Let $X ⊂ \mathbb {A^3}$ be the union of the three coordinate axes. Determine generators for the ideal $I(X)$, and show that $I(X)$ cannot be generated by fewer than three elements.
Call the three coordinate axis $\hat {x} $, $\hat {y}$ and $\hat {z}$; clearly $I(\hat {x}) = (x)\subset \bar {k}[x,y,z]$, where $\bar {k}$ is algebraically closed. A similar observation is true for the other two axis, and since $X = \hat {x} \cup \hat {y} \cup \hat {z}$, we have $I (X) = (x)\cap (y) \cap (z)$. However this should mean that $I (X)=(xyz) $, and this ideal is principal (while the exercise says that it cannot be principal). I really don't see what I am missing, can you give me a hint?
Observe that $(x)$ is not the ideal of the $X$-axis, but of the $YZ$-plane, which are the points where $x=0$ is satisfied.
The ideal of each axis are $(x,y)$, for the $Z$-axis, $(y,z)$, for the $X$-axis, and $(x,z)$, for the $Y$-axis.
The ideal of the union of the axes is the intersection of these ideals. $$(x,y)\cap(y,z)\cap(x,z)=(xz,y)\cap(x,z)=(xz,xy,yz)$$
To prove that it cannot be generated by less than $3$ polynomials, let's look at all initial monomials (largest monomial in the monomial order $x<y<z$) of all elements of this ideal. These are monomials that are multiples of $xy$, $xz$ or $yz$.
If $\mathbb{N}^3$ is the set of possible triples of exponents $(r,s,t)$ in the collection of initial monomials $x^r,y^sz^t$ of the elements of the ideal, there needs to be a generator with initial monomial $(1,1,0)$, another with initial monomial $(1,0,1)$ and another with initial monomial $(0,1,1)$.