Meaning of "freely generated" for an $R$-algebra

Question

Let $A$ be an algebra over some ring $R$. If $a_1,\dots,a_n \in A$, what does the phrase "$a_1,\dots,a_n$ freely generate $A$" mean? Does it mean that $a_1,\dots,a_n$ generate the algebra and are linearly independent? Does it also mean that elements of the form $a_i a_j a_k$ for $i,j,k=1,\dots,n$ are linearly independent?

Answer 1

I assume $R$ is commutative. If you have just one element, say $a$, then the $R$-algebra freely generated by $a$ is $A=R[a]$, the rings of polynomials in the indeterminate $a$.

More generally, the algebra freely generated by $\{a_1,\dots,a_n\}$ is an algebra $A$ containing $\{a_1,\dots,a_n\}$ such that, for any $R$-algebra $B$ and any choice of $b_1,\dots,b_n\in B$, there exists a unique $R$-algebra homomorphism $f\colon A\to B$ such that $f(a_i)=b_i$ $(i=1,2,\dots,n)$.

The algebra $A$ can be described as the set of “noncommuting polynomials”, that is of $R$-linear combinations of monomials, that is, (formal) products of elements on $\{a_1,\dots,a_n\}$, where no commutation is allowed, so $a_1a_2\ne a_2a_1$. However, the coefficients do commute.

Addition is performed in the obvious way, multiplication by using distributivity; for instance $$ (r_1a_1a_2+r_2a_3)(r_3a_1+r_4a_2)= r_1r_3a_1a_2a_1+r_1r_4a_1a_2a_2+r_2r_3a_3a_1+r_2r_4a_3a_2 $$ The multiplicative identity is the “empty monomial”.

Proving that this set is an $R$-algebra is not difficult; the unique morphism after a choice of $B$ and elements $b_1,\dots,b_n$ is obvious.

If we restrict $B$ to be commutative, then we get the “commutative $R$-algebra freely generated” and, in this case, it's just the polynomial ring in $n$ indeterminates.