How to compute the dimension of $\Bbb C[x,y]/I$ on $\Bbb C$? The problem is compute the dimension of $\Bbb C[x,y]/I$ over $\Bbb C$ as vector space where $I=\langle(x+2)^2,(x+2)(y+1),(y+1)^3\rangle$. $\Bbb C[x,y]$ is the polynomial ring over $\Bbb C$.
I have tried in this way.
For $f(x,y)$ in $\Bbb C[x,y]$ it has the form $f(x)=(a_{0}+a_{1}(x+2)+…a_{i}(x+2)^i+…a_{n}(x+2)^n)(b_{0}+b_{1}(y+1)+…b_{j}(y+1)^j+…b_{m}(x+2)^m))=[a_{0}+a_{1}(x+2)][(b_{0}+b_{1}(y+1)+b_{2}(y+1)^2]+f_{1}(x,y)(x+2)^2+f_{2}(x,y)(y+1)^3$
And $[a_{0}+a_{1}(x+2)][(b_{0}+b_{1}(y+1)+b_{2}(y+1)^2]=a_{0}b_{0}+a_{1}b_{0}(x+2)+a_{0}b_{1}(y+1)+a_{0}b_{2}(y+1)^2+f_{3}(x,y)(x+2)(y+1)$
So $f(x,y)=a_{0}b_{0}+a_{1}b_{0}(x+2)+a_{0}b_{1}(y+1)+a_{0}b_{2}(y+1)^2$ in $\Bbb C[x,y]/I$ .
The basis are {$1,x+2,y+1,(y+1)^2$}.Any polynomial's coordinate in $\Bbb C[x,y]/I$  is $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$
The remain is the dimension for $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$ in $\Bbb C^4$ .I am not very sure for it.4 or 5,or other?

A basis of  $C[x,y]$  is $\{(x+2)^m(y+1)^n: m,n\ge 0\}$.
In $C[x,y]/I$, $(x+2)^2$ and $(y+1)^3$ are identified with $1$, hence, just because of them the basis reduces to
$$
(x+2)^m(y+1)^n, \quad m=0,1,\,\,n=0,1,2.
$$
But as $(x+2)(y+1)$ is also identified with $1$, then we are reduced to
$$
1, x+2, y+1, (y+2)^2.
$$
Hence
$$
\dim C[x,y]/I=4.
$$
