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.

• May be, we have to write : $\mathbb{C} [X,Y] / I^n = \mathbb{C} [X,Y] / I \oplus \displaystyle \oplus_{ k = 2}^{n-1} I^k / I^{k+1}$, no ? What is : $\mathrm{dim}_{ \mathbb{C} } I^k / I^{k+1}$ ? – YoYo Sep 7 '16 at 16:52

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.

Choose a basis of monomials $x^ky^l$ with $k$ and $l$ non-negative integers. Obviously you need only consider those with $k+l<n$.

Now let $M:=k+l$. There are $M+1$ ways to write an integer $M$ as a sum of non-negative integers.

Hence in our case there are $1$ monomial of degree $0$, $2$ monomials of degree $1$, ..., $n$ monomials of degree $n-1$.