How Is Symmetric Algebra Isomorphic to a Free Commutative R-Algebra?

[bombo1]

I have this problem that i need to prove and i don't even know where to start. So I have to show that the symmetric algebra ( Sym V ) is isomorphic to free commutative R-algebra on the set {x1, ..., xn}.


Now i know that Sym V could be regarded as +Sym^n V for n>=0. And then we need to show that this is isomorphic to the R-algebra on the set {x1, ..., xn}. What I don't really understand is the +Sym^n V for n>=0
what exactely does the + do? Can i just regard this as the space of homogeneous polynomials of degree n in the variables of e1, e2,... e3, where {ei} is a basis for V.
Every polynomial ring R[x1, ..., xn] is a commutative R-algebra. Do I need to find a map such that it reduces all the polynomials from R to make them homogeneous.. I really don't know how to start this problem.

thnx for everyone who can help me with any ideas:)

 

[matt grime]

You might start be defining what V is, then if you start by writing out the definition of sym(V) we might be able to see what we have to work with. (And, yes I can guess what V is, and I know what the symmetric powers are of, say, a vector space, but that is a suggestion of what you need to start with.)

the plus symbol just means direct sum (as vector spaces) of the degree n parts of the symmetric algebra.

 

[mathwonk]

it helps if you know what the objects mean, i.e. both are the universal commutative algebra over R with n generators, so they are the same.