# minimal number to generate an ideal

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.

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 $$f \mapsto f(z,w)$$.
1. 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 $$J \neq I$$.
2. Similarly, the ideal $$J = (y^2, xy)$$ is not equal to $$I$$.
3. 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 $$\varphi_{1,1}(xy) = 1$$.