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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.

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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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