What does the Grothendieck group of $(\Bbb{N}, x\oplus y := 6xy + x + y)$ look like?

Question

Let $0 \in \Bbb{N}$ and define $\oplus : \Bbb{N}^2 \to \Bbb{N}$ to be $x\oplus y := 6xy + x + y$.

It is cancellative (associative / monoid law) since if $x\oplus y = 6xy + x + y = 6xz + x + z = x\oplus z \implies 6(x(y - z)) + y - z = (6x + 1)(y-z) = 0$ and $6x + 1 \neq 0$ or else $6 \mid 1$ which is absurd. And therefore $y = z$.

Therefore the Grothendieck group $Z$ of $N = (\Bbb{N}, \oplus)$ exists, is non-trivial and in fact embeds a copy of $N$ inside of it.

Therefore, we may work merely with $x \equiv (x,0) \in Z$ by a similar construction to this: Grothendieck group of the integers.

However, was wondering if we can work with the Grothendieck group in the usual algebra-like way, i.e. when we work in the rationals we work with fractions which are very familiar. So is there some more familiar way of working in this Grothendieck group?

You're simply transporting the multiplicative structure of $\Bbb N$ along the bijection $,n\mapsto 6n+1.\ $ We can transport it along a bijection to any set of the same cardinality. — Bill Dubuque, Dec 31 '24 at 04:53

score 5 · Accepted Answer · edited Dec 31 '24 at 04:48

5

Not sure if this is what you are looking for, but note that $$6xy + x + y = \frac{1}{6} ((6x+1)(6y+1) - 1).$$ So if you do a change of variable $x' = 6x + 1$ and $y' = 6y + 1$, we have $$\frac{x' - 1}{6} \oplus \frac{y'-1}{6} = x \oplus y = 6xy + x + y = \frac{x'y' - 1}{6}.$$ So the situation is isomorphic the monoid $6\mathbb{N} + 1$ under usual multiplication, with identity 1. The Grothendieck group just consists of $\frac{x}{x'}$ for $x,x' \in 6\mathbb{N} + 1$.

edited Dec 31 '24 at 04:48

Martin Brandenburg

181,922

answered Dec 31 '24 at 03:19

dummy

814

I already know this. $f: (\Bbb{N}, \oplus) \to (\Bbb{N}, \cdot)$ is an injective monoid homomorphism. Where $f(n) = 6n + 1$ – Daniel Donnelly Dec 31 '24 at 03:38
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Cool, do you consider the group $(6\mathbb{N} + 1) \cup (6\mathbb{N} + 1)^{-1}$ under multiplication not familiar enough? – dummy Dec 31 '24 at 03:44
that is right. You're right! I'm just wondering how I work in $6\Bbb{N} + 5$ since together $6\Bbb{N} + {1, 5}$ another cancellative monoid is formed. – Daniel Donnelly Dec 31 '24 at 03:46
Same argument lead you to consider $(6\mathbb{N} + 1) \cup (6\mathbb{N} + 5) \cup (6\mathbb{N} + 1)^{-1} \cup (6\mathbb{N} + 5)^{-1}$ under multiplication. – dummy Dec 31 '24 at 03:50
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@dummy why union? – Martin Brandenburg Dec 31 '24 at 03:56
@MartinBrandenburg it can actually be upgraded to disjoint union. Or almost disjoint. Sometimes ${1}$. – Daniel Donnelly Dec 31 '24 at 03:57
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No I mean you should talk about fractions $x/y$, but not only about $x$ and $1/y$. – Martin Brandenburg Dec 31 '24 at 03:58
Oops you are correct. – dummy Dec 31 '24 at 04:00
@MartinBrandenburg Oh I guess he's saying take those sets as free monoid generators – Daniel Donnelly Dec 31 '24 at 04:00
All you have to write is $\Bbb{Q}{\gt 0} \cap \langle 6\Bbb{Z} \pm 1\rangle{\Bbb{Q}}^{\times}$. – Daniel Donnelly Dec 31 '24 at 04:01
I fixed the answer accordingly. – Martin Brandenburg Dec 31 '24 at 04:48
@DanielDonnelly Isn't everything in this answer already in your question? What here is new to you? – Bill Dubuque Dec 31 '24 at 04:59

What does the Grothendieck group of $(\Bbb{N}, x\oplus y := 6xy + x + y)$ look like?

1 Answers1