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Oct 9, 2012 at 9:02 comment added Wilberd van der Kallen That means using the map $R\to R/J$.
Oct 8, 2012 at 16:52 comment added Knot I am so sorry because I am not an expert in commutative algebra. What do you mean by factoring $R$ by the ideal $J$ generated by all elements of degrees at least $N$ ?
Oct 8, 2012 at 12:42 comment added Wilberd van der Kallen 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.
Oct 8, 2012 at 8:13 comment added Knot @Wilberd van der Kallen : Sorry, Could you please make it more precisely ?
Oct 8, 2012 at 8:12 comment added Knot Oh, it was in the definition of a homogeneous ideal : A homogeneous ideal $I$ is homogeneous if it is generated by homogeneneous element.
Oct 8, 2012 at 7:46 comment added Wilberd van der Kallen 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.
Oct 8, 2012 at 7:39 comment added Wilberd van der Kallen Presumably your generators are homogeneous?
Oct 8, 2012 at 4:50 comment added Knot Thank you very much, Youngsu. So, what about the degree of the elements in the minimal set of generators ?
Oct 8, 2012 at 2:06 history answered Youngsu CC BY-SA 3.0