Ideal Generated by Three Elements in Polynomial Ring 
How would one prove that the ideal $(xy,xz,yz)$ of $k[x,y,z]$ for some field $k$, cannot be generated by two polynomials. In other words, prove:
  $$(xy,xz,yz) \neq (f,g)\; \forall f,g \in k[x,y,z].$$

Furthermore, can somebody abstract this fact and prove it for $k[x_1,x_2,...,x_n]$? Maybe that the ideal of certain combinations of multiplied variables $x_i$ is not generated by less elements. Please ask if my question is unclear.
Edit: Specifically, I want to say that the ideal generated by all combinatorial products of $n-1$ variables in $k[x_1,x_2,...,x_n]$ cannot be generated by less elements. But if somebody has a different abstraction, please share.
 A: As I have already explained in the comments under this answer, if $I=(xy,xz,yz)$ then it's enough to check the dimension of the $k$-vector space $I/mI$, where $m=(x,y,z)$. It is not hard to show that the (residue classes of) $xy, yz,zx$ are linearly independent over $k$: if $axy+byz+czx\in mI$ with $a,b,c\in k$, then $a=b=c=0$ because $mI$ is a homogeneous ideal generated by homogeneous elements of degree $3$, while $axy+byz+czx$ would have degree $2$ if one of the elements $a,b,c$ is non-zero. Conclusion: $\dim_kI/mI=3$, so the minimal number of generators of $I$ is $3$.
Edit. One can also think like this: $I$ is generated by homogeneous polynomials of degree $2$. If $I_2$ is the set of homogeneous polynomials of degree $2$ in $I$, then $I_2$ is the  $k$-vector space $kxy\oplus kyz\oplus kzx$. On the other hand, if $I=(f,g)$, then the homogeneous components of $f$ and $g$ are of degree at least $2$ (because $f,g\in I$), so $I_2=kf_2+kg_2$. (Here $f_2$ and $g_2$ denote the homogeneous components of degree $2$ of $f$ and $g$.) This is clearly a contradiction.
A: This is the most straightforward version of the canonical proof (i.e. what user26857 posted) that I can think of.  I have run into problems, more than once, explaining this proof to skeptics, so hopefully this adds some clarity, rather than more confusion.

Let $R=k[x,y,z]$.
If $R_{\geq 2}$ is the ideal of all polynomials with lowest degree $\geq 2$, and $R_2$ is the $k$ vector space of all homogeneous polynomials with degree exactly $2$, then we can define a $k$-linear map $\pi: R_{\geq 2}\to R_2$ by throwing out the terms of degree $\geq 3$, e.g. $\pi(x^2 + xz + y^3) = x^2 + xz$.  It is not hard to see that this map is $k$-linear—it is merely a standard projection.
There is only one property of this projection that we really need: If $f\in R_{\geq 2}$, and $g\in R$ has constant term $g_0$, then $\pi(gf) = g_0 \pi(f)$.
(Proof: $(g-g_0)f$ has only terms of degree $\geq 3$, so $\pi((g-g_0)f) = 0$.  Therefore $\pi(cf + (g-g_0)f) = \pi(g_0f) + \pi((g-g_0)f) = g_0\pi(f)$.)
It follows immediately that if $I\subset R_{\geq 2}$ is an ideal, and $I$ is generated by $f_1,\cdots,f_n$, then $\pi(I)$ is generated by $\pi(f_1),\cdots,\pi(f_n)$ as a $k$-vector space.  Explicitly: $\pi(Rf_1 + \cdots + Rf_n) = k f_1 + \cdots + k f_n$.
In the particular case $I=(xy,xz,yz)$, we can easily see that $\pi(I)$, the homogeneous degree $2$ part of $I$, has dimension $3$ over $k$.  By the above, this means that $I$ cannot possibly be generated by $2$ elements, as they would map, under $\pi$, to a generating set that is too small.

In other words, this property that the minimal number of generators of an ideal $I$, generated by homogeneous polynomials of the same degree, equals $\dim_{R/\mathfrak{m}} I/\mathfrak{m} I = \dim_k I/\mathfrak{m} I$, where $\mathfrak{m} = (x,y,z)$, is an immediate and trivial statement once you are comfortable with this operation of throwing out higher-order terms.  And the above generalizes easily.
A: If $(xy,xz,yz)=(f,g) for some f,g∈k[x,y,z]$, $xy \in (xy,yz,xz) \Rightarrow either f=x+c$ or $g=x+c$ or $f=xy+c$ or $g=xy+c $.[the first two cases will be contradicted easily (check!!)].  WLOG let $f=xy+c$ Now take  $yz \in (xy,yz,xz)\Rightarrow g=z$ or $y$ or $yz$ (note if $c=yz$ & $g=xy$ then we will run our arguement with g in stead of f) [here also the first two cases will be contradicted easily], Now consider $xz \in (xy,yz,xz)$ but $xz \notin (xy,yz)$.
Contradiction.
Now extend this idea to n variable.
