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The free commutative ring on a set $X$ is the polynomial ring with variables the elements of $X$. This polynomial ring is the free (additive) abelian group on the free (multiplicative) abelian monoid on $X$.

On the other hand, I've often seen a commutative ring defined as an abelian monoid object in the category of abelian groups, so I might have expected the composition to be the other way around - the free (multiplicative) abelian monoid on the free (additive) abelian group on $X$.

Does this give the same result? Is a ring equivalentely an abelian group object in the category of abelian monoids?

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  • $\begingroup$ A commutative ring is an commutative monoid in the monoidal category of abelian groups (i.e. a triple $(R, e: \mathbb{Z} \to R, m : R \otimes R \to R)$ satisfying suitable axioms, not just an internal commutative monoid). $\endgroup$ – Nex Nov 26 '15 at 17:45
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Intuitively I understand the first definition but not the second. In fact the standard notation for a polynomial with integer coefficients and variables taken from the set X suggests first taking products, then integer linear combinations so that is exactly what a free abelian group on a free abelian monoid means.

The second definition is less intuitive. The free abelian group on X is the collection of homogeneous linear polynomials. The free abelian monoid on that is the collection of all products of such polynomials. Even if X is a singleton {t} this is problematic. Unique decomposition into irreducible factors means that the second definition is a subset of the first one in a canonical way, but in the other direction I would not know what to do with nonlinear polynomials that are irreducible over the integers such as t^2+1; nor with constants, by the way. Generally speaking it is not at all clear how the second definition would even be an abelian group.

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