map between polynomial rings with given image I am supposed to write down an exact sequence $$0\to \mathbb C[x,y]\to \mathbb C[x,y]\to \mathbb C[y]/(y^2)\to 0.$$ To obtain the ring on the right, we should divide the one in the middle by $(x,y^2)$. So the kernel of the second map should be the ideal $(x,y^2)$ and since we want an exact sequence, this should also be the image of the first map and this map should furthermore be injective. I thus tried to define the first map to be $$f\mapsto f\cdot x\cdot y^2,$$ does this work? If not, can you help me doing better?
EDIT: According to the answer I misunderstood something. Still, I want to understand what should be on the left such that we get an exact sequence with $\mathbb C[x,y]\to \mathbb C[y]/(y^2)\to 0$ on the right. I am not sure whether it is correct but does $\mathbb C[x,y]\oplus \mathbb C[x,y]$ work? With the map $f\oplus g\mapsto f\cdot x+g\cdot y^2$.
 A: Such an exact sequence cannot exist (at least with $\Bbb{C}[x,y]$ module homomorphisms). Why not? Well, it's because as a $\newcommand\CC{\Bbb{C}}\CC[x,y]$ module $\CC[x,y]$ is cyclic, generated by $1$. Thus its image under whatever map you choose will also be cyclic (generated by the image of $1$). However $(x,y^2)$ is not cyclic. If there were a single polynomial $f$ generating it, we'd have $f\mid x$ and $f\mid y^2$, which would imply that $f$ is a unit, but units generate all of $\CC[x,y]$.
In particular, your map doesn't work, since $xy^2$ divides every element in the image of your map, so its image doesn't contain either $x$ or $y^2$.
Edit
The only thing that can go on the left to make everything exact on the right is the ideal $(x,y^2)$ itself. Your suggestion of $\CC[x,y]\oplus \CC[x,y]$ with the map $(a,b)\mapsto ax+by^2$ will be exact on the right, but not on the left, since this map is not injective.
If a four term exact sequence is what you're looking for, then its possible you're looking for
$$ 0\to \CC[x,y]\xrightarrow{a\mapsto (-ay^2,ax)} \CC[x,y]\oplus \CC[x,y] \xrightarrow{(f,g)\mapsto fx+gy^2} \CC[x,y]\to \CC[y]/(y^2) \to 0.$$
