Show that $dim_C(m^n/m^{n+1}) = n+1$ Let $R = \mathbb{C}[x,y]$ and $m$ be a maximal ideal. Show that $\dim_\mathbb{C}(m^n/m^{n+1}) = n+1$
I know that every maximal ideal of $\mathbb{C}[x,y]$ is in the form of $(x-a,y-b)$, so I will just let $m = (x,y)$. How does the ring $m^n/m^{n+1}$ look like and what does it mean by the dimension of such ring?
 A: $\mathfrak m^n / \mathfrak m^{n + 1}$ is a vector space over the field $\mathbb C$, and the "dimension" being referred to is its vector space dimension over $\mathbb C$.
Let's look at the case $n = 1$, and without loss of generality, set $a = b = 0$.


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*$\mathfrak m = (x, y)$ contains all polynomials that do not have a constant term, i.e. it contains $\mathbb C$-linear combinations of the degree $\geq 1$ monomials $x, y, x^2, xy, y^2, x^3, x^2y, xy^2, y^3, \dots$

*$\mathfrak m^2 = (x^2, xy, y^2)$ contains all polynomials that do not have constant or linear terms, i.e. it contains $\mathbb C$-linear combinations of the degree $\geq 2$ monomials $x^2, xy, y^2, x^3, x^2y, xy^2, y^3, \dots$
So $\mathfrak m / \mathfrak m^2$ has $\mathbb C$-dimension two. A basis is $\{ x, y\}$.
Can you see how this generalises? 
A: Essentially, $\mathfrak m^n$ is the ideal consisting of polynomials of degree at least $n$. It is generated as an ideal by the monomials $x^ky^{n-k}$ for $0\leq k\leq n$. Similarly, $\mathfrak m^{n+1}$ is generated by the monomials $x^ky^{n+1-k}$. Now, as a general vector space, $\mathfrak{m}^n$ consists of elements $z$ of $\mathbb{C}[x,y]$ such that there are arbitrary polynomials $p_k(x,y)$ with
$$z=\sum_{k=0}^{n}{p_k(x,y)x^ky^{n-k}}$$
If we are taking the quotient by $\mathfrak m^{n+1}$, any positive degree terms in $p_k(x,y)$ will "die" as they will result in a polynomial of degree $n+1$, hence we may actually assume the $p_k$ are complex numbers. Thus we have that
$$z=\sum_{k=0}^{n}{p_kx^ky^{n-k}}$$
for some $p_k\in \mathbb{C}$. This shows that the dimension is at most $n+1$. To see that the dimension is at least $n+1$, note that no polynomial of degree $n$ is sent to $0$ because no such polynomial is contained in $\mathfrak{m}^n$.
