Dimension of $\mathbb{C}$-algebra $\mathbb{C}[x,y]/(x^{3},xy,y^{2})$

Question

I am trying to find the dimension and a basis of $\mathbb{C}$-algebra $\mathbb{C}[x,y]/(x^{3},xy,y^{2})$. I think the easiest way would just be to understand how $\mathbb{C}[x,y]/(x^{3},xy,y^{2})$ looks like. The ideal $I=ax^{3}+xy+cy^{2}$ where $a,b,c \in \mathbb{C}[x,y]$. Am I right in thinking $[ax^{3}+bxy+cy^{2}]=[0]$ and so $\mathbb{C}[x,y]/(x^{3},xy,y^{2})=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}y$ so ${1,x,x^{2},y}$ is the basis. But I haven't defined the maps between rings $\mathbb{C}$ and $\mathbb{C}[x,y]/(x^{3},xy,y^{2})$.

Answer 1

The ideal $(x^3,xy,y^2)$ is a monomial ideal, an ideal generated by monomials. It's easy to get a $\Bbb C$-basis for $\Bbb C[x,y]/I$, where $I$ is a monomial ideal, since one can take the monomials in $x$ and $y$ which are not divisible by the monomials in the generating set for $I$. Here this basis consists of $1$, $x$, $x^2$ and $y$, as you say (or more strictly the of cosets $1+(x^3,xy,y^2)$, etc.)

Answer 2

Indeed, $I=\langle x^3,xy,y^2\rangle$ is a monomial ideal. The only monomials not lying in this ideal are those of $B=\{1,x,x^2,y\}$. These monomials are called the standard monomials of $I$. They form a vector space basis for $A={\Bbb C}[x,y]/I$.

This situation can be generalized to the case when $I$ is a zero-dimensional ideal of ${\Bbb C}[x_1,\ldots,x_n]$. More upon request.