Welcome to MathOverflow

Question

0

Let $I$ be a homogeneous ideal in a graded commutative ring $R$, $S$ be its minimal system of generators.

What is the conclusion that we can say about the element in $S$ ? Is the cardinality of $S$ uniquely determined by $I$ ?

In the book Commutative ring theory of Matsumura, theorem 2.3, page 8 there is a theorem for local ring which we can deduce that the number of generator is unique. So, is there a version of that theorem for the graded ring ?

What about the degree of generator in $S$ ?

edited Oct 8 at 4:30

asked Oct 6 at 2:47

Knot
81●5

2

There's definitely more than one choice of generators (since $\langle x,y\rangle = \langle x, y, x+y, x^2, \dots \rangle$. Do you mean a minimal choice of generators? – Karl Schwede Oct 6 at 13:02

4

I looked up Matsumura page 8 so I think I see what you are asking, you should try googling Graded Nakayama's Lemma. Everything you want is true if you assume that $R$ is $\mathbb{N}$-graded with $R_0$ a field. In particular the same argument goes through from the local case (page 9 of Matsumura). – Karl Schwede Oct 6 at 13:43

Yep, thank you all very much for pointing it out. I mean the minimal set of generator. – Knot Oct 8 at 4:29

Youngsu · Answer 1 · 2012-10-08 02:06:09Z UTC

2

Also, Proposition 1.5.15 in Cohen-Macaulay rings by Bruns and Herzog might help you. This is a bit more general than what Karl mentioned above.

answered Oct 8 at 2:06

Youngsu
212●7

Thank you very much, Youngsu. So, what about the degree of the elements in the minimal set of generators ? – Knot Oct 8 at 4:50

Presumably your generators are homogeneous? – Wilberd van der Kallen Oct 8 at 7:39

If one applies the positive answer to your original question to truncated versions of the ring, one should get the degrees also. Here by a truncated version I mean that you factor out all elements of degree greater than some fixed number. – Wilberd van der Kallen Oct 8 at 7:46

Oh, it was in the definition of a homogeneous ideal : A homogeneous ideal $I$ is homogeneous if it is generated by homogeneneous element. – Knot Oct 8 at 8:12

2

Well, I was in the setting of Karl, not Proposition 1.5.15 of Bruns and Herzog. Actually B & H tell on the same page that if one wants the degrees unique one should restrict to cases like Karl considers. (They call the degrees $\beta_{0i}$.) But their argument is a bit sophisticated. I was simply observing that if one factors $R$ by the ideal $J$ generated by all elements of degree at least $N$, then the minimal system $S$ maps to a minimal system plus zeroes. Now just watch how many generators become zero as $N$ varies. That tells you how many generators there were in each degree. – Wilberd van der Kallen Oct 8 at 12:42

show 3 more comments

YACP · Answer 2 · 2012-11-03 20:09:55Z UTC

1

I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.

A more general answer says the following: if $K$ is a field, $R$ is an $\mathbb{N}$-graded $K$-algebra finitely generated over $K$, and $M$ a $\mathbb{Z}$-graded finitely generated $R$-module, then

$$\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)_j,$$

where $\beta_{ij}(M)$ are the graded Betti numbers of $M$.

edited Nov 3 at 20:09

YACP

community wiki

9 revisions

	Dear @navigetor23, it might be helpful if you would specify what kind of graduations (positive,...) you consider, and which kind of finiteness hypotheses you make. (I guess you need $R$ to be coherent and $M$ to be of finite presentation.) – Fred Rohrer Oct 18 at 12:06
	Dear @navigetor23, you have to have at least 50 reputation to leave comments I believe. – Karl Schwede Oct 18 at 13:34
	Dear @navigetor23, this might be a reasonable abuse of language within a certain book or even community, but in general it is just totally unclear and moreover prone to mistakes. – Fred Rohrer Oct 18 at 14:57
	@navigetor23: Bruns-Herzog, Def. 1.5.1: "A graded ring is a ring together with a decomposition $R=\bigoplus_{i \in \mathbb{Z}}R_i$ such that $R_iR_j\subset R_{i+j}$ for all $i,j \in \mathbb{Z}$." Therefore I doubt that your "kind of standard terminology in commutative algebra" is actually standard. – Ralph Oct 18 at 18:09
	Isn't a graded $K$-algebra also a graded ring ? -:) I actually find this "controvery" useful, as some readers might recognize that using a short additional attribute like "positively graded" or "$\mathbb{N}$-graded" (in contrast to a single "graded") helps to avoid confison. BTW: That's exactly what Bruns does in the liked paper. – Ralph Oct 20 at 9:50

Homogeneous ideal and its system of generators

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.