Show that following modules are isomorphic

Question

I have $M=\mathbb{Z^2/\langle(-2,5)\rangle}$ as $\mathbb{Z}$-module and have to show that it is isomorphic with $\mathbb{Z}$.

I have found an isomorphism $f: \mathbb{Z^2}\rightarrow\mathbb{Z^2}$ which sends $(-2,5)$ to $(0,1)$ and $(1,-2)$ to $(1,0)$. Can I say that $f$ induces an isomorphism $\mathbb{Z^2}/\langle(-2,5)\rangle\rightarrow\mathbb{Z^2}/\langle(0,1)\rangle$ and then conclude that it is isomorphic to $\mathbb{Z}$?

I could say that if $\langle(-2,5)\rangle$ is the kern of $f$ and $\mathbb{Z^2}/\langle(0,1)\rangle$ the image. Why is this the case, and how can I then conclude that it is isomorphic to $\mathbb{Z}$?

Answer 1

As you say $\mathbb{Z}^2/\langle(0,1)\rangle$ is isomorphic to $\mathbb{Z}$. This is pretty clear; you send $(a,b)\mapsto a$ and this is clearly surjective but it's also injective as the kernel of this map is the image of $(0,b)$ in $\mathbb{Z}^2/\langle(0,1)\rangle$. $(0,b)$ is in the ideal that we are quotienting out by so it's the zero element in the quotient.

Now the crux of the question is to show that the first map you wrote down is an isomorphism. Here you want to write down your $\mathbb{Z}$-linear map in terms of its matrix (with respect to a basis) and show that the determinant is $\pm 1$.
Convince yourself that this is indeed the matrix $A=\begin{pmatrix}1 & -2\\-2& 5\end{pmatrix}^{-1}$ with respect to the standard basis. (You can think of as $A\begin{pmatrix}1\\-2\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$ and likewise for the other vector. Now piece these two parts to write down a matrix for $A$.)

Answer 2

It's perhaps better to look for a homomorphism $f\colon\mathbb{Z}^2\to\mathbb{Z}$ such that $\ker f=\langle(-2,5)\rangle$.

You need to choose $f((1,0))=a$ and $f((0,1))=b$ and a requirement is that $$ f((-2,5))=-2f((1,0))+5f((0,1))=-2a+5b=0 $$ Well, a first attempt would be $a=5$, $b=2$.

Suppose now $(x,y)\in\ker f$, which means $ax+by=0$, hence $5x+2y=0$. In particular $5x\equiv0\pmod2$, so $x=2u$, and $2y\equiv0\pmod{5}$, so $y=5v$ and we need $10(u+v)=0$,so $u=-v$ and therefore $$ (x,y)=(-2v,5v)=v(-2,5) $$ OK, $\ker f=\langle(-2,5)\rangle$. Now, is $f$ surjective?

Answer 3

If $f:G\to G$ is a module automorphism and $H$ is a submodule of $G$, then $G/H\simeq G/f(H)$.

In your case the automorphism is $f(a,b)=(5a+2b,2a+b)$ and then $$\mathbb Z^2/\langle(-2,5)\rangle\simeq \mathbb Z^2/\langle(0,1)\rangle.$$ But $\langle(0,1)\rangle=\{0\}\times\mathbb Z$. Now $$\mathbb Z^2/\{0\}\times\mathbb Z\simeq\mathbb Z/\{0\}\times\mathbb Z/\mathbb Z\simeq\mathbb Z.$$

Answer 4

Alterntively, a key observation is that $(-2,5)$ is part of a basis for $\mathbb Z^2$. Indeed, $\{ (-2,5), (-1,2) \}$ is a basis (found by the extended Euclidean algorithm, since $\gcd(-2,5)=1$). Then $$ \frac{\mathbb Z^2}{\langle(-2,5)\rangle} = \frac{\mathbb Z (-2,5) \oplus \mathbb Z(-1,2)}{\mathbb Z (-2,5) \oplus 0\hskip11mm} \cong \mathbb Z(-1,2) \cong \mathbb Z $$