The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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