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We have of course:

$2\Bbb Z-1$, or $2\Bbb N+1$

These I think of as the integers reduced by the congruence $x\cong x+1$ where in one case $x$ is even and in the other $x$ is odd (it makes no difference really).

But I want to think specifically of reducing integers by the congruence $x\cong 2x$ perhaps more clearly the transitive closure of the equivalence relation $x\sim 2x$.

My random stab in the dark is to write it $\Bbb Z/\langle 2\rangle$

Expressed like this, I guess the set includes $0$.

Since this is a multiplicative group modulo a prime, perhaps it's better to exclude $0$ something like: $\Bbb Z^\times/\langle2\rangle$?

What's acceptable / usual here?

it's a hire car baby
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  • Exactly what's your aim? $\Bbb Z/(2)$ has only two elements: (the representatives of) $0$ and $1$ – Berci Jun 21 '18 at 08:14
  • @Berci I intended $\langle2\rangle={1,2,4,8,16,\ldots}$ rather than $(2)$ by which I think you mean something different. My understanding is a quotient is based on equivalence classes. I want to set $\forall x:x\sim2^nx$, so $3\sim6\sim12\sim24\ldots$ and I want a notation for the set of such classes. – it's a hire car baby Jun 21 '18 at 08:33
  • If I understand correctly, you want your equivalence classes to be ${1, 2, 4, 8, \dots }, { 3, 6, 12, 24, \dots }, { 5, 10, 20, 40, \dots }, \dots, { 15, 30, 60, 120, \dots }$, and so on. This is not a usual quotient ring (since, for example $2 + 4 = 6$ and $2 + 2 = 4$ and the left hand sides are sums of equivalent elements but the right hand sides are not equivalent). Your group idea doesn’t work neither because $\Bbb Z^\times = {-1, 1}$. – Eike Schulte Jun 21 '18 at 08:37
  • You might want to look into localization, in particular localization of $\Bbb Z$ by $2$ (written $\Bbb Z_2$). In that ring, to elements are equivalent in the way you want when they are associated, i.e. if one is the other multiplied by a unit. – Eike Schulte Jun 21 '18 at 08:48
  • @EikeSchulte yes. I have rudimentary 2-adic theory in my armoury but I want to restrict closer to $\Bbb {Z}$ and not necessarily go to a field. Maybe the answer is to go to $\Bbb Z[\frac{1}{2}]^\times/\langle2\rangle$ which I believe is then a quotient of the multiplicative group. – it's a hire car baby Jun 21 '18 at 08:57
  • @RobertFrost I don’t think $\Bbb Z[\frac 1 2]^\times/\langle 2 \rangle$ is what you want, because the units of $\Bbb Z[\frac 1 2]$ are the fractions containing only powers of two. I actually don’t think that you can find it as a nice quotient, simply because the set of equivalence classes is basically the free commutative monoid generated by the odd primes, which is neither a group nor a ring, so you can’t expect to find it as a quotient of a group or ring. – Eike Schulte Jun 21 '18 at 09:05
  • @EikeSchulte if the fractions contain only powers of $2$ then if $\langle2\rangle={\ldots\frac14,\frac12,1,2,4,\ldots}$ I think the quotient reduces $\Bbb Z[\frac{1}{2}]^\times$ perfectly to the odd integers and the cosets of the odd integers contain exactly every even multiple and every fraction with power of two denominators? – it's a hire car baby Jun 21 '18 at 09:07
  • @RobertFrost No, the multiplicative group $\Bbb Z[\frac 1 2]^\times$ simply does not contain any odd integers (except $1$) anymore, and you won’t get them back by passing to a quotient. In fact $\Bbb Z[\frac 1 2]^\times/\langle 2 \rangle$ is simply the group of order two, ${-1, 1}$, because every element of $\Bbb Z[\frac 1 2]^\times$ is one of the fractions in $\langle 2 \rangle$ or the negative of one of those fractions. – Eike Schulte Jun 21 '18 at 09:26
  • @EikeSchulte ah yes i see, e.g. $\frac{5}{2}$ would require $\frac{2}{5}$. – it's a hire car baby Jun 21 '18 at 09:28

1 Answers1

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If I understand the question right, we are looking at $\mathbb N$ and the relation $x \sim y \iff $ there exists $i \in \mathbb Z$ such that $x = y \cdot 2^i$. This is an equivalence relation, and - apart from the equivalence class {0} - each equivalence class contains a unique odd integer.

$\mathbb N$ with multiplication is a monoid (like a group, but without the requirement for inverses to exist), and it's easy to show that for any monoid $A$, if you have an equivalence relation $\sim$ such that $a \sim b$ and $x \sim y$ implies $ax \sim by$ - which holds in the case we are considering - then there is a quotient monoid structure on $A\ / \sim$, and it is this quotient monoid structure you are thinking of.

I don't know of any particular standard notation for the equivalence relation $\sim$ or for $\Bbb N \ / \sim$.

Christopher
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  • Thank-you. Not pushing for extensive further information right now, but I guess the quotient monoid has some form of addition - what does that look like and is it a reasonable question to ask whether multiplication and exponentiation can be defined? – it's a hire car baby Jun 21 '18 at 09:31
  • A monoid has a single operation - in this case it is multiplication. Exponentiation - in a monoid $A$ you can define $a^n$ for $a \in A$, $n \in \Bbb N$ but not more general exponentiation. – Christopher Jun 21 '18 at 09:33
  • Multiplication is well definwd in $\Bbb N/\sim$. If you also want addition to be well defined in the quotient, it will become degenerate. – Berci Jun 21 '18 at 11:52
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    Btw, as monoid, $\Bbb N/\sim\cong\Bbb N$, maybe that's a reason why it didn't receive any particular name. (Both are free commutative monoids generated by countably infinite elements, and attached a $0$.) – Berci Jun 21 '18 at 11:55
  • @Berci by "some form of addition" I meant multiplication!! And I was asking whether exponentiation and beyond are defined. If I understand correctly the countably many infinite elements are the primes either with or without $2$. So there's a morphism from $\Bbb N/\sim$ into $\Bbb N$ in which $p_n\to p_{n+1}$ is that correct. – it's a hire car baby Jun 21 '18 at 11:57
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    Yes, exactly, the primes generate it freely. – Berci Jun 21 '18 at 12:09
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    @RobertFrost yes, though the morphism defined by $p_n \mapsto p_{n+1}$ goes $\Bbb N \to \Bbb N \ / \sim$ not the other way around – Christopher Jun 21 '18 at 12:11
  • @Berci I have previously understood "freely generated" in the context of groups, which would imply the set contains multiplicative inverses. I guess it's okay to say freely generated in the context of a monoid and it's assumed every sequence of powers of some prime is well-founded? – it's a hire car baby Jun 21 '18 at 12:54