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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. $\subseteq$?

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"Minimal set of generators" usually means minimal with respect to inclusion (no proper subset generates). To refer to a set of generators that is as small as possible with respect to cardinality, one usually speaks of a "generating set of minimal size/cardinality". Of course, a generating set of minimal cardinality will necessarily be a minimal set of generators, but the converse need not hold. –  Arturo Magidin Dec 29 '11 at 18:41
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If I understand the description of minbase, no. –  Arturo Magidin Dec 29 '11 at 18:49
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(x,y)^2/(x,y)^3 is a 3-dimensional K-vector space on which x and y act as 0, so an R-module generating set must also be a K-module generating set. –  Jack Schmidt Dec 29 '11 at 18:53
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@ArturoMagidin - minimal cardinality does not clearly mean minimal unless minimal cardinality is finite, of course. –  Thomas Andrews Dec 29 '11 at 19:00
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Note that any element of $I$ must necessarily have no terms of degree less than $2$. Now, look at the image of $I$ in $K[x,y]/<x^3,x^2y,y^2x,y^3>$. The image of $I$ in this ring is a vector space over $K$ of dimension $3$, so it cannot be generated by fewer than $3$ elements in this ring, and hence not in the parent ring. (This argument needs refinement, but is basically correct.) –  Thomas Andrews Dec 29 '11 at 19:09

2 Answers 2

up vote 7 down vote accepted

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 $K$.

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That's a tricky solution. The key to notice (for me) was that since multiplying by $x$ and $y$ gives $0$, a $K[x,y]$-module generating subset is also a $K$-module generating subset. Thank you very much. –  Leon Dec 30 '11 at 17:38

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 $\binom{n+1}{2}-n>n$, 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 $j < k$ in order to establish the desired inequality, which seems a pain at best.

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< and > are math operators, so the spacing given around them is that of operators. It's better to use \langle and \rangle, which are delimiters. –  Arturo Magidin Dec 29 '11 at 19:07
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Why $x^2=hf$? Don't we have $x^2=af+bg$ for some $a(x,y),b(x,y)$? –  Leon Dec 29 '11 at 19:09
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@LeonLampret Brain fart on my part, editing. –  Alex Becker Dec 29 '11 at 19:23
    
This "answer" is completely wrong, so I'll give it a downvote. (Actually it is full of mistakes like $m^2/m$ which has no meaning.) –  user26857 Nov 3 '12 at 17:35
    
@navigetor23 I apologize, I meant $\mathfrak m/\mathfrak m^2$ (although I think your comment is a little uncharitable, as I would consider that an obvious typo since I belive my intention is clear). If there are any mathematical mistakes, please list them. –  Alex Becker Nov 5 '12 at 14:48

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