Let $k$ be an algebraically closed field of characteristic $0$ and $A=k[X_1, \ldots, X_n]$ with the grading induced by the total degree. Let $B$ be a graded $k$-subalgebra of $A$, ie, if $(A_k)$ is the grading of $A$, then $(B \cap A_k)$ is the grading of $B$. Suppose $B$ is polynomial, ie freely generated by some $b_1, \ldots, b_r$. Can $B$ be always freely generated by homogeneous elements?
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$\begingroup$ 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. $\endgroup$– Jason StarrCommented Jun 14, 2019 at 19:21
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$\begingroup$ 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? $\endgroup$– keaineCommented Jun 14, 2019 at 21:13
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$\begingroup$ 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. $\endgroup$– Jason StarrCommented Jun 14, 2019 at 22:53
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$\begingroup$ 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. $\endgroup$– keaineCommented Jun 15, 2019 at 1:38
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$\begingroup$ As a maximal ideal in a regular ring of dimensions $n$, also the maximal ideal is generated by a regular sequence of length $n$. $\endgroup$– Jason StarrCommented Jun 15, 2019 at 3:02
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