-3
$\begingroup$

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 $(1,0)$???

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – p.l May 1 '16 at 6:00
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.