Define $\mathbb{Z}^*$ to be the set of integers but with the following operations: $a \circ b = a + b - 1$ and $a * b = a + b - ab$

where $a+b$ and $ab$ are the usual integer addition and multiplication. Assume that these two operations make $\mathbb{Z}^*$ a ring. Are the integers, $\mathbb{Z}$, and $\mathbb{Z}^*$ isomorphic as rings? Demonstrate an isomorphism if so, otherwise demonstrate precisely why not. (hint: every ring is also a group)

I am at a loss. Any help is appreciated, thank you.

  • $\begingroup$ @learner But $(\Bbb Z, \circ)$ is cyclic, generated by the multiplicative identity $0$. Circle adding $0$ to itself gets you the negative integers, and circle adding the additive inverse of $0$ ( which is $2$) gets you the positive integers. $\endgroup$ – rschwieb Apr 29 '16 at 21:51
  • $\begingroup$ @rschwieb, Right, I screwed up earlier. My bad. $\endgroup$ – learner Apr 30 '16 at 5:27

Here is a solution that isn't satisfying because it isn't really deduced from the givens of the problem. Nevertheless, it completely explains this and many more weird operations of the same type.

If you have any bijection of $f:\Bbb Z\to\Bbb Z$ (or from any ring into itself, for that matter) then you can define two new operations

$$ a\oplus b := f^{-1}(f(a)+f(b)) \\ a\odot b := f^{-1}(f(a)\cdot f(b)) $$

And the resulting structure $\Bbb Z^\ast=(\Bbb Z, \oplus,\odot)$ is a ring, and $f$ is a ring isomorphism from $\Bbb Z^\ast$ to the ordinary integers.

In this case, $f(x)=1-x$, which is its own inverse. Compute what $\oplus$ and $\odot$ are with respect to $f$, and you will see that is how your operations are arising.

| cite | improve this answer | |
  • 2
    $\begingroup$ +1, transport of structure. $\endgroup$ – Andreas Caranti Apr 29 '16 at 22:13
  • $\begingroup$ @rschwieb thanks so much. how could I show onto with this function? $\endgroup$ – J. Medley Apr 30 '16 at 15:07
  • $\begingroup$ @J.Medley $x =1-(1-x)=f(1-x)$ you know that invertible functions are both 1-1 and onto, right? $\endgroup$ – rschwieb May 1 '16 at 0:25

Hint: $\mathbb{Z},+$ is generated by $1$. Hence a group morphism $f:\mathbb{Z},+\rightarrow \mathbb{Z}^*,\circ$ is completely determined by $f(1)$.

| cite | improve this answer | |
  • $\begingroup$ I am unsure how to define the isomorphism $\endgroup$ – J. Medley Apr 29 '16 at 18:20
  • $\begingroup$ Suppose $G$ is any group and $f:\mathbb{Z}\rightarrow G$ is a group homomorphism. Suppose you know $f(1)=a\in G$. Then $f(n)= f(\sum_{i=1}^n1)=a\circ a \circ \dots \circ a$ ($n$ times). So suppose that $\mathbb{Z}$ and $\mathbb{Z}^*$ are isomorphic via some isomorphism $f$. To know this isomorphism, you only have to know the value of $f(1)$. Can you determine what $f(1)$ has to be? $\endgroup$ – Mathematician 42 Apr 29 '16 at 18:27
  • $\begingroup$ f(1) would map to a generator of Z* I believe, correct? But for how (a o b) is defined I am unsure how to figure out this generator. $\endgroup$ – J. Medley Apr 29 '16 at 18:39
  • $\begingroup$ Remember, you want $\mathbb{Z}$ and $\mathbb{Z}^*$ to be isomorphic as rings, so you have to map the unit (for the multiplication) of $\mathbb{Z}$ to the unit of the multiplication of $\mathbb{Z}^*$. Can you find that unit? $\endgroup$ – Mathematician 42 Apr 29 '16 at 18:43

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.