The $k$th exterior power as the tensor-power quotient by repeated-vector relations

Let $V$ be a vector space over a field $F$ and let $k\ge 0$. The $k$-fold tensor power is $V^{\otimes 0}=F$, $V^{\otimes 1}=V$, and for $k\ge 2$ the iterated tensor product of $k$ copies of $V$ under any parenthesization. By Finite iterated tensor products represent multilinear maps independently of parenthesization every parenthesization represents the $k$-linear maps out of $V^k$, and different parenthesizations are joined by the unique isomorphism preserving pure tensors, so $V^{\otimes k}$ is determined up to that isomorphism.

Let $W_k\subseteq V^{\otimes k}$ be the subspace spanned by all pure tensors $v_1\otimes \cdots \otimes v_k$ for which $v_i=v_j$ for some pair $1\le i<j\le k$ (so $W_0=W_1=0$). The $k$th exterior power of $V$ is the quotient vector space

${\mathrm{\Lambda }}^{k}V:=V^{\otimes k}\mathrm{/}{W}_k,$

and the basic wedge map is

$\wedge :V^k⟶{\mathrm{\Lambda }}^{k}V,(v_1,\dots ,v_k)⟼v_1\otimes \cdots \otimes v_k+W_k\mathrm{.}$

The construction never divides by $k!$, so ${\mathrm{\Lambda }}^{k}V$ is defined in every characteristic. Its relation to the Alternating $k$-linear maps maps is the universal property of Exterior powers represent alternating multilinear maps and are unique up to unique isomorphism; the convention fixed here is that a repeated entry makes the wedge zero.