# The Unapologetic Mathematician

## Polynomials, take 2

As I said before, if we take the free commutative monoid on $n$ generators, then build the semigroup ring from that, the result is the ring of polynomials in $n$ variables. I hinted at a noncommutative analogue, which today I’ll construct from the other side.

Instead of starting with a set of $n$ generators and getting a monoid, let’s start by building the free abelian group $\mathbb{Z}^n$. This consists of ordered $n$-tuples of integers, and we add them component by component. We can pick out the $n$ generators $x_k=(0,...,0,1,0,...,0)$, where the $1$ shows up in slot $k$. Then every element can be written $a_1x_1+a_2x_2+...+a_nx_n$, where the $a_k$ is entry $k$ in the $n$-tuple form of the element.

So how do we build the tensor product $\mathbb{Z}^n\otimes\mathbb{Z}^n$? First we take all pairs
$(a_1x_1+...+a_nx_n,b_1x_1+...+b_nx_n)$
and use them to generate a free abelian group. Then we impose the linearity relations $(a+a',b)=(a,b)+(a',b)$ and $(a,b+b')=(a,b)+(a,b')$. What does that mean here? Well for one thing we can apply it to the collection of pairs:
$(a_1x_1+...+a_nx_n,b_1x_1+...+b_nx_n)=$
$a_1(x_1,b_1x_1+...+b_nx_n)+...+a_n(x_n,b_1x_1+...+b_nx_n)=$
$a_1b_1(x_1,x_1)+...+a_1b_n(x_1,x_n)+a_2b_1(x_2,x_1)+...a_nb_n(x_n,x_n)$
So we could just as well write the tensor product as the group generated by $x_i\otimes x_j=(x_i,x_j)$.

This same argument goes through as we tensor in more and more copies of $\mathbb{Z}^n$. The tensor power $(\mathbb{Z}^n)^{\otimes d}$ is the free abelian group generated by the elements $x_{i_1}\otimes...\otimes x_{i_d}$, where each index runs from ${}0$ to $n$.

Now we take all of these tensor powers and throw them together. We get formal linear combinations
$\sum\limits_{d=1}^\infty\sum\limits_{i_1,...,i_d=1}^n a_{i_1,...,i_d}x_{i_1}\otimes...\otimes x_{i_d}$
where all but finitely many of the “coefficients” $a_{i_1,...,i_d}$ are zero. These look an awful lot like polynomials, don’t they? In fact, if we only had a commutative property that $x_i\otimes x_j=x_j\otimes x_i$ then these would be exactly (isomorphic to) the polynomials we came up with last time.

To be explicit about the universal properties, any function from the $n$ generators to the underlying abelian group of a ring $R$ with unit has an unique extension to a linear function from $\mathbb{Z}^n$ to $R$. Then this has a unique extension to a ring homomorphism from $R(\mathbb{Z}^n)$ to $R$. From the other side, there is a unique extension of the original function to a monoid homomorphism from the free monoid $M_n$ to the underlying monoid of $R$. Then this has a unique extension to a ring homomorphism from $\mathbb{Z}[M_n]$ to $R$. Since both $R(\mathbb{Z}^n)$ and $\mathbb{Z}[M_n]$ satisfy this same universal property they must be isomorphic. We commonly write this universal ring as $\mathbb{Z}\{x_1,...,x_n\}$, and call it the ring of noncommutative polynomials in $n$ variables.

April 20, 2007 Posted by | Ring theory | 1 Comment