How to show distributivity in a ring, and what is wrong with my algebra?

I am trying to show the following is a commutative ring with unity, however I am encountering a problem.

First, addition and multiplication are defined as:

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

I have shown this "addition" is an abelian group. The identity is $1$ and the negative is $2-a$. Now, I am stuck on distributivity. I found that:

$$a \odot(b \oplus c)$$ $$= a \odot (b+c-1)$$ $$=(a\odot b) + (a \odot c) - (a \odot 1)$$ $$=ab-(a+b)+2+ac-(a+c)+2-(a1-(a+1)+2)$$ $$=ab-a-b+2+ac-a-c+2-a+a+1-2$$ $$=ab+ac-2a-b-c+3$$

Why is this not equal to: $$= a \odot (b+c+(-1))$$ $$=(a\odot b) + (a \odot c) + (a \odot (-1))$$ $$=ab-(a+b)+2+ac-(a+c)+2+a(-1)-(a+(-1))+2)$$ $$=ab-a-b+2+ac-a-c+2-a-a+1+2$$ $$=ab+ac-4a-b-c+7$$

However, the first equation is equal to $$(a \odot b) \oplus (a \odot c)$$ $$=(ab-(a+b)+2) \oplus (ac-(a+c)+2)$$ $$=ab-(a+b)+2+ac-(a+c)+2-1$$ $$=ab-a-b+2+ac-a-c+2-1$$ $$=ab+ac-2a-b-c+3$$

So, why is the first equation not equal to the second, but equal to the third? Which one is correct?

• Is $a\oplus (-1)=a$? No. You have the wrong identity. Commented Nov 28, 2014 at 21:35
• By the way, what is the domain of your operations? It could be $\mathbb Z$, $\mathbb Q$, and $\mathbb R$, or even $\mathbb C$. Commented Nov 28, 2014 at 21:41
• It is $\mathbb Z$ Commented Nov 28, 2014 at 21:41
• The negative is not $-a$, though. $a\oplus(-a)=-1$. You need to find a $b$ so that$a\oplus b=1$... Commented Nov 28, 2014 at 21:49

Note that $(a+1)\oplus (b+1)=(a+b)+1$: This makes us suspicious and we compute \begin{align}(a+1)\odot(b+1)&=(a+1)(b+1)-(a+1)-(b+1)+2\\ &=ab+1.\end{align} We conclude that $x\mapsto x+1$ maps the commutattive ring with unity $(\mathbb Z,+,\cdot)$ to the structure $(\mathbb Z,\oplus,\odot)$, which is therefore also a commutative ring with unity. (And we conclude that $0+1=1$ is its additive neutral and that $1+1=2$ is its unity.

• How did u make note of that from the beginning? Commented Nov 29, 2014 at 18:46

We want to show that $\odot$ distributes over $\oplus$. However, we can NOT assume that $\odot$ distributes over $+$.

For the first part, we have: \begin{align*} a \odot(b \oplus c) &= a \odot (b + c - 1) \\ &= (a)(b + c - 1) - ((a) + (b + c - 1)) + 2 \\ &= (ab + ac - a) - (a + b + c - 1) + 2 \\ &= ab + ac - 2a - b - c + 3 \\ \end{align*} For the last part, we have: \begin{align*} (a \odot b) \oplus (a \odot c) &= (ab - (a + b) + 2) \oplus (ac - (a + c) + 2) \\ &= (ab - (a + b) + 2) + (ac - (a + c) + 2) - 1 \\ &= ab + ac - 2a - b - c + 3 \\ \end{align*} which matches.

• So, my second attempt doesn't hold? Then how come the first does? Commented Nov 28, 2014 at 21:29
• I see now, thank you. But how come my first attempt works? Commented Nov 28, 2014 at 21:39
• I suspect that it was just a coincidence. It shouldn't work in general. Commented Nov 28, 2014 at 21:53

I'm assuming the domain is $\mathbb Z$, but the same argument would work for $\mathbb Q$, $\mathbb R$, or really any ring.

In general, if $f:X\to\mathbb Z$ is $1-1$ and onto, we can define operators on $X$:

$$x\oplus y = f^{-1}(f(x)+f(y))$$ $$x\odot y = f^{-1}(f(x)\cdot f(y))$$

In these cases, the additive identity is $f^{-1}(0)$.

In your case above,this $X=\mathbb Z$ and $f(x)=x-1$, $f^{-1}(n)=n+1$.

Now, with these definitions:

\begin{align} x\odot(y\oplus z) &= x\odot f^{-1}(f(y)+f(z))\\ &= f^{-1}\left(f(x)\cdot(f(y)+f(z))\right)\\ &= f^{-1}(f(x)f(y)+f(x)f(z))\\ &= f^{-1}(f(x\odot y) + f(x\odot z))\\ &=(x\odot y)+(x\odot z) \end{align}

• See also the appendix of math.stackexchange.com/questions/1001186/… Commented Nov 29, 2014 at 1:45