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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?

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

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If I give a set map from X to an A-algebra R, then it extends uniquely to a map of A-algebras from A [X] to R, so you have the universal mapping property right (the description as adjoint to the forgetful functor -- I am not sure what "free" means in this context, but you are probably right about that too).

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  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Nov 14 '13 at 0:38
  • $\begingroup$ I take "ring" to mean "commutative ring with 1" unless otherwise specified. $\endgroup$ – hunter Nov 14 '13 at 9:50
  • $\begingroup$ I should say that I assumed $A$ to be commutative anyway $\endgroup$ – user88576 Nov 14 '13 at 16:35
  • $\begingroup$ Ok, but Mac Lane doesn't use this convention. $\endgroup$ – Martin Brandenburg Nov 17 '13 at 7:35

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