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Hilbert basis theorem states that any ideal of the polynomial ring $k[x_1,x_2,\ldots, x_n]$ is finitely generated. A natural question is then, for given ideal, is the number of generators fixed, or it can vary although always finite?

Of course, you can add an arbitrary element in the ideal into the generator list. But what I mean is the lower bound of the number of generators.

kaiser
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1 Answers1

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It can vary. It can even be arbitrarily large. For example, consider the ring $k[x,y]$ and the ideal $(x,y)^n$. This ideal is generated by all monomials of the form $x^iy^j$ with $i+j=n$. There are $n+1$ of these.

ADDENDUM in response to your edit:

Ah, you are talking about a fixed specific ideal.

Now there are two things you could mean:

(1) Given an ideal $I$ in a noetherian ring $R$, do all generating sets for $I$ that are of minimal cardinality have the same cardinality?

(2) Given an ideal $I$ in a noetherian ring $R$, do all generating sets of $I$ that are minimal as generating sets (i.e. minimal in the sense that if an element is dropped from one of them it will no longer generate $I$) have the same cardinality?

Based on your comments (ed: now deleted), (1) is the one I think you mean. This is true, and doesn't depend on the ring being noetherian; and is true for sort of a vacuous reason. If an ideal has a generating set of cardinality $n$ and no generating set of cardinality lower than $n$, then $n$ is the minimal cardinality of a generating set (by definition!). Thus any generating set of minimal cardinality has this same cardinality $n$.

This may feel like I haven't said anything. The underlying mathematical substance is the fact that the set of natural numbers is well-ordered; any set of natural numbers has a least element. This is why it even makes sense to talk about a minimal-cardinality generating set. But once it makes sense to talk about this, then all generating sets realizing this minimal cardinality must have the same cardinality, because it is this minimal one.

(2) is actually false, even in $\mathbb{Z}$. For example, the ideal $(2)$ equals the ideal $(6,8)$. But the generating set $\{6,8\}$ for this ideal is minimal in the sense that if either element is dropped, the ideal becomes smaller.

Ben Blum-Smith
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  • I mean a fixed ideal. For your example, I guess the minimal number of generators is $n+1$, right? – kaiser Nov 10 '15 at 10:35
  • OH you are asking if all minimal sets of ideal generators for a given fixed ideal have the same cardinality? – Ben Blum-Smith Nov 11 '15 at 14:10
  • yes， exactly! I should have used the terminology cardinality. – kaiser Nov 11 '15 at 14:31
  • What is it supposed to mean in (1) "generating sets ... that are of minimal cardinality"? Is ${1}$ a generating set of minimal cardinality for the ideal $\mathbb Z$ of the ring $\mathbb Z$? What about ${2,3}$? (Anyway, for me the edited question has no meaning.) – user26857 Nov 11 '15 at 17:02
  • @user26857 - I literally meant "minimal cardinality," as in, there exists a generating set of this cardinality and there does not exist a generating set of lower cardinality. In this sense ${1}$ is minimal cardinality for the ideal $\mathbb{Z}$ in $\mathbb{Z}$ but ${2,3}$ is not, as $2>1$. On the basis of the comment discussion (which now seems to be deleted) I had the sense that this is actually the question the OP was asking. – Ben Blum-Smith Nov 12 '15 at 00:10