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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.

Did
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BpZ
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    Do you understand the counting problem you are solving? You need only determine the number of monomials in two variables of total degree less than $n$. Hence you want to count $(i,j)$ that produce $X^iY^j$ with $i+j<n$. This is a triangular lattice and the counting $1+\cdots+n$ gives the desired dimension. – Pedro Mar 11 '18 at 19:20
  • how you do the triangulate lattice? Can you say a example for example with $dim_k( k[x,y]/I^3)$ please – BpZ Mar 15 '18 at 16:12
  • Did you try the case $n=1$ before posting? – hardmath Mar 17 '18 at 18:59

1 Answers1

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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!

Armando j18eos
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