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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 $S$ ?

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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 '12 at 13:02
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 '12 at 13:43
Yep, thank you all very much for pointing it out. I mean the minimal set of generator. – Knot Oct 8 '12 at 4:29

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.

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Thank you very much, Youngsu. So, what about the degree of the elements in the minimal set of generators ? – Knot Oct 8 '12 at 4:50
Presumably your generators are homogeneous? – Wilberd van der Kallen Oct 8 '12 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 '12 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 '12 at 8:12
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 '12 at 12:42

I gave a short answer to this question here:

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


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

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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 '12 at 12:06
Dear @navigetor23, you have to have at least 50 reputation to leave comments I believe. – Karl Schwede Oct 18 '12 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 '12 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 '12 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 '12 at 9:50

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