Is $\mathbb{Z}$ isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5\oplus\mathbb{Z}_7\oplus \cdots$?
This seems like a natural conceptual extension of the Chinese remainder theorem but I'm not sure how it would work with an infinite sum.
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asked Mar 6, 2013 at 18:29
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$\begingroup$ $\mathbb{Z}$ is merely a subring. $\endgroup$– user14972Mar 6, 2013 at 18:33
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1$\begingroup$ Isomorphic in which category? And what does $\mathbb{Z}_p$ mean? $p$-adics or $\mathbb{Z}/(p)$? $\endgroup$– Martin BrandenburgMar 6, 2013 at 18:35
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4$\begingroup$ @Hurkyl $\bf Z$ is a subring of $\prod\limits_p{\bf F}_p$, yes, but $\bigoplus\limits_p{\bf F}_p$, no. $\endgroup$– anonMar 6, 2013 at 18:42
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1$\begingroup$ You should prove for yourself that you can't write $\mathbb Z$ as a non-trivial product. $\endgroup$– JSchlatherMar 6, 2013 at 18:44
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$\begingroup$ @MartinBrandenburg $\mathbb{Z}_p$ here refers to $C_p$ or $\mathbb{Z}/(p\mathbb{Z})$. I meant it as a group isomorphism, but I'd be interested if the answer is different as a ring (or other category). $\endgroup$– Samuel HandwichMar 6, 2013 at 19:02
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4 Answers
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$(1,0,0,0,0,....)$ has order $2$ in $\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5\oplus\mathbb{Z}_7\oplus...$
No element in $\mathbb{Z}$ has order $2$.
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There is a natural map $\mathbb{Z} \to \prod_{p^k} \mathbb{Z}/p^k\mathbb{Z}$, where the product runs over all prime powers, given by taking remainders. The Chinese Remainder Theorem shows that this map is injective. Since the LHS is countable and the RHS is uncountable, it cannot be surjective.
However, the following natural question presents itself: "suppose I describe a number by describing its residues modulo all prime powers in a consistent way, e.g. if the number is $1 \bmod 4$ then it must also be $1 \bmod 2$. What kind of object do I get if I don't get an integer back?" The answer is that you get a profinite integer. The profinite integers $\hat{\mathbb{Z}}$ are a direct product of the $p$-adic integers $\mathbb{Z}_p$ over all primes $p$, which is in some sense the correct salvage of your conjecture.
answered Mar 7, 2013 at 5:04
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Every element in the direct sum has a finite order. To see this, note that an element in the direct sum is a sequence of finitely many non-zero elements, and in the coordinate $n$ the element is less than $p_n$ (the $n$-th prime).
Let $\bar a= \langle a_i\mid i\in\Bbb N\rangle$ be an element, and let $a_{k_1},\ldots,a_{k_n}$ be its non-zero coordinates. Take $m$ to be the $\operatorname{lcm}(k_1,\ldots,k_n)$ then we have that $m\cdot\bar a$ has to be $\bar 0$, the zero sequence, since for every $a_{k_i}$ we have some $t_i$ such that $m\cdot a_{k_i}=t_i\cdot k_i\cdot a_{k_i}\equiv 0\pmod{k_i}$.
So not only the direct sum is not isomorphic, there is no embedding from $\Bbb Z$ into the direct sum or vice versa either, since no element of $\Bbb Z$ has a finite order and no element of the direct sum has an infinite order.
answered Mar 6, 2013 at 18:38
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$\begingroup$ So if I changed this to $\otimes_{p\in \mathbb{P}} \mathbb{Z}_p$, would I be able to embed $\mathbb{Z}\rightarrow \otimes_{p\in \mathbb{P}} \mathbb{Z}_p$ by $1\mapsto (1,1,1,1,1,1,1,1,\ldots)$? $\endgroup$ Mar 6, 2013 at 18:57
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5$\begingroup$ @SamuelHandwich There is a diagonal embedding of rings ${\bf Z}\to\prod_p{\bf F}_p$, yes (and the image is clearly proper, since it is countable while the infinite product is uncountable). The $\LaTeX$ for a product is
\prodfor $\prod$, not\otimesfor $\otimes$; the latter symbol refers to tensor products which are very different. Indeed infinite tensor products are considerably harder to even describe. $\endgroup$– anonMar 6, 2013 at 20:03 -
3$\begingroup$ + Nice Asaf. you killed the poor $\mathbb Z$ $\endgroup$– MikasaMar 7, 2013 at 20:07
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And, after the good points in the comments and other answers, there is something going on here... the projective limit of the quotients $\mathbb Z/n$ is the product of all p-adic integers as p varies over primes. :)
answered Mar 6, 2013 at 18:45
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$\begingroup$ Is that also the projective limit of $\mathbb{Z}/p\mathbb{Z}$? (Also, what does that mean?) $\endgroup$ Mar 6, 2013 at 18:55
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1$\begingroup$ The quotients $\mathbb Z/p$ with p varying over primes don't really map to each other... but $\mathbb Z/p^k$ as $k$ varies have projective limit the p-adic integers $\mathbb Z_p$. "Projective limit" is a standard notion: a proj lim of abelian groups $A_i$ with $i$ in a poset and $f_{ij}:A_i\rightarrow A_j$ is an ab gp $A$ mapping to all $A_i$ compatibly with the $f_{ij}$'s and so that given compatible $F_:A_i\rightarrow B$ for ab gp $B$, there is unique compatible $A\rightarrow B$. (One should see examples and such... more than there is room for here...) $\endgroup$ Mar 6, 2013 at 19:03