How to prove that we need at least three elements to generate the ideal $I=(x^2,y^2,xy)$ in $\mathbb{C}[x,y]$ ? Is this related to dimension of a module? This is not homework, thank you.
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This follows more or less from the definition of being a free variable, but I doubt this would help you in this concrete setting. Thus we will try to show that the maps $\mathbb C^2 \rightarrow \mathbb C$ induced by these polynomials satisfy what you claim.
First, observe that any polynomial in $I$ has no summands of degree $0$ or $1$, since it must be a sum of multiples of degree $2$ polynomials. Since, up to units, the only degree $2$ polynomials in $\mathbb C[x,y]$ are $x^2, y^2, xy$, it suffices to check that no subset of $\{x^2,y^2,xy\}$ generates $I$.
We will make use of the substitution homomorphism $\varphi_{z,w}: \mathbb C[x,y] \rightarrow \mathbb C$ given by $f \mapsto f(z,w)$.
- Consider the ideal $J = (x^2, xy)$. Then, we have that $\varphi_{0, 1}(x^2) = \varphi_{0, 1}(xy) = 0$. Thus, we obtain that $J \subset \ker \varphi_{0,1}$. But $\varphi_{0,1}(y^2) = 1$ and $y^2 \in I$, so $J \neq I$.
- Similarly, the ideal $J = (y^2, xy)$ is not equal to $I$.
- Consider the ideal $J = (x^2, y^2)$. Suppose toward a contradiction that $xy \in J$. Since the degree is additive it follows that $xy = ax^2 + by^2$ for $a, b \in \mathbb C$. (A priori we have that $a,b \in \mathbb C[x,y]$, but by the remark about the degree I just made it follows that $a,b$ are degree-$0$ polynomials.) Note that since $\varphi_{0,1} (xy) = \varphi_{1,0}(xy) = 0$ for all $w$, we obtain that $ax^2 = by^2 = 0$. But this contradicts $\varphi_{1,1}(xy) = 1$.
