I am working on the below problem:
Let $I$ be the ideal generated by $x$ and $y$ in $\mathbb{C}[x,y]$. Prove that there is a unique (up to scalar) polynomial $f(x,y)$ of degree $2$ in $I^2$ which satisfies $f(2,3) = f(-1,5) = 0$.
Here are my thoughts so far:
If $I = \langle{x,y}\rangle$ in $\mathbb{C}[x,y]$, from what I understand, $I^2 = \langle{x^2,y^2,xy}\rangle$ in $\mathbb{C}[x,y]$. Thus, if we have a polynomial of degree $2$ in $I^2$, $f(x,y) = ax^2 + by^2 + cxy$ for some complex scalars $a,b,c$.
Thus, if $f(2,3) = f(-1,5) = 0$, we have $0 = 4a + 9b + 6c$ and $0 = a + 25b -5c$. It remains to show that we can solve this system of $2$ equations in $3$ unknowns uniquely for $a,b,c$.
But, this seems impossible to me. A system of $n$ equations with $n + 1$ unknowns can't be solved uniquely. Where did my thinking go wrong? If I instead take $I^2 = \langle{x^2,y^2}\rangle$, the problem will work out -- we will arrive at a system of $2$ equations in $2$ unknowns, and the corresponding matrix will have full rank (determinant nonzero), so there will be a unique solution. But, this contradicts the notion of a square of an ideal.
With the right notion of the square of an ideal, is there a third equation I can arrive at from the given information?
Thanks!
I note that "up to scalar" allows a third degree of freedom. Are you sure that your parametrized set of solutions are not all the same up to scalar multiple?
(When I solve it, this turns out to be the case.)