# Quotient ring $(\mathbb{Z}_4 \times \mathbb{Z}_6)/S$

Consider the ring $\mathbb Z_4\times \mathbb Z_6$ and $S=\{(0,0),(2,0),(0,3),(2,3)\}$.

Would the elements of the quotient ring $(\mathbb Z_4\times \mathbb Z_6)/ S$ be:

$S+(0,0)$ (trivial set above), $S+(1,1)=\{(1,1),(3,1),(1,4),(3,4)\}$, $S+(2,2)=\{(2,2),(0,2),(2,5),(0,5)\},\cdots S+(1,5)=\{(1,5),(3,5),(1,1),(3,2)\}$. Basically for each $S+(n,n)$, I am adding $(n,n)$ to the original coordinates. Is this correct? Also for the addition and multiplication table of this coset, would it look very similar to the tables for $\mathbb Z_6$? i.e. would $(3+s)+(3+3)=0+s$?

I am just trying to figure out if I am even setting up this problem correctly...

Edit: instead should I be adding $(n,0)+S$, in this case there would be only one coset $(1,0)$???

• My basic question is just: if you are taking the quotient ring with coordinates, are the rings of the form: (n,n) +S or (n,0)+ S – p.l May 1 '16 at 6:00

## 1 Answer

Before I answer your question, let's calculate the cosets. To start with, let's pick some element not in $S$. $(1,0)$ will do for our purposes.

We have $(1,0) + S = \{(1,0),(3,0),(1,3),(3,3)\}$.

Note that $(3,0) \in S$, that is: $(3,0) + S = (1,0) + S$. Perhaps this might dissuade you from the notion that adding $(n,0)$ to $S$ will recover all the cosets. In fact, all of the elements $(n,0)$ have already appeared in just the first two cosets.

Now we need an element that hasn't occured in our two cosets so far. $(1,1)$ will do:

$(1,1) + S = \{(1,1),(3,1),(1,4),(3,4)\}$.

We haven't encountered $(0,1)$ yet, either, so our fourth coset can be:

$(0,1) + S = \{(0,1),(2,1),(0,4),(2,4)\}$

$(2,2)$ has yet to occur, so we have a fifth coset:

$(2,2) + S = \{(2,2),(0,2),(2,5),(0,5)\}$

The last coset has to be "whatever is left over", so we have:

$(1,2) + S = \{(1,2),(3,2),(1,5),(3,5)\}$.

Now, on to your question-is it true that adding $(n,n)$ will yield all the cosets? For this to be true, we need exactly one element of the form $(n,n)$ in each coset....but-there's a catch. $n$ can only cycle up to $3$ in the first coordinate, but can go up to $5$ in the second. That is, instead of:

$(4,4)$ we get $(0,4)$, and instead of $(5,5)$ we get $(1,5)$.

Indeed, we find that:

$(0,0) + S = S$

$(1,1) + S \neq S$

$(2,2) + S \neq S,(1,1) + S$

$(3,3) + S = (1,0) + S \neq S, (1,1) + S, (2,2) + S$

$(0,4) + S = (0,1) + S \neq S, (1,1) + S, (2,2) + S, (3,3) + S$

and, of course, $(1,5) + S = (1,2) + S$, the only coset not yet accounted for.

The deeper question you should be asking yourself, here, is:

If $R = R_1 \times R_2$, is is true that if $I$ is an ideal of $R_1$ and $J$ is an ideal of $R_2$,

that $I \times J$ an ideal of $R$; and do we have:

$R/(I \times J) \cong R_1/I \times R_2/J$?