# The Unapologetic Mathematician

## Polynomials

Now we come to a really nice example of a semigroup ring. Start with the free commutative monoid on $n$ generators. This is just the product of $n$ copies of the natural numbers: $\mathbb{N}^n$. Now let’s build the semigroup ring $\mathbb{Z}[\mathbb{N}^n]$ on this monoid.

First off, an element of the monoid is an ordered $n$-tuple of natural numbers $(e_1,e_2,...,e_n)$. Let’s write it in the following, more suggestive notation: $x_1^{e_1}x_2^{e_2}...x_n^{e_n}$. We multiply such “monomials” just by adding up the corresponding exponents, as we know from the composition rule for the monoid. Now we build the semigroup ring by taking formal linear combinations of these monomials. A generic element looks like $\sum\limits_{(e_1,e_2,...,e_n)\in\mathbb{N}^n}c_{(e_1,e_2,...,e_n)}x_1^{e_1}x_2^{e_2}...x_n^{e_n}$

where the $c_{(e_1,e_2,...,e_n)}$ are integers, and all but finitely many of them are zero.

Assuming everyone’s taken high school algebra, we’ve seen these before. They’re just polynomials in $n$ variables with integer coefficients! The addition and multiplication rules are just what we know from high school algebra. The only difference is here we specifically don’t think of $x_i$ as a “placeholder” for a number, but as an actual element of our ring.

But we can still use it as a placeholder. Let’s consider any other commutative ring $R$ with unit and pick $n$ elements of $R$. Call them $r_1$, $r_2$, and so on up to $r_n$. Since $R$ is a commutative monoid under multiplication there is a unique homomorphism of monoids from $\mathbb{N}^n$ to $R$ sending $x_i$ to $r_i$. That’s just what it means for $\mathbb{N}^n$ to be a free commutative monoid. Now there’s a unique homomorphism of rings from $\mathbb{Z}[\mathbb{N}^n]$ to $R$ sending $x_i$ to $r_i$, because $\mathbb{Z}[\mathbb{N}^n]$ is the semigroup ring of $\mathbb{N}^n$.

The upshot is that $\mathbb{Z}[\mathbb{N}^n]$ is the free commutative ring with unit on $n$ generators. Because of this, we’ll usually omit the intermediate step of constructing $\mathbb{N}^n$ and just write this ring as $\mathbb{Z}[x_1,x_2,...,x_n]$.

There are similar constructions to this one that I’ll leave you to ponder on your own. What if we just constructed the free monoid on $n$ generators (not commutative)? What about the free semigroup? What sort of rings do we get, and what universal properties do they satisfy?

April 16, 2007 - Posted by | Ring theory

## 6 Comments »

1. […] take 2 As I said before, if we take the free commutative monoid on generators, then build the semigroup ring from that, […]

Pingback by Polynomials, take 2 « The Unapologetic Mathematician | April 20, 2007 | Reply

2. One nice and easy to check property of the polynomials is that K[X] = Z[X] + K, where “+” is the categorical sum (co-product) in the category of rings in both commutative and non-commutative cases. Comment by furia_krucha | December 12, 2007 | Reply

3. furia_krucha, I don’t think that’s quite right for the non-commutative case, since X commutes with elements of K in K[X]. (We have K[X] = K tensor Z[X] in general, but the coproduct in the noncommutative case is not tensor, as you probably know already.) Comment by Todd Trimble | December 12, 2007 | Reply

4. Thanks for pointing this out Todd, of course it doesn’t work in the non-commutative case, because we are only looking at the homomorphisms K -> R, where image lies in the center of R. Comment by furia_krucha | December 12, 2007 | Reply

5. […] into linear algebra. Specifically, we’ll need to know a few things about the algebra of polynomials. Specifically (and diverging from the polynomials discussed earlier) we’re talking about […]

Pingback by Polynomials I « The Unapologetic Mathematician | July 28, 2008 | Reply

6. […] should be no surprise that the symmetric algebra on the dual space is isomorphic to the algebra of polynomial functions on , where the grading is the total degree of a monomial. If has finite dimension , we […]

Pingback by Tensor and Symmetric Algebras « The Unapologetic Mathematician | October 26, 2009 | Reply