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Similar question here Let $R$ be the set of all integers with alternative ring operations defined below. Show that $\Bbb Z$ is isomorphic to $R$. The difference is that in attempting to answer my own problem, I can't.

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 $R$.

Let $f: \mathbb{Z} \to R$.

Suppose we have $0, 1, a \in \mathbb{R}$.

Then we have $f(a\otimes 1) = a + 1 - a = 1$.

For $f(a\oplus 1) = a + 1 - 1 = a$.

For $f(a \otimes 0) = a + 0 = a$.

Thus, the multiplicative identity is $0$ and the additive identity is $1$.

I DO NOT KNOW HOW TO DEDUCE THAT THE FUNCTION IS $f(x) = -x + 1$. I need serious help in that. Isn't the identity for multiplication in the second ring "0" where it's 1 in the integers, thus $f(e_G) \neq e_H$ so it isn't a homomorphism?

Now, we check $f(ab) = f(a) \otimes f(b). f(ab) = (-ab + 1); f(a)f(b) = (-a + 1)(-b + 1)$ which is obviously not homomorphic. In a homomorphism, $f(e_\mathbb{Z}) = e_R$. What did I do wrong??

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Don Larynx
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  • Your third line is weird on the light of your first two: "This is a..." What is a homomorphism ? – DonAntonio Oct 24 '13 at 12:13
  • Duplicate of your own question? http://math.stackexchange.com/questions/535587/let-r-be-the-set-of-all-integers-with-alternative-ring-operations-defined-belo – lhf Oct 24 '13 at 12:14
  • @lhf: it's not a duplicate. – Don Larynx Oct 24 '13 at 12:21
  • @DonAntonio: Fixed. – Don Larynx Oct 24 '13 at 12:22
  • Ok, @DonLarynx...but still it looks odd: you're trying to define a function not on single elements but on their sum/multiplication $;\otimes;$ ...how come?! – DonAntonio Oct 24 '13 at 12:26
  • Close enough to a duplicate that it makes sense to merge the two questions --- or at least you could have included a link to the previous question, and an explanation of just what the alleged difference is. – Gerry Myerson Oct 24 '13 at 12:27
  • @DonAntonio I am just showing it is homomorphic?? – Don Larynx Oct 24 '13 at 12:29
  • @DonLarynx, I suspect you wanted to say that "suppose $;f:\Bbb Z\to R;$ is a isomorphism, then..." , but you did not say that and it was pretty confusing. Besides this, if you really wanted to do that, what is $;f(a\otimes b);$ ? The operation $;\otimes;$ is defined on $;R;$ , not on $;\Bbb Z;$ , so either you confused things even further or who knows what's going on here... – DonAntonio Oct 24 '13 at 12:40
  • @DonAntonio $f(a\otimes b)$ is not well-defined on $\mathbb{Z}$, as I just learned...... – Don Larynx Oct 24 '13 at 12:49
  • Exactly my point, @DonLarynx ... – DonAntonio Oct 24 '13 at 13:09

1 Answers1

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It is important to keep in mind where the different elements are. If $f:\Bbb Z \to R$, then there is no sense in applying $\oplus$ and $\odot$ to the arguments of $f$. Writing something like $f(a\odot b)$ is not very productive, since $a, b$ are considered to be elements of $\Bbb Z$, and not elements of $R$ where these operations makes sense. Rather, you want to apply those operations to the resulting elements, like so: $f(a)\odot f(b)$.

The fact that they are both rings defined on the set of integers (showing that $R$ is in fact a ring is important as well) does make it difficult to keep $\Bbb Z$ and $R$ apart, but it is necessary to do so in order to understand what is going on. If you want to show to show that $f$ is a homomorphism, you want to show the following:

  1. $f(a + b) = f(a)\oplus f(b)$
  2. $f(ab) = f(a)\odot f(b)$

Remember, if $a, b\in \Bbb Z$, then $\oplus$ and $\odot$ have no business being near them. Similarily, if $c, d \in R$, then you need to be very careful about $+$ and $\cdot$, and remember that they are only a means for calculating $\oplus$ and $\odot$, not "real" operations.

I will do point 1 above, I hope you can follow it to show number 2. $$ f(a + b) = -(a + b) + 1 = -a -b + 1 = (-a + 1) + (-b + 1) - 1 \\= f(a) + f(b) - 1= f(a) \oplus f(b) $$

Edit
Here is a small addendum on how one might deduce $f$. First of all, we need to identify the additive and multiplicative identities $0_R$ and $1_R$. That can be done by solving the equations $a\oplus 0_R = a$ and $b \odot 1_R = b$. This will result in $1_R = 0$ and $0_R = 1$.

Now, assume there is a (non-trivial) homomorphism $f$. We will use $f(1) = 0$ to deduce what $f(a)$ must be for a general $a$ if it does exist. We can start by deducing $f(2)$: $$ f(2) = f(1+1) = f(1)\oplus f(1) = 0\oplus 0 = 0 + 0 - 1 = -1 $$ Now $f(3)$ and $f(4)$ are calculated similarily to be $-2$ and $-3$, and a pattern seems to emerge; the definition $f(a) = -a + 1$ fits so far. To show that this actually is a homomorphism, we prove the two steps above, and since any homomorphism from $\Bbb Z$ is completely determined by what its value on $1$ is, this is the only one.

Once we show that it is bijective, we know that it is an isomorphism, and therefore, $\Bbb Z$ and $R$ are isomorphic.

Arthur
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  • Just adding, that OP should also see which are the neutral elements and inverse for the "sum" in $R$ and check that $f(0) = 0_R$, $f(1) = 1_R$ – bernatguillen Oct 24 '13 at 12:38
  • I checked for that @Duronman; $f(0 + 0) = 0 - 0 - 1 = -1 \neq f(0) \oplus f(0)$? @duronman – Don Larynx Oct 24 '13 at 12:40
  • @DonLarynx We have $$ f(0 + 0) = -(0 + 0) + 1 = 1\ f(0)\oplus f(0) = (-0 + 1)\oplus(-0 + 1) = 1\oplus 1 = 1+1-1 = 1 $$ and they are equal. – Arthur Oct 24 '13 at 12:48
  • @Arthur I am most confused on how you arrived at $f(x) = -x + 1$. That is the most confusing part of the problem. Showing it is a homomorphism is straightforward. – Don Larynx Oct 24 '13 at 12:52
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    @DonLarynx See my edit. Finding a candidate for $f$ in this case is more or less a guided trial and error (most of the time it is), but once you have a candidate, checking whether it's a (the) correct one is a matter of applying machinery. – Arthur Oct 24 '13 at 13:10
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    @DonLarynx In short, you should see that $1$ is the additive identity in the new ring, $0$ is the multiplicative identity in the new ring, and a ring homomorphism needs to preserve identities. So it would have to map $1\mapsto 0$ and $0\mapsto 1$. At this point, one might begin to suspect that $1-x$ might do the trick. – rschwieb Oct 24 '13 at 13:47
  • So algebra is more systematic and numerical, instead of formal and general, like analysis? I seem to have to prove something isn't something in algebra by contradiction or by showing a specific example, which is pretty lame. @Arthur – Don Larynx Oct 24 '13 at 16:09