're talking about, but it seems intuitively clear why it should also define the direct limit. Say we have $G_1\to G_2\to G_3\to\cdots$. We want to interpret the arrows as "inclusions" (any noninjectivity means we pretend elements were the same to begin with in order to maintain this interpretation), in which case the direct limit is the "union" of all the $G_i$s. So our direct limit needs elements from all the things, so we can start off with $\bigoplus_i G_i$ and identify things that are supposed to be equal by quotienting by their differences.

In particular, $g\in G_i$ should represent the same element in the direct limit as $\alpha_i(g)\in G_{i+1}$ (where $\alpha_i:G_i\to G_{i+1}$), so we want their difference $(\cdots,0, g,-\alpha_i(g),0,\cdots)$ within the direct sum to be zero, and thus we must quotient by the subobject comprised of elements of the form

$$(g_1,g_2-\alpha_1(g_1),g_3-\alpha_2(g_2),\cdots)$$

(where of course all but finitely many coordinates are zero). Indeed, the kernel of the homomorphism $\bigoplus_i\mathbb{Z}\to\mathbb{Z}[1/p]$ where the $i$th factor gets multiplied by $\frac{1}{p^i}$ and included is generated by things that looks like $(z_1,z_2-pz_1,\cdots,z_n-pz_{n-1},-pz_n,0,\cdots)$.

$\endgroup$
2
  • 1
    $\begingroup$ One possible reason to present this colimit as in your top row is to avoid using an ambient $\mathbb Q_p$, especially if the point of the exercise is to construct $\mathbb Q_p$ as an abelian group (rather than declaring it the field of fractions of $\mathbb A_p$, although it certainly is). $\endgroup$ Dec 6, 2015 at 22:55
  • $\begingroup$ Wasn't expecting such a clear and in depth reply, thanks! $\endgroup$ Dec 6, 2015 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.