Dimension of $\dim_{\mathbb C}\mathbb C[X,Y]/I(Y^2-X^2,Y^2+X^2)$

Question

I'm trying to solve the question 1.36 from Fulton's algebraic curves book:

Let $I=(Y^2-X^2,Y^2+X^2)\subset\mathbb C[X,Y]$. Find $V(I)$ and $\dim_{\mathbb C}\mathbb C[X,Y]/I$.

Obviously $V(I)=\{(0,0)\}$ and by a corolary in the same section we know that $\dim_{\mathbb C}(\mathbb C[X,Y]/I)\lt \infty$, but I don't know how to calculate $\dim_{\mathbb C}(\mathbb C[X,Y]/I)$.

I need help in this part.

Thanks in advance

Answer 1

Given a ring $R$, every element of $R[x]$ is a sum of elements $a_0+a_1 x + a_2 x^2 + \ldots +a_n x^n$ where $a_i \in R$. Using this twice, we have that a basis for $\mathbb C[x,y]$ is the monomials $x^iy^j$ where $i,j\in \mathbb N$ Since $I=(x^2+y^2,x^2-y^2)=(x^2,y^2)$, we have that every monomial where either $i$ or $j$ is at least $2$ will be in $I$, and so every element of $\mathbb C[x,y]/I$ is of the form $a + bx + cy + dxy +I$. In fact, the elements of $I$ are spanned by the monomials where either $i$ or $j$ are at least $2$, and so $\{1,x,y,xy\}$ is a basis for $\mathbb C[x,y]/I$, and so the dimension is $4$.

Answer 2

Hint: Note that $Y^2=\bigl((Y^2-X^2)+(Y^2+X^2)\bigr)/2$ belongs to $I$, which in turn implies that $X^2=(Y^2+X^2)-Y^2\in I$. In other words $(X^2,Y^2)\subseteq I$, and a similar reasoning shows the reversed inclusion. Having this, what can you say about the higher order terms of an arbitrary polynomial modulo $I$?