Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there any consensus on what the projective dimension of the zero module should be? Here are three statements one commonly encounters in textbooks, sometimes with or without the condition $M\neq 0$:

(1) $\mbox{pd}(M)\leq n$ iff $\mbox{Ext}^{n+1}(M,-)=0$

(2) $\mbox{pd}(M)=0$ iff $M$ is projective

(3) $\mbox{grade}(M):=\infty$ if $M=0$

If one attempts to define $\mbox{pd}((0))$ by extending one of these results, (1), (2), (3) suggest $\mbox{pd}=-1, 0, \infty$, respectively.

share|improve this question
I can't see the point in worrying about this. Is there any application where assigning the zero module a projective dimension is significant? –  Robin Chapman Sep 4 '10 at 17:21
For example, change of rings formulas like $\mbox{pd}_{A/(a)}(M/aM)\leq\mbox{pd}_{A}(M)$ would be false if one defined $\mbox{pd}(0)=\infty$; Take $A=\mathbb{Z}, a=3, M=\mathbb{A}/5$. –  ashpool Sep 4 '10 at 17:30
What is the dimension of the empty manifold? –  Tom Goodwillie Sep 4 '10 at 20:01
There are many, many more such questions that can be asked (and some already have been asked, cf mathoverflow.net/questions/31621): one can replace projective dimension with injective or free dimension; or ask about the Krull dimension of the trivial ring; or the dimension of the trivial and empty simplicial complexes (trival: the only simplex is the empty simplex; empty: the set of simplices is the empty set). But does this mean that they $\textit{should}$ be asked? Then I would prefer to have one such catch-all trivial exceptions question and get done with it for good. –  Victor Protsak Sep 4 '10 at 21:22
@Victor: I think that this question makes a lot more sense than a question about modules over the zero ring, so your comparing this to the question you linked is completely unfair. Modules over arbitrary rings can be zero without you knowing and so it makes sense to expect that applying a reasonable definition would give something reasonable for the zero module. –  Sándor Kovács Nov 20 '12 at 4:37

3 Answers 3

up vote 10 down vote accepted

Although I agree that one can easily decide to not worry about the case of the zero module, but as ashpool points out, it happens that sometimes we end up with the zero module whether we want or not and then each time we need to say (using ashpool's example) if $M/aM\neq 0$, then bluh and if $M/aM=0$ than something else happens.

So, I think there is actually something to be gained from making a definition that makes sense for the zero module (or the zero object in a more general situation). Of course, sometimes the definition that makes one (in)equality work does not work for another. However, one could still say in a paper (less likely in a book I suppose) that we are using the following definition for whatever which is the usual one if the object is not zero and gives this or that when it is zero and makes the following inequality work.

So having philosophized about this let me give a definition of projective dimension that gives $-\infty$ for the zero module.

Definition Let $(R,\mathfrak m,k)$ be a noetherian local ring and $M$ a finite $R$-module. Define the projective dimension of $M$ as $$ \mathrm{proj\\, dim}_R M:=\sup \left\{ i\in \mathbb{Z} \ \vert \ \mathrm{Ext}_R^i (M,k)\neq 0 \right\}, $$ where $\sup$ is taken in $\mathbb{Z}\cup\{\pm\\,\infty\}$.

This is actually essentially ashpool's definition (1), except that for $M=0$ it takes the $\sup$ of the empty set. (This may have been what samantha's professor told her). It also makes the change of rings formula to work.

In fact, I would argue that this is the "right" definition anyway, because the point is those Ext groups that are non-zero, not those that are.

Regarding adding the $\{\pm\\,\infty\}$ possibilities: We definitely need to allow $+\infty$, so it makes sense to allow $-\infty$ as well, especially because we need it for $M=0$.

Comment Of course one can start wondering what to do with non-local and/or non-noetherian rings, but I will leave that meditation to the reader.

