
> 
> Fix a field $\mathbf{k}$ and an $\mathbb{N}$-graded commutative $\mathbf{k}$-algebra $A = \bigoplus\limits\_{n = 0}^{\infty} A\_n$ of finite type. ("Finite type" means that each $A\_n$ is a finite-dimensional $\mathbf{k}$-vector space. All algebras are understood to contain $1$.) Let $B = \bigoplus\limits\_{n = 0}^{\infty} B\_n$ be an $\mathbb{N}$-graded $\mathbf{k}$-subalgebra of $A$ (so that $B\_n \subseteq A\_n$ for all $n$). Assume that $A$ is free as a $B$-module.
> 
> 
> **Question 1.** Does it follow that the $B$-module $A$ has a basis consisting of homogeneous elements?
> 
> 
> **Question 2.** If yes: Can we omit the requirement that $A$ is of finite type?
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> **Question 3.** If yes: Does this still hold if $\mathbf{k}$ is a commutative ring rather than a field?
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> **Question 4.** If no, does it help to assume that $A$ is a graded subalgebra of a polynomial ring over $\mathbf{k}$?
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> **Question 5.** Does it help if $A\_0 \cong \mathbf{k}$ and the $\mathbf{k}$-algebra $A$ is generated by $A\_1$ ?
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> 


Sorry for the onslaught of questions -- I am hoping that an answer to one will likely solve most of the others, which is why I prefer not to split them across several topics.


The question cloud originates from working with Vic Reiner on [cyclic quasisymmetric functions](https://arxiv.org/abs/1811.05440), but I find it more fundamental. I originally thought these would be easy exercises in graded linear algebra (using projection maps, linear combinations and recurrent constructions), but I see no obvious point of vantage for such tactics.



If a commutative graded algebra is free over a graded subalgebra, then must it have a graded basis?
Q1: no (this makes Q2, Q3 obsolete)


Q4, Q5: yes (for $k$ a field)




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**Example for Q1:** Let $B = k \oplus k$ (componentwise operations) be concentrated in degree zero and take
$$A = A\_0 \oplus A\_1 \oplus A\_2 := B \oplus k \oplus k$$
$B$ acts on $A\_i=k$ via the projection $p\_i:B \twoheadrightarrow k$. This makes $B \cong k \oplus k$ as $B$-modules. Make $A$ into a graded ring by letting the products of positive degree be zero. As $B$-module, $A\cong B \oplus B$ is free. But it has no homogeneous base (otherwise the homogeneous components would be direct sums of $B$).




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In general we can say:



> 
> Let $B$ be a non-negatively graded ring (not necessarily commutative) such that each projective $B\_0$-module is free. Then each bounded below, graded $B$-module $M$ which is free as $B$-module has a homogenous base.
> 
> 
> If $M$ is of finite type, it's enough that each *finitely generated* projective $B\_0$-module is free.
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> 
> 


Sketch of proof: Let $B\_+ = \oplus\_{n > 0}B\_n$ be the irrelevant ideal. Since $M$ is free as $B$-module, $M/B\_+M$ is free as $B\_0$-module. It's also a graded $B\_0$-module. Hence the homogeneous summands of $M/B\_+M$ are direct summands of a free $B\_0$-module and hence projective. By the assumptions on $B\_0$ the homogeneous summands are free $B\_0$-modules. Now the result follows verbatim from Manny Reyes' answer in <https://math.stackexchange.com/questions/557402/graded-free-is-stronger-than-graded-and-free>. I should add that the projective-free argument is due to Eric Wofsey taken from the same link. q.e.d.


In Q4, Q5 we have $A\_0 = B\_0 = k$ so the result applies, if $k$ is a field.


