Every ideal of $K[x_1,\ldots,x_n]$ has $\leq n$ generators? Is this true: Every ideal of $K[x_1,\ldots,x_n]$ is generated by some subset with $\leq n$ elements?

It is true when $n=1$, since $K[x]$ is a PID.

I'm trying to prove it is not true for $n\geq2$, via the example $I:=\langle x^2,xy,y^2\rangle\unlhd K[x,y]$.

Does the SINGULAR code below confirm that $I$ is not generated by $1$ or $2$ polynomials?

ring R=0,(x,y),ls;
ideal I=x2,xy,y2;
minbase(I);


From the SINGULAR manual: 


  5.1.76 minbase
  Syntax:  minbase ( ideal_expression )  minbase ( module_expression )
  Type:  the same as the type of the argument
  Purpose:  returns a minimal set of generators of an ideal, resp. module, if the input is either homogeneous or if the ordering is local.


What does a "minimal set of generators" mean? Minimal w.r.t. cardinality, or w.r.t. $\subseteq$?
Expanding on my comment above:

Let $J=\langle x^3,x^2y,xy^2,y^3\rangle$.  Since $J\subset I$, if $I$ is generated by two elements in $K[x,y]$ then its image, $I&#39;\subset K[x,y]/J$, is generated by two elements as a $K[x,y]$-module.

But $I&#39;$ in $K[x,y]/J$ is just $\{ax^2+bxy+cy^2: a,b,c\in K\}$.  As a $K[x,y]$-module, $x$ and $y$ act as zero on $I&#39;$, and, as a vector space over $K$, $I&#39;$ is 3-dimensional.  So there cannot be two generators for $I&#39;$, and hence there cannot be two generators in $K[x,y]$.

In general for a graded ring, $R=\oplus_{i\geq 0} R_i$ with $R_0=K$ a field, any generating set for the ideal $I_j=\oplus_{i\geq j} R_i$ has at least as many elements as the dimension of $R_j$ as vector space over $K$.
Edit: I finally remembered how to do algebra, I hope. Below is a completely different answer than my original (and hopefully a correct one).

Your example works just fine. In fact, if we let $R = K[x_1,\ldots,x_n]$ and $\mathfrak{m}=\langle x_1,\ldots,x_n\rangle$ (which is maximal) then we get that $\mathfrak{m}^2$ requires more than $n$ generators for any $n\geq2$. To see this, note that we do not change the generators by localizing at $\mathfrak{m}$, so we can assume $\mathfrak{m}^2 = \langle f_{1}\ldots f_{n}\rangle$ for $f_{1},\ldots,f_{n}\in R_\mathfrak{m}$ to derive a contradiction. Any polynomial $p\in \mathfrak{m}^2$ must have degree at least $2$, so each $f_{1},\ldots,f_{n}$ must have degree at least $2$ and any linear combination (over $R_\mathfrak{m}$) of these with coefficients outside $K$ must have degree more than $2$, thus of the $\binom{n+1}{2}$ monomials of degree $2$ or less, at most $n$ are contained in $\mathfrak{m}$. Hence $R_\mathfrak{m}/\mathfrak{m}^2$ has dimension at least $\binom{n+1}{2}-n$. Yet $K\cong (R_\mathfrak{m}/\mathfrak{m}^2)/(\mathfrak{m}/\mathfrak{m}^2)$ has dimension $1$ and $\mathfrak{m}/\mathfrak{m}^2$ has dimension $n$ because $R_\mathfrak{m}$ is a regular local ring, hence $R_\mathfrak{m}/\mathfrak{m}^2$ has dimension $n$. But for $n\geq 2$ we have $\binom{n+1}{2}-n&gt;n$, a contradiction, hence $\mathfrak{m}^2$ has more than $n$ generators. 

I believe this can actually be generalized further to $\mathfrak{m}^k$ for all $k\geq 2$, but this requires computing the dimension of $\mathfrak{m}^k/\mathfrak{m}$ inductively using the exact dimensions of $R/\mathfrak{m}^j$ for $j &lt; k$ in order to establish the desired inequality, which seems a pain at best.