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For any integers $a,b$, define $a\oplus b=a + b - 1$ and $a\odot b=a + b - ab.$ Let $R$ be the ring of integers with these alternative operations. Show that $\Bbb Z$ is isomorphic to $R$.

What I thought about doing was showing $f(a + b) = a + b - 1$, meaning it isn't isomorphic since it doesn't equal $f(a)f(b)$. True?

Martin Sleziak
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Don Larynx
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  • Why would $f(a+b)\neq f(a)f(b)$ have anything to do with isomorphism? I think you are confused :) How about you write out the things you need to verify in your post: this might help you organize your thoughts. – rschwieb Oct 22 '13 at 13:25
  • @rschwieb that proves it's a homomorphism....... – Don Larynx Oct 22 '13 at 13:27
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    Let's denote your new addition with $\oplus$ and new multiplication with $\otimes$. You would have to find an $f$ such that $f(a+b)=f(a)\oplus f(b)$ and $f(ab)=f(a)\otimes f(b)$ to show they are isomorphic. – rschwieb Oct 22 '13 at 13:28
  • @DonLarynx: To begin with, you should probably confirm that $R$ is in fact a ring with those new operations, so that you can get familiar with their workings. In particular, what is the new additive identity? The new multiplicative identity? That should help you in figuring out what your function $f:R\to\Bbb Z$ ought to be, satisfying the properties of ring homomorphism as rschwieb described. – Cameron Buie Oct 22 '13 at 13:34
  • Hey, I was able to complete what you did @rschwieb. But doesn't that just show it is homomorphic (operation preserving under +, *)? – Don Larynx Oct 22 '13 at 13:35
  • @DonLarynx: What function $f$ did you use, out of curiosity? – Cameron Buie Oct 22 '13 at 13:36
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    f(x) = -x + 1.. – Don Larynx Oct 22 '13 at 13:37
  • @DonLarynx: Very nice! Note then that $f:R\to\Bbb Z$ is a bijection, so since it's a ring homomorphism, then we're done! I'd encourage you to actually write up your proof in an answer. After $3$ days, you'll be able to accept it, and in the meantime, I (for one) will gladly upvote it. – Cameron Buie Oct 22 '13 at 13:39
  • @DonLarynx Very nice indeed: the hardest part is coming up with $f(x)=1-x$. Cameron's suggestion for self-answer is good: look forward to seeing it. – rschwieb Oct 22 '13 at 13:40
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    Note that using the integers as underlying set for $R$ is bound to lead to confusion, since the symbols $0$ and $1$ are used as constants in the language of rings; they normally designate the neutral elements for addition and multiplication. The elements $0$ and $1$ of the underlying set of $R$ do not have these properties, respectively, so you will need to distinguish between the element represented by $0\in R$ and the neutral element for $\oplus$, whic is another element; – Marc van Leeuwen Oct 22 '13 at 13:48
  • That, or this is a good exercise in not being overly inflexible with notation. – rschwieb Oct 22 '13 at 14:42
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    Sorry for the offtopic comment: I always wonder what's the purpose of these exercises (which seem to be very common). This kind of ring does never, never appear in real mathematics. The creator of this exercise just took an arbitrary permutation of $\mathbb{Z}$ and has written down the induced twisted ring operations. Now the students have to find this permutation. So what? Why doesn't he show them interesting, important or cute rings, appearing in all areas of pure mathematics? – Martin Brandenburg Oct 22 '13 at 14:47
  • Perhaps I should mention that sometimes these twisted operations do appear, but I find this confusing. For example, the Segre embedding for projective spaces is $\mathbb{P}^n \times \mathbb{P}^m \hookrightarrow \mathbb{P}^{nm+n+m-1}$. Why on earth $nm+n+m-1$?! Well, the reason is $\mathbb{P}^n$ is actually $\mathbb{P}(k^{\textbf{n+1}})$, and the Segre embedding should be seen as the coordinate-free embedding $\mathbb{P}(V) \times \mathbb{P}(W) \hookrightarrow \mathbb{P}(V \otimes W)$, so the vector space dimension multiplies as expected. No twist! Same for Veronese, Plücker etc. – Martin Brandenburg Oct 22 '13 at 14:55
  • @MartinBrandenburg +1 I totally agree with the offtopic comment – rschwieb Oct 22 '13 at 14:56
  • Related to http://math.stackexchange.com/questions/513045/need-to-prove-that-s-defined-by-the-binary-operation-ab-abab-is-an-abel and http://math.stackexchange.com/questions/373731/showing-that-g-is-a-group-under-an-alternative-operation. – lhf Oct 22 '13 at 15:14
  • @rschwieb doesn't that just show it is homomorphic (operation preserving under (+, *)? – Don Larynx Oct 22 '13 at 15:36
  • @DonLarynx: Yes, as I said, it does just show that $f$ is a ring homomorphism (if we already know that $R$ is a ring under the operations $\oplus$ and $\odot$). We do still have to show that $f$ is bijective. – Cameron Buie Oct 22 '13 at 15:57
  • @MartinBrandenburg Because I have a sadistic algebra professor. – Don Larynx Oct 24 '13 at 02:51
  • @MartinBrandenburg They're one of the most convenient examples of finding isomorphisms of a ring to a familiar ring, since they don't involve any operations that students aren't already familiar with themselves. They're not a good practical example by any means, but they're a useful one. – Steven Stadnicki Oct 25 '13 at 20:52

3 Answers3

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To establish an isomorphism, you would have to find a mapping $f:\Bbb Z \to R$ satisfying the definition of an isomorphism:

  1. $f(a+b)=f(a)\oplus f(b)$
  2. $f(ab)=f(a)\odot f(b)$
  3. $f$ one-to-one and onto (you could just find an inverse homomorphism)

You're going to have to come up with a candidate for $f$.

