This topic suggested me the following question:

If $R$ is a commutative graded ring and $F$ a graded $R$-module which is free, then can we conclude that $F$ has a homogeneous basis (that is, a basis consisting of homogeneous elements)?

In general the answer is negative, and such an example can be found in Nastasescu, Van Oystaeyen, Methods of Graded Rings, page 21. But their example is not quite usual, and

I'd like to know if however the property holds for positively graded $K$-algebras, for example.

If not, then maybe someone can provide a generic example when the property holds.

• It almost never holds. Twist the grading by any (non-graded) automorphism. Commented Nov 9, 2013 at 21:28
• @Martin Brandenburg: What do you mean by "almost never" ? Is there a precise measure for this quantity in the current context?
– tj_
Commented Nov 14, 2013 at 10:33
• @MartinBrandenburg - I'm not quite getting your suggestion, can you elaborate? What's $F$? Automorphism of $R$ as a ring or $R$ as an $F$-module? And what do you use the automorphism for? To define the action of $R$ on $F$? Please clarify? Commented Dec 11, 2015 at 23:28

It is not clear to me in the previous answer what assumptions are being made on the $\mathbb{N}$-graded algebra $R$ (particularly, what assumptions are made on $R_0$), but I'll try to be very careful here. Suppose that $R = \bigoplus_{n=0}^\infty R_n$ and that $R$ is connected, meaning that $R_0 = K$. Let's say that a graded $R$-module $M = \bigoplus_{n \in \mathbb{Z}} M_n$ is bounded below if $M_n = 0$ for $n \ll 0$.

Then I claim that if $R$ is connected, then every bounded-below graded $R$-module that is free has a homogeneous basis.

This can be deduced, even in a noncommutative setting, in the very informative Section 2 of the paper The structure of AS regular algebras by Minamoto and Mori, especially Lemma 2.6. (In fact, one can try to drop further assumptions; I suspect that it's enough to assume $R_0$ is a perfect local ring, not necessarily commutative, in which case every bounded-below graded projective left module would probably be free with a homogeneous basis.)

I'll sketch an argument here; it's essentially the same as the one in the answer already given by user26857. The key observation is that the graded Nakayama's lemma works for bounded-below modules, of which finitely generated graded modules form a special case.

Proof of claim: Supposing that $F$ is graded, bounded-below, and free, fix a set of homogeneous elements $\{x_i\} \subseteq F$ whose images $\overline{x_i} \in F/R_+ F$ form a homogeneous basis for the graded vector space. We get a graded map from a bounded-below graded free module $\phi \colon G = \bigoplus R(-l_i) \to F$, where each $l_i = \deg(x_i)$, sending the $i$th basis element to $x_i$. Nakayama implies that this is a "minimal projective cover," in the sense that $\phi$ is surjective and $\ker(\phi) \subseteq R_+ G$. Since $F$ is free we have $G \cong F \oplus L$ for the graded submodule $L = \ker(\phi) \subseteq G$. But now $$L = \ker(\phi) \subseteq R_+ G = R_+(F \oplus L) = R_+ F \oplus R_+ L.$$ Inspecting this direct sum decomposition, we must in fact have $L \subseteq R_+ L$. Nakayama now implies that $\ker(\phi) = L = 0$. Thus $F \cong G$ as graded modules, so that $F$ has a homogeneous basis. QED

What I would really like to know is what happens if one drops the bounded-below assumption. Is there an example of a graded module over a connected graded (commutative) ring $R$ that is not bounded-below, and which does not have a homogeneous basis?

• Thanks for the answer! I think the key step I was missing when thinking about this before is the fact that $L$ is a direct summand of $G$, so $L\subseteq R_+G$ implies $L\subseteq R_+L$. Commented Dec 1, 2015 at 1:23
• @BenBlum-Smith: The term "connected" comes rather from algebraic topology. If $R$ is the cohomology graded ring of a space $X$ with coefficients in $K$, then X is path-connected iff $R_0=K$. Commented Dec 26, 2015 at 2:57
• @MannyReyes - the situation is even better than that. If $R_0$ has connected spectrum, then $R$ has connected spectrum. (I.e. no need to distinguish a concept of "graded-connected" from "connected" in the sense of connected spectrum.) Pf: Let $x = x_0 + \dots + x_n$ be an idempotent, split into homogeneous components. Looking component by component at $x^2 = x$ lets you conclude first that $x_0=0$ or $1$ and then by induction (starting with $i=1$) that $x_i=0$ for $i > 0$. Thus a graded connected $A$-algebra is connected in both senses, if $A$ is connected. Commented Feb 27, 2017 at 13:02
• @MannyReyes : how do you say that $\phi$ is surjective and $\ker \phi \subseteq R_+G$ ?
– user
Commented Jun 23, 2017 at 14:05
• @users - The homomorphism is constructed to induce an isomorphism between $G/R_+ G$ and $F/R_+ F$, so its kernel must be contained in $R_+ G$. The fact a homomorphism inducing a surjective map $G/R_+ G \to F/R_+ F$ must itself be surjective is a common corollary of Nakayama's Lemma; see the "ungraded" version here: en.wikipedia.org/wiki/Nakayama%27s_lemma#Module_epimorphisms Commented Jun 23, 2017 at 15:43

Only a partial result for now.

If $R$ is an $\mathbb N$-graded $K$-algebra and $F$ is a $\mathbb Z$-graded $R$-module which is free of finite rank, then $F$ is gr-free.

Let $x_1,\dots,x_n$ be a minimal system of homogeneous generators for $F$. Then their images $\bar x_1,\dots,\bar x_n$ in $F/R_+F$ generate the $K$-vector space $F/R_+F$, and therefore we can suppose that $\bar x_1,\dots,\bar x_m$, $m\le n$, form a $K$-basis in $F/R_+F$. Thus we have $F=R_+F+\langle x_1,\dots,x_m\rangle$, and the graded Nakayama's lemma gives us $F=\langle x_1,\dots,x_m\rangle$ hence $m=n$.
This shows that $\dim_KF/R_+F=n$, so $n=\operatorname{rank}F$. But a system of generators consisting of $\operatorname{rank}F$ elements is necessarily a basis, and we are done.

• For a bit of perspective, the key way in which $K$ being a field is used here is fact that every graded $K$-vector space (in particular, $F/R_+F$) has a homogeneous basis (i.e., that the result holds for $R=K$). Commented Nov 23, 2015 at 21:16
• More generally, the same argument works whenever $K$ is a ring with invariant basis number such that every f.g. projective $K$-module is free (since the graded parts of $F/R_+F$ are direct summands of the free f.g. $K$-module $F/R_+F$ and hence f.g. projective). Commented Nov 23, 2015 at 21:23
• @EricWofsey Hello could you please explain the implication: "This shows that $\dim_KF/R_+F=n$, so $n=\operatorname{rank}F$". I dont see how the dimension of vector space implies the rank of $F$. Regards Commented Feb 18, 2022 at 10:46
• Ah is see. Let $r$=rank(F). Then $r\geq n$ (i.e the basis elements will generate $F/R_+F$ as a $K$-vector space.). Since $x_1,..,x_n$ generate $F$ this implies there exists surjective module homomorphism from $R^n$ to $R^r$ which implies $n\geq r$. Commented Feb 18, 2022 at 15:52