Definition of the symmetric algebra in arbitrary characteristic for graded vector spaces 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.
 A: Symmetric algebras (aka free commutative associative unital algebras) are given by a functor, and they satisfy a universal property: If M is a module over a commutative ring k and R is a commutative k-algebra, then k-algebra homomorphisms from Symk(M) to R are in bijection with k-module maps from M to R.  This bijection should be functorial with respect to R (i.e., ring homomorphisms in the target).  More succinctly, the symmetric algebra functor is left adjoint to the forgetful functor from commutative k-algebras to k-modules.  This is the description in the "categorical properties" section of the Wikipedia article.
The universal construction normally yields definition 1, but in the derived world, you might have to do something extra, like take homotopy coinvariants (this means taking some kind of resolution to get a free symmetric group action).
A: My understanding is that the "right" way to define the symmetric algebra comes from a braiding that tells you how the symmetric group acts on tensor products.  And an easy and general way to get such a braiding is to consider the category of representations of a quasi-triangular Hopf algebra; this is a short way to define the category of supervector spaces and recover the usual notion of graded-commutativity there.
The problem is that when you say "graded" (say, with respect to a group) you're only specifying at best a Hopf algebra, not a quasi-triangular structure.  So the answer depends on what possible quasi-triangular structures are floating around (in the $\mathbb{Z}/2\mathbb{Z}$ case on the group algebra of $\mathbb{Z}/2\mathbb{Z}$) and of course for a group of order $|G|$ terrible things are going to happen in characteristic dividing $|G|$.
A: In simple terms, it seems to me to depend on your definition of "graded commutative k-algebra".  I would take this mean that 
$xy =(-1)^{|x||y|}yx$
where |x| denotes the degree of the homogenous element x.  So I would divide by the ideal generated by 
$xy -(-1)^{|x||y|}yx$
for homogeneous elements x, y in V.  This does capture the polynomial algebra if the vecotr space is in degree 0, and the exterior algebra if the vector space is in degree 1 and the characteristic is not 2, but the polynomial algebra if the characteristic is 2. This is the right definition for algebraic topology.  
But of course, there are other possible definitions, as others have said. 
A: 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).
A: This is not an answer, as I think Scott did a better job than I could have.  Another algebra that generalizes the symmetric algebra in characteristic non-zero is the algebra of divided polynomials.  Let V be a finite-dimensional vector space over k a field, and let V* be its dual space.  For each n, write the space of n-linear maps from V* to k, and take the subspace of maps that are invariant under the natural Sn-action.  The direct sum of all of these spaces is a k-algebra, where the multiplication is as follows.  To multiply a symmetric m-linear map by a symmetric n-linear map, for each subset of size m of a set of size m+n consider the (m+n)-linear map that sends the subset through the m-linear multiplicand and the rest through the n-linear one; then add up all the m-choose-n many ways to do this.  Then this algebra agrees with the symmetric algebra over V when k is characteristic-zero, but not otherwise.  For example, when V is one-dimensional with basis vector x, then the symmetric algebra is the polynomial algebra k[x], whereas the above construction yields the algebra k[x,x2/2,x3/6,x4/24,...].  In fact, the symmetric algebra of V* and the above algebra are both naturally graded Hopf algebras, and in fact they are duals as graded Hopf algebras.
A: Which one is the right definition depends on what you want to do with it. 
No matter how much technology you throw at the question, including homotopy coinvariants and quasi-triangular Hopf algebras, "right" is a relative notion :D
