# $a\odot b = ab+a+b$ over $\Bbb Z$ no zero divisors?

Define the operations $\oplus$ and $\odot$ on $\Bbb Z$ by $a\oplus b = a+b-1$ and $a\odot b = ab+a+b$. Prove that $(\Bbb Z,\odot,\oplus)$ is an integral domain.

Surely they first want me to prove its a ring, so I'll just take that for granted here.

I just want to show that it is an integral domain, given that it is a ring here:

So we need to show that $a\odot b=0\implies a=0$ or $b=0$ I imagine. Okay, so:

$ab+a+b=0$, then if $ab=0$, since $\Bbb Z$ is an integral domain, it follows that either $a=0$ or $b=0$ and we are done. So assume not, let $ab\ne 0$, then $a\ne 0$ and $b\ne 0$ and we get $ab+a+b=0 \implies a=b=-2$ as the only valid solution when $ab\ne 0$.

I.e we have shown:

$a\odot b=0 \implies a=b=-2$ or $ab=0\implies a=0$ or $b=0$. Doesn't that mean in the case that $a=b=-2$ that this isn't an integral domain?

We have zero divisor $-2\odot -2=4-4=0$

• You need to first figure out what the 0 is for the ring. 0 isn't always 0. – Paul Nov 2 '15 at 13:56
• That is not a ring. Note that $$1 \oplus 1 = 1, \quad 1 \odot 1 = 3$$ as we have $$3 \oplus 3 = 6 - 1 = 5$$ we have $$(1 \oplus 1) \odot 1 = 3 \ne 5 = (1 \odot 1) \oplus (1 \odot 1)$$ – martini Nov 2 '15 at 13:56
• Isn't it $a\oplus b = a+b+1$ ? – lhf Nov 2 '15 at 14:00
• Yeah, you're right. Or if the product was defined as $a+b-ab$. – Thomas Andrews Nov 2 '15 at 14:11
• @ThomasAndrews If so it's a typo on the lecturers behalf, the notes ask us for $a\oplus b = a+b -1$ and $a\odot b = ab+a+b$. I'll try the problem with the modified operator though – Functional Analysis Nov 2 '15 at 14:13

Hint. $$a \odot b = ab + a + b = (a + 1)(b+ 1) - 1.$$ and $(\mathbf Z, +, \cdot)$ is an integral domain.