# Question regarding the multiplication table of $\mathbb Z/6\mathbb Z$

I'm sorry if this is a dumb question, but I'm honestly confused. I was reading my modern algebra textbook and it started talking about the multiplication and addition table of $$\mathbb Z/6\mathbb Z$$. I understood the addition table, but the multiplication table had me confused. For example, why is the multiplication of $$0+6\mathbb Z$$ and $$0+6\mathbb Z$$ equal to $$0+6\mathbb Z$$? My initial impression was that it would equal $$36\mathbb Z$$ because $$6*0=0$$, $$6*6=36$$, $$6*12=72$$, etc. What aspect of my reasoning is faulty? Thank you in advance.

• By definition,$\Bbb Z/6\Bbb Z$ is the set of cosets $a+6\Bbb Z$, but $0+36\Bbb Z$ is not such a coset, so isn't an element of $\Bbb Z/6\Bbb Z$. – Lord Shark the Unknown Oct 29 '18 at 7:49
• Thank you for your response. I recognize that $0+36\mathbb Z$ doesn't make sense to be an element of $\mathbb Z/6\mathbb Z$. I guess my main question is how you would perform multiplication of the elements of $\mathbb Z/6\mathbb Z$ because it seems counter-intuitive to me. – KronZ Oct 29 '18 at 7:58
• My assumption is that in order to multiply $0+6\mathbb Z$ with $0+6\mathbb Z$, you need to multiply every element of these sets together and see what set comes out. But when I do this, I get $0+36\mathbb Z$ which doesn't make sense. – KronZ Oct 29 '18 at 8:08

That is precisely the beauty of ideals, thez are stable under arbitrary multiplication. hence your cosets are also stable under multiplication. this is actually a really nice exercise to prove that a $$R\backslash I$$ is a ring if and only if $$I\subset R$$ is an Ideal (well definedness of multiplication here then actually boils down, to $$r0=0$$ for all $$r \in R$$. So in your concrete case that means that $$(2 + 6\mathbb{Z})*(2+6\mathbb{Z})= 4 + 6\mathbb{Z} + 6\mathbb{Z} +36\mathbb{Z}$$ but now since all the factors containing $$\mathbb{Z}$$ are contained in $$6\mathbb{Z}$$ this actually is $$4 + 6\mathbb{Z}.$$ Hence you can actually just multiply as in $$\mathbb{Z}$$ and then taking modulo $$6$$.