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We know that $(\mathbb{Z},+)$ is an abelian group. And under usual multiplication, $(\mathbb{Z},+,\times)$ become a ring. Is there another binary operation $\square$ on $\mathbb{Z}$ that can make $(\mathbb{Z},+,\square)$ to be a ring?

PS: The question comes with a more philosophical argument: junior high students often ask why we should define $(-3)\times(-5)=15$, why not define $(-3)\times(-5)=-15$ or just another number, say $(-3)\times(-5)=155555$? I think a good motivation and reason for operations with negative numbers defined like so, is because only when we defined like that, the integers $\mathbb{Z}$ with $+$ and $\times$ can have associativity, distribution law and so on(i.e. forms a ring). If we didn't choose to define so, then we don't have such good properties!

Eric
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3 Answers3

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Yes, there is a different ring structure on $(\mathbb{Z}, +)$: letting $\cdot$ denote the usual product, define $*$ as $$a*b=-(a\cdot b).$$

Why does this work? Well, let $(G, \oplus, \otimes)$ be any ring, and $\alpha$ any automorphism of $(G, \oplus)$. Then we can "push" $\otimes$ through $\alpha$ to get a new operation, $\otimes_\alpha$, such that $(G, \oplus,\otimes_\alpha)$ is also a ring. Namely, $\otimes_\alpha$ is defined as $$a\otimes_\alpha b=\alpha^{-1}(\alpha(a)\otimes\alpha(b)).$$ In our case, we have the automorphism $\alpha:x\mapsto -x$, and $\alpha(a)\cdot \alpha(b)=a\cdot b$, so this yields the operation described above.

In fact, $\cdot$ and $*$ are the only operations which yield a ring structure: suppose $k$ is a unit of $\odot$ with $k\not\in\{1, -1\}$. WLOG, $k$ is positive - but then $1\odot 1+1\odot 1+...$ ($k^2$ times) equals $k$, but no integer has this property if $k>1$.

It's now enough to show that any ring expansion with unit $1$ is the usual one, and with unit $-1$ is the "twisted" version described above; and this isn't too hard.

Noah Schweber
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  • The twisted ring is still isomorphic to the untwisted $\mathbb{Z}$ under $x \mapsto -x$. – Ted Oct 23 '16 at 03:35
  • @Ted That's true - but the OP didn't ask about rings up to isomorphism. (Note that this satisfies the OP's hope for $-1*-1=-1$.) – Noah Schweber Oct 23 '16 at 03:36
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IMPORTANT: I have changed my answer taking into accounts the comments.

A) Unitary ring:
Let us call $*$ the second law, $e$ the corresponding neutral element, and $\cdot$ the usual multiplication.
Suppose $e > 1$ then take $k \in \mathbb{Z}$ prime with e (according to the usual multiplication law): On the one hand we have $k*e=k$ and on the other hand we have $k*e=k*(1+1+...+1) =k*1+k*1=...+k*1=e \cdot (k*1)$. We therefore have $k=e \cdot (k*1)$ which contradicts e prime with k ($(k*1) \in \mathbb{Z}$).
If we suppose $e < 1$, we get the same contradiction with a similar reasoning.
If we suppose $e=0$ ,for $n \neq 0$ we would have $n*e=n*(1-1)=n*1-n*1=0 \neq n$
Therefore $e=1$ or $e=-1$

Now for $a \in \mathbb{Z}$ and $n \in \mathbb{N}$:
$a*(n \cdot e)=a*(e+e+\cdots+e)=a*e+a*e+\cdots+a*e=(a*e)\cdot n$
If $e=1$ we get $a*n=a\cdot n$ (for any $a \in \mathbb{Z}$ and $n \in \mathbb{N}$), so that * and $\cdot$ are identical
If $e=-1$ we get $a*(-n)=a\cdot n$ (for any $a \in \mathbb{Z}$ and $n \in \mathbb{N}$), so that * is "$\cdot (-1) \cdot$"
So there are two ways to extend $(\mathbb{Z},+)$ as a unitary ring

B) Non unitary ring:
We can then define the second law * as "$\cdot k \cdot$" for any $k \in \mathbb{Z}$ (I think the distributivity imposes a "linearity" for the second law)

Edouard L.
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  • @N.S. You're absolutely right, and thank you for pointing out my mistake in pointing out the mistake! That calculation is correct, and gives us $a(n\cdot e)=(ae)\cdot n=a\cdot n$. However, this does not give us $an=a\cdot n$ - it just gives us $a(n\cdot e)=a\cdot n$! These are only the same if $n\cdot e=n$, that is, if $e=1$; but if $e\not=1$, then these differ. – Noah Schweber Oct 23 '16 at 03:09
  • Thank you for the comments, there were useful for me to understand the problem deeper! – Edouard L. Oct 23 '16 at 09:49
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I think you can answer this to your students :

Let $\mathbb{Z}[1/2] = \{ \frac{a}{2^b}, a \in \mathbb{Z},b \in \mathbb{N}\}$. With the usual operation it is a (commutative) ring.

But you can define another multiplication $\ast$ on it : $$a \ast b = 2 ab$$ This way $1/2$ is the multiplicative identity, and $(\mathbb{Z}[1/2],+,\ast)$ is a (commutative) ring where

$$(-1) \ast (-1) = 2$$

reuns
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