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

Question

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.

Answer 1

That is precisely the beauty of ideals: they are stable under arbitrary multiplication. Hence your cosets are also stable under multiplication. This is actually a really nice exercise: prove that $R/ I$ 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 take modulo $6$.