How is the ideal $(x,y)$ isomorphic to $k[x, y]$ as $k[x, y]$-modules? 
Let $R = k[x, y]$ where $k$ is a field and consider the ideal $I = (x, y)$ as a $R$-module.


Consider the $R$-module homomorphism $\varphi : R^2 \to I$ given by $\varphi(a, b) = ax + by$.


Prove that the kernel of $\varphi$ is the set $\{(−cy, cx) \mid c ∈ R\}$, and show that $\ker \varphi$ is isomorphic to $R$ as a $R$-module.


Deduce an isomorphism $R^2/R \cong I$.


I've figured everything out, but I'm bothered by the last statement.
We have $I \cong R^2/ \ker \varphi \cong R^2/R$ since $R \cong \ker \varphi$.
However, isn't $R^2/R \cong R$?
This would mean $I \cong R$, but $I$ is an $R$-module generated by two elements whereas $R$ is an $R$-module generated by one element.
Am I not understanding something?
 A: Certainly, it is true that there are isomorphisms between $R^2/M$ and $R$ where $M$ is a free rank one submodule of $R^2.$ For example, $R^2/(R\oplus 0)$ or $R^2/(0\oplus R)$ are both isomorphic to $R.$ You are also correct that $I\not\cong R.$
The issue here is that there are many different ways to embed $R$ as a submodule of $R^2$ (equivalently, there are many $R$-submodules $M\subseteq R^2$ which are isomorphic to $R$), and the quotients of $R^2$ by different such $M$ need not be isomorphic. Indeed, as you've shown, there exists such an $M$ with $R^2/M\cong I\not\cong R$! So, writing $R^2/R$ is a little misleading.
An example of this phenomenon which might be more familiar is the following. Consider the ring $\Bbb{Z},$ and note $$\Bbb{Z}/\Bbb{Z}\cong 0$$ while on the other hand we have $$\Bbb{Z}/2\Bbb{Z}\not\cong0.$$ However, $\Bbb{Z}\cong2\Bbb{Z}$ as $\Bbb{Z}$-modules, via $n\mapsto 2n.$
So, given two $R$-modules $M$ and $N$ such that there exist $R$-submodules $N'$ and $N''$ of $M$ with $N\cong N'\cong N'',$ we cannot unambiguously write $M/N.$ We need to specify the embedding $N\to M$ which we are using to consider $N$ as a submodule of $M$!
