Verify that a set is a basis for a ring

Question

Consider the ring $\mathbb{C}[x,y]/(x^2-x,y^2-y)$. How would I go about proving that $(1,x,y,xy)$ is a $\mathbb{C}$-basis for it? It seems obvious as I could reduce every term of a polynomial of that ring according to those $x^2=x$ and $y^2=y$ and yet this doesn't lead me to something rigorous enough.

Answer 1

Every polynomial $P$ in $\mathbb{C}[x,y]$ can be written as $P(x,y)= \alpha + \sum_{i \geqslant 1} a_i x^i + \sum_{j \geqslant 1} b_j y^j + \sum_{k,l \geqslant 1} c_{k,l} x^k y^l$, and that gives, in $\mathbb{C}[x,y]/(x^2-x,y^2-y)$, $\bar{P}=\alpha + \sum a_i x + \sum b_j y + \sum c_{k,l}xy$, so the family $(1,x,y,xy)$ generates $\mathbb{C}[x,y]/(x^2-x,y^2-y)$.

To show that the family is free, suppose that $\alpha + \beta x + \gamma y + \epsilon xy =0$. That means that in $\mathbb{C}[x,y]$ you have $\alpha + \beta x + \gamma y + \epsilon xy \in (x^2-x, y^2-y)$ From there you can show that $\alpha$, ... are zero.

Answer 2

In general, the basis $\Bbb C[x,y]$ over $\Bbb C$ as a vector space is $$\{(x)^m(y)^n; n,m\ge0 \}$$ so you can prove that basis for $\Bbb C[x,y]/(y^2-y,x^2-x)$ is $\{1,x,y,xy \}$