This question was first asked on MathSE but nobody answered.

In his proof of Lemma A2.5 in his book Commutative Algebra with a View towards Algebraic Geometry, Prof. Eisenbud writes something like this:

Let R be a commutative ring, $M$ an $R$-module, $S(M)$ the symmetric algebra (the quotient of the tensor algebra by the ideal of commuting relations) on $M$ considered as a coalgebra with $\Delta$ being the comultiplication, defined here as the algebra homomorphism induced by the diagonal map from $M$ to $M\oplus M$.

To prove that for all $x\in M$, there exists an element $y\in S(M)^{\otimes d}$ such that $$\Delta^d x\ =\ \Sigma_{\sigma \in G}\ \sigma(y)$$ where $G$ is the symmetric group on $d$ letters acting on $S(M)^{\otimes d}$ by the action coming from its action by permutation of factors on the $R$-module $M\oplus \cdots\oplus M$ ($d$ times), one can do it on elements that are product of elements in $M$, then use induction on the number of factors of the product. For the first step of the induction, he writes it is quite easy to see that if $x\in M$ and $y=x\otimes 1\otimes 1\cdots \otimes 1$, the symmetrization of $y$ is $\Delta^d x$.

I cannot easily see that since for me $(d-1)!\Delta^d x = \Sigma_{\sigma \in G}\sigma(y)$. Am I misunderstanding something obvious ? And if not how to do this first step ?

Edit: This lemma is used in the proof that the dual of the symmetric algebra has a system of divided powers the following way: to show that for $u\in S_n(M)^*$, $u^d$ is divisible by $d!$, one look at the action of $u^d$ on an element $x\in S(M)$, which is exactly the action of $u^{\otimes d}$ on $\Delta^d(x)\in S(M)^{\otimes d}$. But $u^{\otimes d}$ acts nontrivially only on tensor products of $d$ elements all of degree $n$ (which implies that the degree of $x$ is $nd$).

For this, there is a weak version of the lemma that works fine and can be proven by induction on $n$ which is the following: if $x\in S_{nd}(M)$, then $\Delta^dx$ is a sum of a symmetric terms like $\Sigma_{\sigma \in G}\ \sigma(y)$ where $y\in S_n(M)^{\otimes d}$ and other terms not in $S_n(M)^{\otimes d}$.

To conclude my too long question, I believe the lemma "Symmetry of Diagonalization" in not true on a general commutative ring, but the weak version is sufficient for the use of the Lemma in the book of Prof. Eisenbud. Can anybody give me an opinion here ?

Edit 2: When Prof. Eisenbud writes $\Delta^d$ it is actually $\Delta^{d-1}$. I sticked to his convention.

  • 1
    $\begingroup$ I think Eisenbud is wrong about this. Good catch, which you should forward to Eisenbud. $\endgroup$ – darij grinberg May 27 '14 at 9:39

I think this is a mistake in Eisenbud's book.

It seems, however, that Eisenbud only uses Lemma A2.5 in two places: in the proof of Proposition A2.4, and in the proof of Proposition-Definition A2.6.

DISCLAIMER: All that I am saying below is meant to refer to the case when $M$ is an even $R$-module. I am not making any statements about superalgebra.

In the latter place, the reference to Lemma A2.5 can be replaced by an application of the following weaker (but correct) statement: In the situation of Lemma A2.5, we have $\Delta^d \left(x\right) \in \mathrm{sym}_d \left( \mathcal{S}\left(M\right)^{\otimes d} \right) + U$, where $\mathrm{sym}_d : \mathcal{S}\left(M\right)^{\otimes d} \to \mathcal{S}\left(M\right)^{\otimes d} $ is the symmetrization map (sending every $p$ to $\sum\limits_{g \in G} gp$), and where $U$ is the $R$-submodule $\sum\limits_{\substack{i_1, i_2, \ldots, i_d; \\ \text{at least one }k \text{ satisfies } i_k = 0}} \mathcal{S}_{i_1}\left(M\right) \otimes \mathcal{S}_{i_2}\left(M\right) \otimes \cdots \otimes \mathcal{S}_{i_d}\left(M\right)$ of $\mathcal{S}\left(M\right)^{\otimes d}$. This can be proven combinatorially: take $x = x_1 x_2 \ldots x_k$ for $x_1, x_2, \ldots, x_k \in M$, and write $\Delta^d \left(x\right)$ as a sum over ordered set partitions of $\left\{1,2,\ldots,k\right\}$ into $d$ possibly empty sets; the addends containing at least one empty set are in $U$, whereas the other addends correspond to ordered set partitions into $d$ nonempty sets, and the symmetric group $G$ clearly acts freely on these addends, so they can be combined into orbits whose sums belong to $\mathcal{S}\left(M\right)^{\otimes d}$. And as far as I understand, this is all that is needed in the proof of Proposition-Definition A2.6, because the $n$ there is positive and so $u^{\otimes d}\left(U\right) = 0$.

The proof of Proposition A2.4 only seems to be using Lemma A2.5 in the supercase.

| cite | improve this answer | |
  • $\begingroup$ thank you very much for your precise answer which agrees perfectly with my own analysis. Our two "weak" versions of the lemma agree too which conforts me very much. I will accept your answer as such but remains one slight detail: I believe the lemma as written by Prof. Eisenbud is not true, meaning the image of the diagonal map is not usually contained in the symmetrization of the symmetric algebra, right ? $\endgroup$ – brunoh May 27 '14 at 10:57
  • 1
    $\begingroup$ Yes, $\Delta^3 x = x \otimes 1 \otimes 1 + 1 \otimes x \otimes 1 + 1 \otimes 1 \otimes x$ is not generally a symmetrization. $\endgroup$ – darij grinberg May 27 '14 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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