Let $I = (X,Y) \subset k[X,Y]$ then $\dim_k(k[X,Y]/I^n) = \frac{n(n+1)}{2}$

Question

Why we have this:

Let $I = (X,Y) \subset k[X,Y]$ then $\dim_k(k[X,Y]/I^n) = 1+2+3+...n = \frac{n(n+1)}{2}$

This is no clear for me.

Answer 1

Simply $$ \forall n\in\mathbb{N}_{\geq1},\,I^n=(x,y)^n=(x^iy^j\mid i,j\in\{0,1,\dots,n\},\,i+j=n), $$ so $$ \mathbb{K}[x,y]_{\displaystyle/I^n}=\mathbb{K}[x,y]_{<n}=\{f\in\mathbb{K}[x,y]:\deg f<n\}; $$ a base of this $\mathbb{K}$-vector space is $\{x^iy^j\mid i,j\in\{0,1,\dots,n-1\},\,i+j<n\}$, and its size is $\displaystyle\binom{(n-1)+2}{2}=\binom{n+1}{2}=\frac{(n+1)n}{2}$.

For some proof of this formula, you can read here!