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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.

Community
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    It almost never holds. Twist the grading by any (non-graded) automorphism. – Martin Brandenburg Nov 09 '13 at 21:28
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    @Martin Brandenburg: What do you mean by "almost never" ? Is there a precise measure for this quantity in the current context? – tj_ Nov 14 '13 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? – Ben Blum-Smith Dec 11 '15 at 23:28

2 Answers2

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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?

Manny Reyes
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  • 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 $L\subseteq R_+L$. – Eric Wofsey Dec 01 '15 at 01:23
  • @MannyReyes - does "graded vector space" just mean a vector space with a distinguished direct sum decomposition indexed by $\mathbb{Z}$? – Ben Blum-Smith Dec 12 '15 at 01:20
  • @BenBlum-Smith yes that's right. Equivalently, it is a graded module over the field $K$, the latter viewed as a graded algebra concentrated in degree zero. – Manny Reyes Dec 12 '15 at 14:16
  • @MannyReyes - thank you! Two more questions for you if you don't mind. (1) do you know if the terminology connected as you're using it is related to usage I've heard before that the Spec is connected as a top. space? I can't see how it would be related. (2) I'm trying to understand to what extent this proof still works if $R_0$ is a p.i.d. rather than a field. The use I can see of the fieldness of $R_0$ is to guarantee that $F/R_+F$ is free as a module over $R_0$, right? – Ben Blum-Smith Dec 14 '15 at 01:30
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    @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 $R_0=K$. – Eric Wofsey Dec 26 '15 at 02:57
  • @EricWofsey - Thank you, got it. – Ben Blum-Smith Dec 27 '15 at 03:49
  • @BenBlum-Smith I think I agree with what you say in (2), that $R_0$ being a field is useful because it guarantees $F/R_+ F$ is free over $R_0$. And thanks to Eric Wofsey for addressing question (1). I would also point out that if $R_0$ has no idempotents (as in the case when $R_0 = K$), then $R$ is "graded-connected" in the sense that there is no graded isomorphism $R \cong R_1 \times R_2$ for graded rings $R_i$. – Manny Reyes Dec 30 '15 at 17:40
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    @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. – Ben Blum-Smith Feb 27 '17 at 13:02
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    @MannyReyes : how do you say that $\phi$ is surjective and $\ker \phi \subseteq R_+G$ ? – user Jun 23 '17 at 14:05
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    @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: https://en.wikipedia.org/wiki/Nakayama%27s_lemma#Module_epimorphisms – Manny Reyes Jun 23 '17 at 15:43
  • @MannyReyes : By $R(l_i)$ where $l_i :=\deg x_i$ , you mean the homogenous part of $R$ in which $x_i$ lies , right ? And what do you mean by " the $i$-th basis element " ? – user Jun 23 '17 at 15:51
  • @users The notation $R(l)$ refers to the "twist" of the graded module $R$, so that the graded parts become $R(l)i = R{l+i}$. Your comment makes me realize that I meant $R(-l_i)$, which I'll correct shortly. Now the $i$th basis element of $\bigoplus R(-l_i)$ that I referred to is just the identity of the direct summand $R(-l_i)$, which lies in degree $l_i$. For the definition of the twist, see: http://stacks.math.columbia.edu/tag/00JL – Manny Reyes Jun 23 '17 at 18:34
  • @MannyReyes You say that you have the isomorphism between $G/R_+G$ and $F/R_+F$, but how do you prove that? You need to show that the kernel of the map $G \to F \to F/R_+F$ is $G/R_+G$ -- why is that? And if you don't mind answering, why does $L \subseteq R_+F \oplus R_+L$ imply $L \subseteq R_+L$? Thank you! – windsheaf Aug 14 '19 at 08:13
  • @windsheaf - if helpful: $L\subset R_+F \oplus R_+L$ implies $L\subset R_+L$ because $L$ intersects $F$ only trivially (because $G = F\oplus L$ to begin with), thus $L$ intersects $R_+F\subset F$ only trivially. For the other question, note that in the answer itself Manny is arguing the stronger claim that the map from $G$ to $F$ is an isomorphism. The proof goes through the argument that $G \rightarrow F/R_+F$ is a surjection as an intermediate step, and uses the graded Nakayama lemma to conclude that $G\rightarrow F$ is surjective. The rest argues that $G\rightarrow F$ is injective. – Ben Blum-Smith Apr 09 '21 at 21:09
  • cont'd: (This involves the thing about $L\subset R_+L$, followed by a second application of Nakayama.) Anyway it all arrives at the conclusion that $G\rightarrow F$ is already an isomorphism. From this it follows that $G/R_+G \rightarrow F/R_+F$ is an isomorphism, but that is not really its own step in the argument; it is moreso a guiding intuition. The point is that we built $G$ as a free $R$-module with a basis designed to map to the lifts of a basis of $F/R_+F$ to $F$. Cont'd: – Ben Blum-Smith Apr 09 '21 at 21:11
  • Let me know if this was helpful -- I can try to explain more clearly. – Ben Blum-Smith Apr 09 '21 at 21:17
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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.

user26857
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  • 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 $R=K$). – Eric Wofsey Nov 23 '15 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). – Eric Wofsey Nov 23 '15 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 – Jhon Doe Feb 18 '22 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 $n\geq r$. – Jhon Doe Feb 18 '22 at 15:52