How to compute $ \mathrm{dim}_{ \mathbb{C} } \big( \mathbb{C} [X, Y] / I^n ) $? 
Can you show me how to compute $ \mathrm{dim}_{ \mathbb{C} } \big( \mathbb{C} [X, Y] / I^n  \big) $ where $ I = (X,Y) \subset \mathbb{C} [X,Y] $ ? 

I have to find : $ \mathrm{dim}_{ \mathbb{C} } \big( \mathbb{C} [X, Y] / I^n ) = 1 + 2 + \dots + n = \dfrac{n(n+1)}{2} $, but i don't know how.  
Thank you in advance.

As a vector space, $\mathbb{C}[x,y]/I^n$ is spanned by the monomials in $x$ and $y$ of total degree at most $n-1$. Notice that these monomials are linearly independent. If you enumerate by degree $d$, it is easy to count them:


*

*$d=0$ gives the monomial $1$ (1 term)

*$d=1$ gives the monomials $x, y$ (2 terms)

*$d=2$ gives the monomials $x^2, xy, y^2$ (3 terms)

*$d=3$ gives the monomials $x^3, x^2y, xy^2, y^3$ (4 terms)

*...

*$d=n-1$ gives the monomials $x^{n-1}, x^{n-2} y, ..., x y^{n-2}, y^{n-1}$ ($n$ terms)


So the basis consists of $1+2 + \cdots + n = \frac{n(n+1)}{2}$ terms, as desired.
