Rings of multivariable Laurent polynomials

The commutative group ring of the free commutative group $F^{\circ}(X)$ over a field $k$ is a ring extension of the commutative semigroup ring of the free commutative monoid $F(X)$ over $k$. Although, $F^{\circ}(X)$ is a natural semigroup extension of $F(X)$, rings of multivariable Laurent polynomials don't have the same level of importance as polynomial rings in commutative algebra.

A distinguishing property of ordinary polynomials is that they can be cast into functions, so that over affine space $\mathbb{A}^n$ they are functions $f: \mathbb{A}^n \to k$ and this works for any commutative ring. On the other hand, for Laurent polynomials to be cast into functions $f : \mathbb{A}^n \to k$ we need to have some notion of division to deal with negative degree exponents, so we need to work over a field.

In those cases when we take the commutative group ring of a commutative ring $R$ which is not a field, with respect to $F^{\circ}(X)$ then what we get is certainly still a commutative ring, it is just not a commutative ring of functions. Consider the commutative group ring $\mathbb{Z}(\mathbb{Z}^n,+)$ consisting of polynomials with integer exponents and integer coefficients. Then this is a valid commutative ring, but its elements are not functions because $\mathbb{Z}$ is not a field.

So in some sense we ought to have our free commutative group ring be over a field $k$. Then we get terms like $5\frac{x}{yz} + \frac{6y}{xz}$ consisting of fractional monomials, but then given any term like this we can add to get $\frac{5x^2 + 6y^2}{xyz}$ which is always a polynomial with a monomial in the denominator. So we see that multivariable Laurent polynomials are simply rational functions with monomials in the denominator, so we have an embedding. \[ kF^{\circ}(x,y,z) \subseteq k(x,y,z) \] The ring of Laurent polynomials is simply the localisation of the polynomial ring by the multiplicative set of monomials, and so they are merely a special case of rational functions. Now it is clear why we don't see Laurent polynomials so much in commutative alebra. They are simply part of the far more important and general concept of fields of rational functions $k(x,y,z)$.

Just as we don't tend to restricted localisations of the integers like the dyadic rationals that much, we won't see rings of multivariable Laurent polynomials showing up as much. In general, we always want to deal with the largest localisation of a domain, which is its field of fractions.

So although the group completion of the free commutative semigroup $F(X)$ is an important and natural concept in commutative semigroup theory, it doesn't have the same role and level of importance in commutative algebra, as determined by commutative semigroup rings. This demonstrates that not everything in commutative algebra is a consequence of commutative semigroup rings, but the use of semigroup rings is still an infinitely powerful technique in the construction of commutative rings.