dimension of ideal of polynomials

Question

I would like to determine the dimension as $\mathbb C$-vector space of the set of polynomials of degree $n$ that lie in $(x_0x_1-x_2^2)=I$ where $I$ is the principal ideal of$\{x_0x_1-x_2^2\}$ in $ \mathbb C[x_0,x_1,x_2]$. My idea is that $I$ should be similar to the set of polynomials of dimension $n-2$. Any suggestion to compute this dimension?

Answer 1

Your intuition is exactly right: for every polynomial $p \in I$, we can write $p = (x_0 x_1 - x_2^2) \cdot q$ for some polynomial $q$. If $\deg(p)\leq n$, then we must have $\deg(q) = n-2$. The question now becomes, what is the dimension of the vector space of polynomials in $3$ variables with degree less than or equal to $n-2$?

A nice basis for this space is the monomials with degree at most $n -2$, so we just need to count the number of monomials $x_0^a x_1^b x_2^c$ with $a + b + c \leq n-2$. Defining $d: = n - 2 - a - b - c$, we get that the quadruple $(a,b,c,d)$ is a weak composition of $n - 2$, i.e. some parts may be zero. The number of these is well known: there are $$\binom{n-2 +4 - 1}{4 - 1} = \binom{n+1}{3}$$ many of them, implying that this is the dimension.