Polynomial ring is a free algebra I'm trying to prove that the polynomial ring $F[X]$ (where $F$ is a field and $X$ is a set of commuting variables) is a free algebra in the class of associative, commutative and unitary $F$-algebras.
I'm rather confused about the definition of free algebra and the polynomial ring. I have to find a set that generates $F[X],$ so should it be $\{1\}\cup\{x \,\ |\,\  x\in X\}$ or just $X$?
If it is just $X,$ then how do I define the unity in $F[X],$ as $x_i^0$?
Thanks in advance for any help!
 A: You have to find $S\subset F[X]$ such that $S$ generates $F[X]$ (as an associative commutative $F$-algebra) such that for any $A$ associative commutative $F$-algebra, and any map $S\to A$, there is an $F$-morphism $F[X]\to A$ extending the map. 
$S=X$ sounds like a good candidate indeed. I don't know what you mean by "define unity"; but I guess your question is about why $X$ generates $F[X]$, in particular why $1$ is in the subalgebra generated by $X$ (if that's not the question, please tell me so). Well for this remember what "the subalgebra generated by $S$" means for $S\subset F[X]$ : it is the smallest subalgebra of $F[X]$ that contains $S$. In particular it is a subalgebra; so it contains $1$ (here "algebra" means unitary algebra).
Now it happens that the subalgebra generated by $S$ is the set of polynomials with coefficients in $F$, evaluated at points of $S$, and it so happens that $1$ is such a polynomial, but the real reason why $1$ is in this subalgebra is that the term "subalgebra" is defined so that it would work.
