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Let $\mathbb{Z}_x$ denote the ring of integers with the operations $\odot, \oplus $ defined as $a \odot b$ = $ a+b -a b$ and $a \oplus b$ = $a+b -1$. Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$.

Not sure how to start.

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  • $\begingroup$ Show that the defined product and sum satisfies the properties of isomorphism: 1) It is bijective. 2) It is a homomorphism. $\endgroup$ – Tyler Oct 14 '13 at 20:44
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    $\begingroup$ Dear tausch86, Already the way you phrase the question shows a certain confusion; more precisely, there are two things wrong with your statement "prove $f$ [...] is isomorphic". (1) It should read "is an isomorphism". (2) There is no function $f$ given (at least in the way you phrased the problem); one of the major parts of solving this problem will be to find the candidate function $f$. Regards, $\endgroup$ – Matt E Oct 14 '13 at 20:45
  • $\begingroup$ I understand the axioms of an isomorphism but the new operations are confusing me. like what would the value of f(1) be. would it be one or would it be $f(1 + 0) = 1 + 0 - 1 = 0$? $\endgroup$ – tausch86 Oct 14 '13 at 20:46
  • $\begingroup$ @MattE got rid of the "prove $f$..." but the other wording is straight out of my textbook. $\endgroup$ – tausch86 Oct 14 '13 at 20:49
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    $\begingroup$ Dear tausch86, Okay. Then one of the main parts of solving this problem will be to find a candidate for $f$. Regards, $\endgroup$ – Matt E Oct 14 '13 at 20:49
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So we want to find a bijection $f \colon \def\Z{\mathbb Z}\Z \to \Z$ such that $$ f(a+b) = f(a) \oplus f(b) , \quad f(ab) = f(a) \odot f(b) $$ holds for all $a,b \in \Z$. Let's suppose we have such an $f$ and want to find $f(0)$, for all $a$ we must have \begin{align*} f(a) &= f(a+0)\\ &= f(a) \oplus f(0)\\ &= f(a) + f(0) - 1\\ \iff f(0) &= 1 \end{align*} Using the same idea, we must have \begin{align*} f(a) &= f(a\cdot 1)\\ &= f(a) \odot f(1)\\ &= f(a) + f(1) - f(a)f(1)\\ \iff f(1) &= f(a)f(1) \end{align*} As $f$ is onto, we must have $f(1) = 0$. Now we can get an idea what to do for a general $n$: \begin{align*} f(n+1) &= f(n) \oplus f(1)\\ &= f(n) \oplus 0\\ &= f(n) - 1 \end{align*} So if there is such an $f$, we must have: $f(1) = 0$, $f(n+1) = f(n)- 1$ for any $n$, so we must have $f(n) = 1-n$ for every $n$. Now check if this $f$ is the desired isomorphism.

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  • $\begingroup$ This was just such a tremendous help. Thank you $\endgroup$ – tausch86 Oct 14 '13 at 20:56

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