TITLE: Definition of the symmetric algebra in arbitrary characteristic for graded vector spaces
QUESTION [8 upvotes]: What is the right definition of the symmetric algebra over a graded vector space V over a field k?
More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?
Two possible definitions come to my mind:
1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.
2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).
The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.
The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.
Finally, both definitions have a shortcoming in that they don't commute well with base change.

REPLY [2 votes]: The right definition is: take the free associative (tensor) algebra generated by $V$; divide out the ideal generated by the elements $xy-(-1)^{|x||y|}yx$ for all homogeneous $x$, $y\in V$ and $z^2=0$ for all odd $z\in V$.  This should commute with the base change well (when $V$ is flat over your base).