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"Define a new addition and multiplication on $\mathbb Z$ by the rules: $a(+) b = a + b – 1$ and $a(*) b = ab – (a + b) + 2$. Prove that with these new binary operations $\mathbb Z$ is an integral domain. You may assume that under these new operations $\mathbb Z$ is a ring."

I can show that $\mathbb Z$ is a commutative ring, I'm not sure how to find the identity element of $\mathbb Z$ to show that it's an integral domain.

Thanks in advance for any help.

Stefan Hansen
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mike
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3 Answers3

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First of all, we usually call the additive identity the "zero" and the multiplicitive identity just the "identity" for clarity. I believe you are looking for zero to show that the ring is an integral domain. To do this, you need to find some $x \in \mathbb{Z}$ such that for any $a \in \mathbb{Z}$, $a(+)x = a+x-1 = a$. You can do this by algebra. A similar method can be used to find the (multiplicitive) identity.

To show that the ring is an integral domain, you need to show that, if we denote the zero element by $z$, $a\ast b = z \implies a = z $ or $b = z$.

Tom Oldfield
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This ring is constructed via a standard trick, which allows us to transport the structure of an arbitrary ring to any set with the same cardinality.

Suppose $(A, +, *)$ is a ring. Let $f : A \to B$ be a bijection, where $B$ is an arbitrary set. Define operations on $B$ via $$ x(+)y = f(f^{-1}(x) + f^{-1}(y)), \qquad x(*)y = f(f^{-1}(x) * f^{-1}(y)). $$

Then it is immediate that $(B,(+),(*))$ is a ring, and $f$ an isomorphism.

In your case, $A = B = \Bbb{Z}$, and $f(x) = x+1$, and thus $f^{-1}(x) = x - 1$. In fact $$ x(+)y = (x - 1) + (x-1) + 1 = x + y -1, \quad x(*)y = (x-1)(y-1) + 1 = x y -(x+y) +2. $$

Andreas Caranti
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  • I thought it was strange that this HW was assigned without covering bijections or isomorphisms in lecture, but I think I was able to prove it without those concepts. Thank you! – mike Feb 28 '13 at 13:58
  • @mike, you're welcome. I tried to show you what goes on behind the wings, that is, how such an exercise has been constructed by your instructor. Note that my explanation provides a proof for the required result, as $1 = x(\star)y = (x-1)(y-1) + 1$ if and only if $x = 1$ or $y = 1$. Here $1$ is the zero of $(\Bbb{Z},(+), (\star))$. – Andreas Caranti Feb 28 '13 at 14:05
  • Transport of structure is jargon in wide use for such isomorphisms. One can find less trivial examples in prior posts here, e.g. transporting the class group structure of quadratic fields from ideals to primitive binary quadratic forms - which greatly simplifies Gauss's presentation of composition of binary quadratic forms (e.g. this immediately yields associativity, which for Gauss took many pages of obfuscated calculations). – Math Gems Feb 28 '13 at 15:34
  • @MathGems, thanks for the nice references. I am in the middle of some calculations which might be interpreted in terms of transport of structure, except that I am well in the obfuscated phase. – Andreas Caranti Feb 28 '13 at 15:38
  • @Andreas Good luck with your calculations. Perhaps we will be so lucky to learn about that in a future post. Of course the comment was intended for readers (for algebra professors, no doubt it is well-known). – Math Gems Feb 28 '13 at 15:47
  • @MathGems, trust me, there is a lot of stuff I do not know, even if we are just talking algebra, but at least I try not to stop learning. – Andreas Caranti Feb 28 '13 at 16:03
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This exercise is best understood as a special case of the following trivial observation, which also explains how to come up with these rather exotic (ring) operations (which are, of course, useless, and this "exercise" is just an end in itsself).

Let $R$ be a ring with underlying set $|R|$. If there is a bijection $f : X \to |R|$ from a set $X$, then there is a unique ring $S$ with $|S|=X$ such that $f$ becomes an isomorphism of rings. Namely, one has $0_S = f^{-1}(0_R)$ and $s+t = f^{-1}(f(s)+f(t))$, the same with the multiplicative structure. In fact, the same works for arbitrary algebraic structures. Since $f$ is an isomorphism, every axiom or property of $R$ is inherited to $S$. For example, if $R$ is an integral domain, the same is true for $S$.

Now, take the bijection $\mathbb{Z} \to \mathbb{Z}$, $a \mapsto a-1$. The induced addition is $a+'b=(a-1)+(b-1)+1=a+b-1$, the zero is $1$, the multiplication is $a*'b = (a-1)*(b-1)+1=a*b-a-b+2$, the unit is $2$.

Thus, this ring is nothing else than the usual ring $\mathbb{Z}$, but with a different notation for its elements.

Martin Brandenburg
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