MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Atiyah and Bott's paper "The Moment Map and Equivariant Cohomology", they say that for any exact sequence of modules over $\mathbb{C}[u_1,...,u_l]$ $$D \to E \to F,$$ we have that

Supp $E \subset$ Supp $D \cup$ Supp $F$,

where Supp$E$ is defined to be $\cap V_f$ where $f$ runs over all polynomials such that $fE=0$. This must be quite elementary, but I can't manage to prove it.

I tried to show that if some $f$ annihilates $E$, then it anihilates either $F$ or $D$ (which, if I'm not mistaken, implies what we want), but I didn't manage this either. Is it too strong ? Is the fact that the ground ring is $\mathbb{C}[u_1,...,u_l]$ that important ?

share|cite|improve this question
What is $V_f$ ? – Ralph Apr 24 '13 at 14:24
Have a look at Serre's book Local algebra. – Liviu Nicolaescu Apr 24 '13 at 14:28
$V_f=f^{-1}(0)$. – Liviu Nicolaescu Apr 24 '13 at 15:39
You can find this in most standard books in Commutative Algebra. – Mahdi Majidi-Zolbanin Apr 24 '13 at 17:22
up vote 3 down vote accepted

If I understand the definitions properly, the following should work:

Let $D \xrightarrow{\alpha} E \xrightarrow{\beta} F$ be exact and suppose $\text{Supp} E \not\subseteq \text{Supp}D \cup \text{Supp}F$. Hence there is $a \in \text{Supp} E$ such that $a \notin \text{Supp}D$ and $a \notin \text{Supp} F$. This means:

  1. $\forall f: f\cdot E = 0 \Rightarrow f(a)=0$
  2. $\exists g: g \cdot D=0 \wedge g(a)\neq 0$
  3. $\exists h: h \cdot F=0 \wedge h(a)\neq 0$

For $x\in E$ we have $\beta(hx)=h\beta(x)=0$, i.e. $hx\in \ker \beta=\text{im}\;\alpha$. Thus there is $y\in D$ such that $hx=\alpha(y)$. So $ghx=g\alpha(y)=\alpha(gy)=\alpha(0)=0$ and consequently $ghE=0$. Now, by 1., $0=(gh)(a)=g(a)h(a)$, contradicting 2. and 3. q.e.d.

share|cite|improve this answer

This is more or less a classical result but I believe some comments are warranted. Atiyah and Bott are first and foremost geometers/topologists and they opted for a more geometric description of the concept of support better adapted to the purposes of that (wonderful) paper.

One can define the support of a module over an arbitrary ring $R$. It is a subset of $\mathrm{spec} (R) =$ the set of all prime ideals of $R$. When $R$ is the ring of polynomials over $\mathbb{C}$ (more generally an algebraically closed field) a remarkable accident happens, and it goes by the name Hilbert Nullstellensatz. This theorem establishes a correspondence between varieties of $\mathbb{C}^n$ and ideals of $\mathbb{C}[z_1,\dotsc,z_n]$. Each variety decomposes into irreducible components and the irreducible components correspond to prime ideals.

As for the localization theorem, I taught this subject a while ago in a graduate course. I only looked at the special case of Hamiltonian $S^1$-actions which already has many nontrivial consequences, and all the ideas needed in the general case are needed in this special as well.

You might want to look over section 3.5 of these course notes where I go through the proof in great detail and at a slower speed than Atiyah-Bott. In the case $n=1$ the question you ask has a more familiar interpretation and the spectrum can be identified with the spectrum of a matrix, whence the the word spectrum. As I point in the notes, the key technical result of the localization theorem was known to A. Borel, almost two decades prior to Atiyah-Bott's work.

share|cite|improve this answer

As Ralph mentioned, if I understand the definitions properly, there is another way to see this by using a short exact sequence. I believe that $V_f = V(f)$ where $R = \mathbb{C}[u_1, \dots, u_l]$ and $V(f) = \{ p \in \hbox{spec(R)} \mid f \in p \}$ since we are over an algebraically closed field. If so we have $\hbox{Supp(-)} = \cap V_f = V(I)$ for some ideal $I$.

Lemma. Let $R$ be a commutative Noetherian ring. Let $N,M,L$ be $R$-modules such that $$0 \to N \to M \to L \to 0$$ is exact. Then we have $\hbox{Supp} (M) = \hbox{Supp} (L) \cup \hbox{Supp} (N)$. Here $\hbox{Supp} (M) = \{ p \in \hbox{Spec}(R) \mid M_p \neq 0 \}$.

We give a name to the maps in your exact sequence, i.e,. $$ D \stackrel{d}{\to} E \stackrel{e}{\to} F. $$ Then $$ 0 \to \hbox{im}(d) \to E \to \hbox{coker}(d) \to 0$$ is exact where $\hbox{im}(-)$ denotes the image of a map. Hence $\hbox{Supp}(E) = \hbox{Supp}(\hbox{im}(d)) \cup \hbox{Supp}(\hbox{coker} (d))$ by the lemma. Since the sequence $ D \stackrel{d}{\to} E \stackrel{e}{\to} F$ is exact, we have $\hbox{im} (d) = \hbox{ker} (e)$, and this implies that $\hbox{coker}(d) = E / \hbox{im}(d) = E / \hbox{ker}(e)$. Applying the lemma again to the exact sequences $D \to \hbox{im}(d) \to 0$ and $0 \to \hbox{coker}(d) \to F$ we have $\hbox{Supp}(\hbox{im}(d)) \subseteq \hbox{Supp}(D)$ and $\hbox{Supp}(\hbox{coker}(d)) \subseteq \hbox{Supp}(F)$, respectively. This shows the containment in the question.

share|cite|improve this answer
I'm having trouble understanding that : how can one hope to have 0 -> ker(d) -> E exact ? For instance, if e:E->F is an isomorphism, then d=0, and D=ker(d) can be arbitrarily big, obviously not fitting inside E... – Samuel Tinguely May 7 '13 at 15:43
Samuel Tinguely: I corrected the modules. Thank you. – Youngsu May 9 '13 at 0:21

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.