# Polynomial ring and the free algebra

In the Algebra book of Mac Lane there is an exercise in Chap. IV which tells me to construct a polynomial ring $A[X]$ for any set (not necessarily finite) $X$ ($A$ a ring), and to give correct the universal property. As far as the construction is concerned, I have no clue, but I suggest the following UMP: $A[X]$ is the free object in $\mathbf{A-alg}$ on $X$ (a fancier way to say this is to say that the "polynomial" functor is left adjoint to the forgetful functor $\mathbf{A-alg}\rightarrow\mathbf{Set}$, innit?). Do you agree with me?

• It is left adjoint to the category of commutative algebras if thats what you asking for. – user52045 Nov 14 '13 at 0:11

This is indeed so; in fact Bourbaki (Algebra II, chap. 4, sect. 1 'Polynomials and rational functions) defines the polynomial ring $A[(X_i)_i\in I]$ to be the free commutative algebra on $I$.
• This is not completely correct unless you assume commutativity or you use the free algebra $A\langle X \rangle$ instead. In the universal property of the polynomial algebra $A[X]$, $X$ should map to the center. – Martin Brandenburg Nov 14 '13 at 0:38
• I should say that I assumed $A$ to be commutative anyway – user88576 Nov 14 '13 at 16:35