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 Starr Jun 14 at 19:21

$\begingroup$ 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? $\endgroup$ – keaine Jun 14 at 21:13

$\begingroup$ 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 positivedegree 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 Starr Jun 14 at 22:53

$\begingroup$ 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. $\endgroup$ – keaine Jun 15 at 1:38

$\begingroup$ 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 Starr Jun 15 at 3:02