5
$\begingroup$

In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ One does this with the first isomorphism theorem. With the same idea I proved for example that $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by $$ \phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right) $$ is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.

What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?

Thanks in advance for any help/hint/comment!

$\endgroup$
5
$\begingroup$

Another approach is to solve $mx-ny=k$ and then show that $r(m/k,n/k)+s(x,y)$ is all of $\mathbb Z\times \mathbb Z$, and each element of $\mathbb Z\times\mathbb Z$ is expressible in this way only one way.

Then it is clear that $r(m/k,n/k)+s(x,y)\sim r'(m/k,n/k)+s'(x,y)$ if and only if $k\mid r-r'$ and $s=s'$.

So, how do we show the above? If $mx-ny=k$ then $$\begin{pmatrix}m/k&n/k\\y&x\end{pmatrix}$$ has determinant $1$, and thus has an inverse with integer coefficients.

$\endgroup$
  • $\begingroup$ Thanks a lot, this looks more similar to what I am trying to do! I kind of understand what you say. However, I cannot see exactly how this solves the problem! Could you elaborate a little more please? For example what do you mean by $\sim$ and once I know that every $(a,b)$ in $\mathbb{Z} \times \mathbb{Z}$ has a unique expression of that form, how does that yield a surjective homomorphism $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}_k$ with kernel $\langle (m,n) \rangle$? Thanks again! $\endgroup$ – Leo Sera Feb 20 '17 at 5:45
3
$\begingroup$

As SpamIAm mentioned, Smith Normal Form is a useful tool here. With $(m,n)=(4,6)$ we get $(4,6)\rightarrow (4,2)\rightarrow (2,2)\rightarrow (2,0)$ which means we get $\mathbb{Z} \times \mathbb{Z}_2$. The steps are to turn your $m$ and $n$ into a matrix and you can perform operations such as subtracting an integer multiple of a column from another and re-ordering the columns. Try this for the cases you used above to get a sense of the pattern. Then see if you can get a general homomorphism from these examples. Best of luck!

$\endgroup$
  • $\begingroup$ Thanks! I've never heard of Smith normal form before. I will look into @SpamIAm link right now. What do you mean by turning $m$ and $n$ into a matrix? $\endgroup$ – Leo Sera Feb 20 '17 at 3:16

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.