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Jun 15, 2019 at 5:44 comment added keaine Nope it's ok, I found what I wanted. Thank you very much!
Jun 15, 2019 at 3:27 comment added keaine Thank you very much. I think I will need some more insight about this. Do you have any references?
Jun 15, 2019 at 3:02 comment added Jason Starr As a maximal ideal in a regular ring of dimensions $n$, also the maximal ideal is generated by a regular sequence of length $n$.
Jun 15, 2019 at 1:38 comment added keaine I don't quite understand where this goes. With what you said, we prove that our minimal set of homogeneous elements generates the ring. But I don't get why they are algebraically independent.
Jun 14, 2019 at 22:53 comment added Jason Starr This is a standard result included in commutative algebra textbooks. It suffices to prove that all homogeneous elements of the ring are contained in the subring generated by the ideal generators. Every positive-degree element of the ring is in the ideal, hence is a linear combination of ideal generators whose coefficients are homogeneous of strictly smaller degree. Now use the induction hypothesis.
Jun 14, 2019 at 21:13 comment added keaine I don't really understand why a minimal set of homogeneous generators for the ideal would be a minimal set of generators of the ring. Of course, it is a minimal set of homogeneous generators for the ring, but is it a minimal set of generators for the ring?
Jun 14, 2019 at 19:21 comment added Jason Starr If the ring is abstractly a polynomial ring, then it is a regular ring. Thus the maximal ideal generated by all homogeneous elements of positive degree is regular. A minimal set of homogeneous generators for this ideal is also a minimal set of generators of the ring as a $k$-algebra.
Jun 14, 2019 at 17:00 review First posts
Jun 14, 2019 at 18:44
Jun 14, 2019 at 16:56 history asked keaine CC BY-SA 4.0