Determining the generators of $I(X)$ Let $ X \subset \mathbb A^3 $ be the union of three coordinate axes.
How do I determine the generators of $I(X)$?
Also, how do I show it has at least 3 elements?
 A: Here is a completely elementary and short proof that, given   an arbitrary  field $k$, the ideal $I=(xy,xz,yz)\in k[x,y,z]$ cannot be generated by two polynomials.  
Suppose, by way of contradiction, that we had $(xy,xz,yz)=(f,g)$.
The two polynomials $f,g\in (xy,xz,yz)$ obviously contain  only monomials of degree $\geq 2$ and can thus be written as $f=f_2+f_+,\: g=g_2+g_+$, where $f_2,g_2$ are homogeneous of degre $2$ and $f_+,g_+ $ only contain monomials of  degree $\geq 3$.   
But then taking the $k$- vector space of homogeneous degree $2$ polynomials in  $(xy,xz,yz)=(f,g)$ we get the equality of $k$-vector spaces $$k\cdot xy\oplus k\cdot xz\oplus k\cdot yz=k\cdot f_2+k\cdot g_2$$ Since the vector space on the left has dimension $3$ and that on the right dimension $\leq 2$, we have our contradiction.
A: Let $K$ be an infinite field; let $X_1,X_2,X_3$ be indeterminates; for $i=1,2,3$ let $L_i\subset K^3$ be the $i$ th coordinate axis; let $V\subset K^3$ be the union of the $L_i$; let $\mathfrak a_V\subset A$ be the ideal attached to $V$; and let $B$ be the quotient $A/\mathfrak a_V$.
Then $V$ is the limit of the inductive system 
$$
\begin{matrix}
L_1&\leftarrow&\{0\}&\to&L_2\\ \\
&&\downarrow\\ \\
&&L_3
\end{matrix}
$$
(where the maps are the obvious ones). This is true at least in the category of sets. I suspect that this holds also in some appropriate category of schemes, but I don't know enough about schemes to be able to say anything more about this. I hope some friendly help here!
Anyway, we'll see that, as we hope, $B$ is the limit of the projective system of $K$-algebras
$$
\begin{matrix}
K[X_1]&\to&K&\leftarrow&K[X_2]\\ \\
&&\uparrow\\ \\
&&K[X_3],
\end{matrix}
$$
where each map is the evaluation at $0$. As easily seen, this limit can be written as 
$$
C:=K\oplus x_1K[x_1]\oplus x_2K[x_2]\oplus x_3K[x_3],
$$
where $x_i$ is transcendent over $K$ and $x_i\,x_j=0$ for $i\not=j$.
The main point we'll check is that the obvious morphism from $A$ to $C$ induces an isomorphism of $B$ onto $C$.
Let us drop temporarily the assumption that $K$ is infinite. We claim that $C$ has no nonzero nilpotent element. Indeed, let 
$$
f(x_1,x_2,x_3)=a+x_1\ f_1(x_1)+x_2\ f_2(x_2)+x_3\ f_3(x_3)
$$
satisfy $f(x_1,x_2,x_3)^n=0$. By spelling out the equalities $f(0,0,0)^n=0$, $f(x_1,0,0)^n=0$,\dots, we get $f(x_1,x_2,x_3)=0$.
Let $\mathfrak a\subset A$ be the ideal generated by the $X_iX_j$ with $i\not=j$, let $\mathfrak m\subset A$ be the ideal (clearly maximal) generated by the $X_i$, and note the inclusion $\mathfrak a\subset\mathfrak a_V$.
We'll use tacitly the (easily proved) fact that $\mathfrak a$ and $\mathfrak m$ are generated over $K$ by their monomials. The monomials in $\mathfrak a$ are precisely the ones which involve at least two of the three indeterminates, and you know, dear reader, what the monomials in $\mathfrak m$ are.
As $\mathfrak a/\mathfrak m\mathfrak a$ is three dimensional, $\mathfrak a$ cannot be generated by two elements.
