# If a commutative graded algebra is free over a graded subalgebra, then must it have a graded basis?

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?

Question 3. If yes: Does this still hold if $$\mathbf{k}$$ is a commutative ring rather than a field?

Question 4. If no, does it help to assume that $$A$$ is a graded subalgebra of a polynomial ring over $$\mathbf{k}$$?

Question 5. Does it help if $$A_0 \cong \mathbf{k}$$ and the $$\mathbf{k}$$-algebra $$A$$ is generated by $$A_1$$ ?

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, 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.

• Yes to 1, 2, 3, and you only need $A$ to be a $B$-module (not an algebra). See Bourbaki's Algèbre X (not yet translated, I am afraid), §8, Proposition 8. – abx Feb 21 at 21:02
• @abx: Nice reference, but I'm not quite seeing it. You're talking about Proposition 8 a) presumably, specifically the "projectif" track (as the "gradue libre" track would require more assumptions than I have), and probably the (ii) $\Longrightarrow$ (i) direction in particular. And I agree: (ii) follows from what is given. But does (i) directly imply that $M$ has an $A$-module basis consisting of homogeneous elements? – darij grinberg Feb 21 at 21:17
• The connected case (i.e. $B_0 = k$) is answered in math.stackexchange.com/questions/557402/…, where, in accordance with abx, it suffices that $A$ is a graded $B$-module. – tj_ Feb 21 at 23:51
• @darij grinberg: I had actually in mind the case $B_{0}=\mathbf{k}$. Then (i) tells us that our graded module is of the form $B\otimes _{\mathbf{k}}N$, where $N$ is a graded $\mathbf{k}$-vector space. Now just take a basis of each component of $N$. – abx Feb 22 at 6:26
• @abx: Q1doesn't hold in general: Let $B= k \oplus k$ (componentwise operations) be concentrated in degree zero and take as graded $B$-modul $A = A_0 \oplus A_1 := k \oplus k$ where $B$ acts on $A_i$ by projection on the first resp. second component. Then $A \cong B$ as $B$-modules, but there is no free base because $\dim_k A_i < \dim_k B$. – tj_ Feb 22 at 6:57

Q1: no (this makes Q2, Q3 obsolete)

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

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 $$B$$).

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.

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.

• Nice counterexample for Q1! (I would add that $B = k \times k$ and $A \cong B\left[\varepsilon\right]/\left(\varepsilon^2\right)$ as non-graded $k$-algebras in this example; this makes verifying things much easier.) I'll take a look at your positive claims tomorrow. When you say "homogeneous summands", you mean "homogeneous components", right? – darij grinberg Feb 22 at 7:06
• Yes, I use homogeneous summands as synonym for homogeneous components. In the notation I tried to express the key idea that $B$ is decomposable and to use the components as homogeneous components. But the other structures are surely easier to observe in your notation. – tj_ Feb 22 at 7:21
• Nice proof! Thanks a lot. So the proof definitely works when $B_0 \cong \mathbf{k}$ and every projective $B_0$-module is free. This covers all the examples I can name :) – darij grinberg Feb 22 at 18:15