Let $C$ \mathcal{C}$ be a cocomplete $R$-linear tensor category. Many notions in of commutative algebra may be internalized to $C$. \mathcal{C}$. For example an algebra is an object $A$ in $C$ \mathcal{C}$ together with morphisms $e : 1 \to A$ (unit) and $m: A \otimes A \to A$ (multiplication) satisfying the usual laws. The $n$th n$-th symmetric power $\text{Sym}^n(X)$ of an object $X$ is the quotient of $X^{\otimes n}$ by $x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of two maps the $(X^{\otimes n})^{n!n!$ symmetries $X^{\otimes n} \to X^{\otimes n}$. Then $\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$ is an algebra object and $\text{Sym}$ is in fact left adjoint to the forgetful functor from algebras in $C$ \mathsf{CAlg}(\mathcal{C}) \to $C$.\mathsf{CAlg}(\mathcal{C})$.
But now what about the exterior power $\wedge^n(X)$? \Lambda^n(X)$? It is clear how to define $X^{\otimes n}$ modulo $x_1 \otimes ... \otimes x_n = \text{sign}(\sigma) text{sgn}(\sigma) \cdot x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$ in this context, which one might call the skew-symmetric anti-symmetric power. $\mathrm{ASym}^n(X)$. The correct definition of the exterior power also has to mod out $... \otimes a \otimes ... \otimes a \otimes ... = 0$. But I have no idea how to internalize this to $C$. \mathcal{C}$, even for $n=2$. The reason is that there is no morphism $X \to X \otimes X$ which acts like $a \mapsto a \otimes a$. Another idea would be to define $\wedge(X)$ \Lambda(X)$ as a graded-commutative algebra object with the usual universal property, classifying maps from morphisms $f$ on $X$ which satisfy something like $f(x)^2=0$, but again it is unclear how to formulate this in $C$.\mathcal{C}$.
If this is not possible at all, which additional structure on $C$ \mathcal{C}$ do we need in order to define exterior powers within them? Is this some categorified $\lambda$-ring structure? This structure should be there in the case of usual module categories .
PS: I am interested in exterior powers since they should reduce locally free objects to line objects via the Plücker embedding(over rings or even ringed spaces). The $d$th anti-symmetric power of $1^n$ Of course there is not no problem when $1^{\binom{n}{d}} = 2 \wedge^d 1^n$ in generalR^*$, therefore it is not because then the correct notion hereexterior power equals the anti-symmetric power.
EDIT: The question was also discussed in a blog post.
Here is a more specific (and a bit stronger) formulation: Is there an some $R[S_n]$-module R[\Sigma_n]$-module $T$, such that for every $R$-module $X$, we have that $T \otimes_{R[S_n]} otimes_{R[\Sigma_n]} X^{\otimes n}$ equals $\wedge^n n} \cong \Lambda^n X := X^{\otimes n}/(... \otimes x ... \otimes x ...)$? Probably not, because $T$ should be a quotient of $R[S_n]$, and the corresponding ideal cannot generate squares.
EDIT:
Concerning the "hidden extra structure" in the case of modules, I have made the following observation: Let the ground base ring be $\mathbb{Z}$ (\mathbb{Z}$, or more generally a ring $R$ in which $r^2 - r \in 2R$ for all $r \in R$; this indluces includes boolean rings such as $\mathbb{F}_2$ and also $\mathbb{Z}/n$). Then \mathbb{Z}/n$. If $X$ is an $R$-module, then there is a well-defined(!) map homomorphism $f_n : X^{\otimes nX^{\otimes~ n-1} \to \text{Asym}(X), text{ASym}^n(X), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$in the case of modules. The , and its cokernel of $\oplus_{n \geq 2} f_n$ is $\wedge(X)$.
EDIT: If there won't be any definition in the general case, one might define $\lambda$ tensor categories as categorified $\lambda$-rings. But I'm still curious why have quite often a distinguished $\lambda$ structure, coming from a specific quotient of the tensor power ...\Lambda^n(X)$.