Consider the sub-$K$-vector space of $A$ defined by 
$$
W:=K\oplus X_1K[X_1]\oplus X_2K[X_2]\oplus X_3K[X_3].
$$
(The sum is clearly direct.) We have $A=\mathfrak a\oplus W$, from which we infer that the obvious morphism from $A$ to $C$ induces an isomorphism of $A/\mathfrak a$ onto $C$, and that $\mathfrak a_V=\mathfrak a$ if and only if $K$ is infinite.
A: It is easy to see that the ideal $I(X)$ is generated by $yz,xz,xy\; $, that is $I(X)=(yz,xz,xy)$.
The difficulty is to show that no two polynomials suffice to generate this ideal.
For that there is a fine invariant, aptly called the embedding dimension, which tells you  whether you can locally embed your variety in affine space.   
This embedding dimension  for a local noetherian ring  $(A,{\mathfrak m},k)$ is defined as $$\mu(A)= dim_k({\mathfrak m}/{\mathfrak m^2})$$   
This number is the minimum number of generators for  ${\mathfrak m}$ .
 ( Atiyah-Macdonald,  Commutative Algebra  , Theorem 11.22)   
In your case $A= k[x,y,z]_{(x,y,z)}/I(X) \;$  and    ${\mathfrak m}=(\bar x,\bar y,\bar z) \; $   so that ${\mathfrak m}/{\mathfrak m^2}=k\bar x \oplus k\bar y \oplus k\bar z \;$ has dimension $3$ and thus $\mathfrak m$ cannot be generated by less than $3$ elements.
A: @Mohan In my opinion, Georges second answer cannot be topped by a more elementary argument. But facing the opposite direction, one may consider the problem as an entry point to the subject of Commutative Algebra and some more advanced methods:
If $I$ denotes an ideal in a Noetherian ring  $R$, then the ring $R/I$ is a complete intersection iff 
$$ht(I) =  \mu (I).$$ 
One may take this as definition. Here the corang $\mu (I)$ is the number of generators of $I$ while the height $ht(I)$ is the minimal codimension $codim_R (R/p)$ with respect to all prime factors $p$ of $I$.
In our case: $ht(I) = 3 - 1 = 2$ because each $R/p$ is the coordinate ring of one coordinate line. We want to prove $\mu(I) = 3$. Hence we claim: $R/I$ is not a complete intersection. 
Proof: It suffices to show that the localisation $A := (R/I)_{(\bar x,\bar y,\bar z)}$ is not a complete intersection. Whether a Noetherian local ring $A$ is a complete intersection can be checked by a homological criterion (Matsumura, Hideyuki: Commutative ring theory): $A$ is a complete intersection iff 
$$h_1(\mathfrak m, A) = emb \ dim A - dim A.$$
If ${\mathfrak m}=(\bar x,\bar y,\bar z) \subset A$ denotes the maximal ideal, then $A$ has embedding dimension $emb \ dim A := dim_k (\mathfrak m/\mathfrak m^2), k = A/\mathfrak m$. In our case $emb \ dim \ A = 3, dim \ A = 1$, see Georges first answer. 
The numerical value $h_1(\mathfrak m, A)$ is the $k$-dimension of the 1-st homology group of the Koszul complex of A with respect to $\mathfrak m$, see Matsumura, p. 127.
In our case a maximal basis of ${\mathfrak m}$ is the set $\{\bar x,\bar y,\bar z\}$. An explicit computation of the morphisms of the Koszul complex
$d_1: E \longrightarrow A$ and $d_2 : \Lambda^2(E) \longrightarrow E$ with $E := A^{\oplus 3}$ shows 
$$h_1(\mathfrak m, A) = dim_k \frac{ker \ d_1}{im \ d_2} = 6 - 3 = 3.$$
Hence 
$$3 = h_1(\mathfrak m, A) > emb \ dim A - dim A = 2, q.e.d.$$
