1

For example. Everyone hates $3$, so let's remove it all-together from $\Bbb{Z}$: Let $\Bbb{Z}' = \Bbb{Z} - 3\Bbb{Z}$. Then is there a way to keep it a ring, i.e. $\Bbb{Z}'$ forms a ring by redefining $+$ somehow? By that I mean not too much as obviously any infinite set can be turned into an additive group. But for instance, something like, if $a, b \in \Bbb{Z}', a + b \in 3\Bbb{Z} \implies a + b := 0$. Something simple like that.

Daniel Donnelly
  • 21,342
  • 1
    Well, you've killed the additive identity, which isn't a good sign. Are you sure you don't want to consider the quotient ring?

    Edit: okay, you definitely want the quotient ring. Do you know how that's defined?

     – Ian Coley Apr 10 '14 at 23:10
  • @IanColey no way, needs to be infinite. You're telling me that a related structure to $\Bbb{Z}$ doesn't reside in $\Bbb{Z} - 3\Bbb{Z}$. That's shocking to me.... – Daniel Donnelly Apr 10 '14 at 23:12
  • 4
    What is your recommendation for the new additive identity element in $\mathbb Z\setminus 3\mathbb Z$? – Ian Coley Apr 10 '14 at 23:13
  • 1
    One can transport the structure of any ring onto any set of the same cardinality. So if you want an answer less trivial than that, then you need to specify what you desire preserved in the process. – Bill Dubuque Apr 10 '14 at 23:15
  • Ian Coley: You're literally asking him what he's asking you. I feel like that's not very helpful. – syusim Apr 10 '14 at 23:59
  • 1
    @SamuelYusim No other element acts multiplicatively the way zero does. – Karl Kroningfeld Apr 11 '14 at 01:28

2 Answers2

4

Here is a (rather clunky) way of putting an additive structure on ${\Bbb Z}-3{\Bbb Z}$: $$m\oplus n=\Bigl\lceil\frac{m+n+1}{3}\Bigr\rceil +\Bigl\lceil\frac{2m+2n+2}{3}\Bigr\rceil-1\ .$$ Doesn't make a ring structure with ordinary multiplication though, you need to redefine multiplication too, and I think it will look even worse than addition ;-)

David
  • 82,662
1

Let $p_i, i=1...n$ be a subset of the set $P$ of all primes such that that the set of the other primes $P'=P-\{p_i, i=1...n\}$ is infinite. Let $\mathbb Z'=\mathbb Z-\cup \mathbb p_iZ$. Then you can turn $\mathbb Z'$ into a ring while retaining $\mathbb Z$'s multiplication structure and redefining $+$ as follows:

Since $P$ and $P'$ have the same cardinality there exists a bijection $f: P \to P'$. Extend $f$ to map $0\to 0$, $1\to 1$, and to all integers by mapping $\Pi_j p_j^{k_j}\to \Pi_j f(p_j)^{k_j}$. This obviously preserves multiplication structure.

Then introduce the addition operation: for $a,b\in \mathbb Z'$ define $a+_{\mathbb Z'}b=f(f^{-1}a+_{\mathbb Z}f^{-1}b)$.

It only remains to check whether $\times_{\mathbb Z'}$ is associative relative to $+_{\mathbb Z'}$:

$(a+_{\mathbb Z'}b)\times_{\mathbb Z'} c=f(f^{-1}a+_{\mathbb Z}f^{-1}b)\times_{\mathbb Z'} f(f^{-1}c)=f((f^{-1}a+_{\mathbb Z}f^{-1}b)\times_{\mathbb Z}f^{-1}c)=f(f^{-1}a\times_{\mathbb Z}f^{-1}c+_{\mathbb Z}f^{-1}b\times_{\mathbb Z}f^{-1}c)=f(f^{-1}(a\times_{\mathbb Z'}c)+_{\mathbb Z}f^{-1}(b\times_{\mathbb Z'}c))=a\times_{\mathbb Z'}c+_{\mathbb Z'}b\times_{\mathbb Z'}c$

Michael
  • 1,649
  • 10
  • 16