My first hint would be to look at $a\odot 1$ and $a\odot 0$ and $a\oplus 1$. $0$ and $1$ are "special" in $R$, and if you sort out what they are doing in the new operation, you can deduce the right $f$ to pick.

rschwieb
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  • Any specific reason why we must pick out operations with 0, or 1? – Don Larynx Oct 24 '13 at 11:49
  • @DonLarynx In reality, there's no objective reason for checking $0$ and $1$. It just turns out that in this case they are special, and I wanted to make sure you didn't overlook them :) The thing you should take away is that it's usually a good idea to think about what the identity elements in the ring might look like. – rschwieb Oct 24 '13 at 12:35
  • f(ab) = -(ab + 1) = (-a + 1)(-b + 1) but THESE ARE NOT EQUAL. @rschwieb I am running into problems? – Don Larynx Oct 24 '13 at 12:36
  • @DonLarynx there are a lot of typos in that. It should be: $f(ab)=1-ab=(1-a)\odot(1-b)=f(a)\odot f(b)$. Are you computing $(1-a)(1-b)$ and not $(1-a)\odot (1-b)$ by accident? – rschwieb Oct 24 '13 at 12:41
  • Yes. So we get (1-a) + (1-b) - (1-a)(1-b), which equals $1 - a + 1 - b - (ab - (a + b) + 1) = 1 - ab$? – Don Larynx Oct 24 '13 at 12:54
  • @DonLarynx Sure, if you work the algebra out the equality holds, yes. – rschwieb Oct 24 '13 at 12:58
  • How did you find out they are special in $R$? it is still unclear how I have $f(x) = -x + 1, because I considered all three possibilities. I did arrive to a conclusion, as shown here, but not yours. http://math.stackexchange.com/questions/538115/help-on-abstract-algebra-proof – Don Larynx Oct 24 '13 at 13:02
  • dear @DonLarynx : I found out they were special by just trying them out. I had a hunch. I don't understand the rest of your question. Arthur's explanation there is exactly what I meant, and I don't think I can improve upon his description much. His, yours, and my conclusions are apparently all the same... – rschwieb Oct 24 '13 at 13:36
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You officially need to first show that $R$ is indeed a ring, before you can go about showing anything involving $R$ is a ring homomorphism. However, if you have a candidate bijection $f$ with another (known) ring, in this case $\Bbb Z$, which means it satisfies $f(x+y)=f(x)\oplus f(y)$ and $f(x\times y)=f(x)\odot f(y)$ as indicated in the answer by rschwieb, then you can cheat a bit because $f$ translates all the operations of $\Bbb Z$ into those of$~R$, and their properties come with them. For instance checking the left distributive law in $R$ can be done as $$ \begin{aligned} a\odot(b\oplus c) &=f(x)\odot(f(y)\oplus f(z)) =f(x)\odot(f(y+z)) =f(x\times(y+z))\\ &=f(x\times y+x\times z) =f(x\times y)\oplus f(x\times z)\\ &=(f(x)\odot f(y))\oplus(f(x)\odot f(z)) =(a\odot b)\oplus(a\odot c), \end{aligned} $$ where $x,y,z\in\Bbb Z$ are such that $f(x)=a,f(y)=b,f(z)=c$ (which is well defined since $f$ is a bijection). While it looks a bit complicated, it just transfers the responsability for the distributive law from $R$ to $\Bbb Z$ (for which it is used in the middle of the computation). So nothing is really going on. The other axioms can be checked similarly without effort.

This assumes you have a candidate $f$, but also that you know the neutral elements for $\oplus$ and for $\odot$, which must be chosen as $f(0)$ and $f(1)$ respectively (it is part of the requirement for homomorphisms). You can easily deduce from the definitions of $\oplus$ and $\odot$ which these neutral elements are. Once you got them, you know $f(0)$ and $f(1)$, and other values follow, like $f(2)$ which must be $f(1+1)=f(1)\oplus f(1)$. You will see it is downhill from there.

Marc van Leeuwen
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  • The first sentence is not really true, but I guess in the context of a learner it is the best advice. – rschwieb Oct 22 '13 at 14:55
  • The first sentence is correct because ring homomorphisms are defined only between rings. But the idea is to use $(+,*,0,1)$-homomorphisms instead. – Martin Brandenburg Oct 22 '13 at 14:59
  • @MartinBrandenburg The point is that if you define a bijection and verify it preserves operations, then the ring axioms for the second ring follow from those of the first, and the bijection automatically becomes a ring homomorphism. So no, there is no absolute necessity to prove the second thing is a ring first. But I agree that it is easier to teach it the other way around. – rschwieb Oct 24 '13 at 12:38
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Well OP choose $f(x) = -x + 1$. To prove the isomorphism, we shall show $$f(a*_1b) = f(a)*_2f(b)$$ under addition and multiplication. Thus, we have for $a, b \in \mathbb{Z}$,

$$f(a)*_2f(b) = (-a + 1) + (-b + 1) - 1 = -(a + b) + 1 = f(a*_1b)$$ and $$f(a)*_2f(b) = (-a + 1) + (-b + 1) - (ab-(a+b)+1) = -(a + b) + 2 - ab + (a + b) - 1 = -ab + 1 = f(a*_1b)$$

As you can see OP it is all about finding a function by trial-and-error, finding out if it is operation preserving, and performing simple high-school algebra. In case you are unsure you can check if $f(e_G) = e_H$ for example.

Don Larynx
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