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. $\subseteq$?

Solution 1:

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, $I'\subset K[x,y]/J$, is generated by two elements as a $K[x,y]$-module.

But $I'$ 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 $I'$, and, as a vector space over $K$, $I'$ is 3-dimensional. So there cannot be two generators for $I'$, 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 $K$.