Finding Dimension of quotient ring

Question

Let $I=(X, Y) \subset k[X,Y]$. show that $\mbox{dim}_k(k[X,Y] / I^n) = 1+2+...+n=n(n+1) /2 $

Here k is algebracally closed field. And $(X,Y) $ is ideal in polynomial ring $k[X,Y]$ generated by $ X, Y$. Actually I can't figure out how to find a basis practically. Please help.

Answer 1

Hints: Justify the following simpler observations.

1. If $I=(X,Y)$, then $I^n$ is generated by all the monomials of degree $n$, i.e. $\{X^aY^b|a,b\geq 0,a+b=n,\}$. In addition, every element of $I^n$ has degree $n$ or greater.

2. So when you quotient $R=k[X,Y]$ by $I^n$, every coset can be represented by polynomials of degree $<n$. These are distinct because the second remark of 1. implies that no polynomial of degree less than $n$ can be in $I^n$.

3. Now count the dimension of the space of all 2 variable polynomials of degree less than $n$. This is spanned by all monomials of degree less than $n$ so count these instead. Another hint here is that $$\frac{n(n+1)}{2}=\binom{n+1}{2}.$$

I rest my case.