Deligne's exterior power

In "Catégories Tannakiennes", Deligne defines the $n$th exterior power of an object $A$ of an abelian tensor category $\mathcal{C}$ as the image of the morphism $$p : A^{\otimes n} \to A^{\otimes n}, a_1 \otimes \dotsc \otimes a_n \mapsto \sum_{\sigma \in \Sigma_n} \mathrm{sgn}(\sigma) \cdot a_{\sigma(1)} \otimes \dotsc \otimes a_{\sigma(n)}.$$ Let's call this $\Lambda^n_{Deligne}(A)$. Notice that $p^2=n! p$. If $n!$ acts invertibly on $\mathcal{C}$, this is certainly the correct definition of the exterior power (but then we could also define it as suggested here). But what happens in general?

If $\mathcal{C}=\mathsf{Mod}(R)$ for some commutative ring $R$, then it is easy to see that the map $(a_1,\dotsc,a_n) \mapsto p(a_1 \otimes \dotsc \otimes a_n)$ is alternating and thus induces a surjective homomorphism $$\Lambda^n(A) \to \Lambda^n_{Deligne}(A).$$ Here, $\Lambda^n(A)$ is the usual exterior power. Is this map an isomorphism? In other words, is the map $\Lambda^n(A) \to A^{\otimes n}$ induced by $p$ injective?

The answer is yes if $A$ is free. It follows formally that it also holds when $A$ is locally free. By Lazard's Theorem we get the result if $A$ is flat. It also holds when $A$ is cyclic, because then $\Lambda^n(A)$ vanishes for $n>1$. I've also tried some other examples. Are there any counterexamples?

• In Darij Grinberg's answer $R$ is not an integral domain. What happens if we assume that $R$ is an integral domain? – Martin Brandenburg Oct 8 '14 at 10:28
• What happens if you replace my $k$ by the polynomial ring $\mathbb{F}_2\left[\alpha,\beta,\gamma\right]$, change the definition of $L$ to $L = M / \left< \alpha x + \beta y + \gamma z, \alpha^2 x, \alpha^2 y, \alpha^2 z, \beta^2 x, \beta^2 y, \beta^2 z, \gamma^2 x, \gamma^2 y, \gamma^2 z \right>$, and change the image of $f$ to $k / \left(\alpha^2, \beta^2, \gamma^2\right)$ ? – darij grinberg Oct 8 '14 at 13:32
• Well, that sounds good. And why do we need 3 variables, why not just 2? (Haven't checked, but the trick in your answer seems to work...) – Martin Brandenburg Oct 8 '14 at 13:54
• In fact, if the module $A$ is generated by two elements then the map $\Lambda^2A\to A\otimes A$ is split injective. – Tom Goodwillie Oct 9 '14 at 11:14
• Yes. $f(a\otimes b)=a\wedge b$, $f(a\otimes a)=f(b\otimes a)=f(b\otimes b)=0$. It works because if $pa+qb=0$ then $p(a\wedge b)=0=q(a\wedge b)$. – Tom Goodwillie Oct 9 '14 at 12:58

It is not. Here is a counterexample. (I have essentially copied the setting from https://mathoverflow.net/a/87958/2530 , which in turn goes back to a PBW counterexample by P. M. Cohn.)

Let $k$ be the commutative ring $\mathbb F_2 \left[\alpha,\beta,\gamma\right] / \left(\alpha^2,\beta^2,\gamma^2\right)$.

Let $M$ be the $k$-module $\left\langle x,y,z\right\rangle$. Let $L$ be the $k$-module $M / \left\langle \alpha x + \beta y + \gamma z\right\rangle$. We shall abusively write $x$, $y$ and $z$ both for the elements $x$, $y$ and $z$ of $M$ and for their projections on $L$ as well.

Define a $k$-linear map $f : M \otimes M \to k$ by

$f\left(y \otimes z\right) = \alpha$, $f\left(z \otimes y\right) = \alpha$, $f\left(x \otimes x\right) = 0$,

$f\left(z \otimes x\right) = \beta$, $f\left(x \otimes z\right) = \beta$, $f\left(y \otimes y\right) = 0$,

$f\left(x \otimes y\right) = \gamma$, $f\left(y \otimes x\right) = \gamma$, $f\left(z \otimes z\right) = 0$.

It is easy to see that $f$ vanishes on $\left\langle \alpha x + \beta y + \gamma z\right\rangle \otimes M + M \otimes \left\langle \alpha x + \beta y + \gamma z\right\rangle$. Hence, $f$ descends to a $k$-linear map $L \otimes L \to k$ (since $L \otimes L \cong \left(M \otimes M\right) / \left(\left\langle \alpha x + \beta y + \gamma z\right\rangle \otimes M + M \otimes \left\langle \alpha x + \beta y + \gamma z\right\rangle\right)$ as a consequence of the right-exactness of the tensor product). This latter map is also alternating, and therefore descends to a $k$-linear map $\wedge^2 L \to k$. This latter map sends $\beta y \wedge \gamma z + \gamma z \wedge \alpha x + \alpha x \wedge \beta y$ to $3 \alpha \beta \gamma = \alpha \beta \gamma \neq 0$. Hence, $\beta y \wedge \gamma z + \gamma z \wedge \alpha x + \alpha x \wedge \beta y \neq 0 \in \wedge^2 L$.

But in $L \otimes L$, we have

$p \left( \beta y \wedge \gamma z + \gamma z \wedge \alpha x + \alpha x \wedge \beta y \right)$

$= \beta y \otimes \gamma z - \gamma z \otimes \beta y + \left(\text{cyclic permutations}\right)$

$= \underbrace{\alpha x \otimes \beta y - \alpha x \otimes \gamma z}_{= - \alpha x \otimes \beta y - \alpha x \otimes \gamma z = \alpha x \otimes \left( -\beta y - \gamma z\right)} + \left(\text{cyclic permutations}\right)$

$= \alpha x \otimes \underbrace{\left( -\beta y - \gamma z\right)}_{=\alpha x \text{ (since } \alpha x + \beta y + \gamma z = 0 \text{ in } L \text{)}} + \left(\text{cyclic permutations}\right)$

$= \underbrace{\alpha x \otimes \alpha x}_{=\alpha^2 x\otimes x} + \left(\text{cyclic permutations}\right)$

$= \underbrace{\alpha^2}_{=0} x \otimes x + \left(\text{cyclic permutations}\right) = 0$.

So $p$ fails to be injective.

• Thank you for your answer, Darij! I think that we can simplify the calculation, since $\gamma z \wedge \alpha x + \alpha x \wedge \beta y = 0$ and therefore your element in $\Lambda^2 L$ is just $\beta y \wedge \gamma z$. The image under $p$ is then $\beta y \otimes \gamma z - \gamma z \otimes \beta y$. – Martin Brandenburg Oct 8 '14 at 8:02
• ... and $\beta y \otimes \gamma z + \gamma z \otimes \beta y = \alpha x \otimes \gamma z + \gamma z \otimes \gamma z + \alpha x \otimes \beta y + \beta y \otimes \beta y = \alpha x \otimes \alpha x = 0$ – Martin Brandenburg Oct 8 '14 at 8:19
• Nice. I tried to simplify the calculation like that, but I failed, so I decided to keep the symmetric form. – darij grinberg Oct 8 '14 at 13:28