share|improve this answer
What needs to be defined is the sup of a subset $Y$ of a given (totally) ordered set $X$. It depends on both $Y$ and $X$. Specifically, the sup of the empty subset of $\mathbb{N}$ is clearly $0$; the sup of the empty subset of $\mathbb{Z}\cup\{-\infty,+\infty\}$ is $-\infty$, and the sup of the empty subset of $\mathbb{Z}$ doesn't exist. –  Laurent Moret-Bailly Nov 20 '12 at 7:43
Dear Laurent, you are absolutely right, so I edited the definition. At the same time I would say that the essential part of your comment is whether one should allow negative numbers or not. One argument for allowing negative numbers is that that makes this definition give $-\infty$ for $M=0$ which seems to be the best choice. Another, perhaps better, argument is that if we view everything in the derived category, then we might encounter negative $i$'s such that $\mathrm{Ext}^i\neq 0$. So, in order to be consistent we have to allow $i$ to be negative. Then we obviously have to allow $-\infty$. –  Sándor Kovács Nov 20 '12 at 8:24
I would add that it is perhaps strictly speaking incorrect, but customary to allow $\pm\infty$ for values of $\sup$ and $\inf$ for a subset of $\mathbb{R}$ without explicitly saying that we extend the overset to include those. –  Sándor Kovács Nov 20 '12 at 8:26

I came online to ask this question myself, as it came up today during a reading course on homological algebra.

According to my professor, the zero module, even though it is trivially projective (as a trivial direct summand of any free module), does not have projective dimension 0, but rather $-\infty$. I cannot remember the precise reasoning he used. I believe it had something to do with the supremum of the empty set equaling negative infinity, although I don't remember how an empty set even came into the picture.

The issue had come up from one of his papers in which a theorem he stated would turn out to be false for the zero module, as a reader pointed out. He responded that the theorem held only for modules of finite projective dimension and as such the zero module would be excluded.

I also recognize that this question is over 2 years old, but for what it's worth, we have another candidate for the projective dimension of the zero module.

And while I personally despise treating trivial cases, I acknowledge that a really good definition should always be able to account for them, making this question somewhat worthwhile?

share|improve this answer
Personally, I'd like more details (e.g. how the empty set came into play). Could you tell us which of his papers you are referencing above? Perhaps the reasoning is in there for why $-\infty$ is the right answer. By the way, welcome to MathOverflow –  David White Nov 19 '12 at 21:29
I meant to say, the paper he may have been referring to was simply a paper on Auslander-Buchsbaum type formulas, and the setting was that of a commutative noetherian local ring ($R$, $m$, $k$) and the theorem is for $R$-modules $M$ of finite projective dimension , the projective dimension of M plus the depth of M is equal to the depth of R. The hypothesis has a footnote reading: We consider the zero module to be of projective dimension $\infty$ –  Samantha May 22 at 2:43

Let me explain a definition of projective dimension which gives the same result as the one given by Sándor Kovács, but without any restriction on the ring or the module we are talking about. This is, by the way, the one chosen by Bourbaki (A.X.8.1).

Let $A$ be a ring.

0) We write $\overline{\mathbb{Z}}=\mathbb{Z}\cup\{-\infty,\infty\}$ and furnish $\overline{\mathbb{Z}}$ with the ordering that extends the canonical ordering on $\mathbb{Z}$ and has $\infty$ as greatest and $-\infty$ as smallest element. We convene that suprema and infima of subsets of subsets of $\overline{\mathbb{Z}}$ are always understood to be taken in $\overline{\mathbb{Z}}$.

1) If $C$ is a complex of $A$-modules and $C_n$ denotes its component of degree $n\in\mathbb{Z}$, then we set $$b_d(C)=\inf\{n\in\mathbb{Z}\mid C_n\neq 0\}$$ and $$b_g(C)=\sup\{n\in\mathbb{Z}\mid C_n\neq 0\},$$ and we call $$l(C)=b_g(C)-b_d(C)$$ the length of $C$. Note that if $C$ is the zero complex then we have $b_d(C)=\infty$ and $b_g(C)=-\infty$, hence $l(C)=-\infty$.

2) If $M$ is an $A$-module and $(P,p)$ is a left resolution of $M$, then the length $l(P)$ of the complex $P$ is called the length of $(P,p)$. Note that if $P$ is the zero complex (which may be the case if and only if $M=0$) then the length of $(P,p)$ is $-\infty$.

3) If $M$ be an $A$-module, then the infimum of the lengths of all projective resolutions of $M$ is called the projective dimension of $M$. Hence, if $M=0$ then we have a projective resolution of length $-\infty$, and thus the projective dimension of $M$ is also $-\infty$. Conversely, if $M$ has projective dimension $-\infty$ then - since every $A$-module has a projective resolution - it necessarily has a projective resolution of length $-\infty$, and thus it follows $M=0$.

Note: This clearly makes sense in every abelian category with enough projectives, and there are obvious variants of the above that yield analogous definitions of injective or flat dimensions.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.