The Free Ring on an Abelian Group

Last week I talked about how to make a ring out of a semigroup by adding an additive structure. Now I want to do the other side. Starting with an abelian group we’ll build a ring by adding a “free” multiplication.

The main tools will be what I was saying about tensor products and abelian groups. Specifically, we have isomorphisms
$(A\otimes B)\otimes C\cong A\otimes(B\otimes C)$
$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes\left(\bigoplus\limits_{j\in\mathcal{J}}B_j\right)\cong\bigoplus\limits_{(i,j)\in\mathcal{I}\times\mathcal{J}}\left(A_i\otimes B_j\right)$

The first of these lets us unambiguously talk about “the” tensor product of any finite list of abelian groups. The particular case we’re interested in here is when all of them are the same group. We define the $n$ th “tensor power” $A^{\otimes n}=A\otimes...\otimes A$ with $n$ copies of $A$ tensored together on the right. In the case $n=0$ we define $A^{\otimes0}=\mathbb{Z}$ . Then we see that for all natural numbers $m$ and $n$ we have $A^{\otimes m}\otimes A^{\otimes n}\cong A^{\otimes(m+n)}$ .

Now we can take all these tensor powers of $A$ indexed by $\mathbb{N}$ and form the direct sum
$R(A)=\bigoplus\limits_{n\in\mathbb{N}}A^{\otimes n}$
I claim that this abelian group carries the structure of a ring. Remember that a multiplication on an abelian group $R$ that distributes over addition is equivalent to a linear function $R\otimes R\rightarrow R$ . So I want to exhibit such a function for $R(A)$ .

This is where the second isomorphism comes in. We consider the tensor product
$\left(\bigoplus\limits_{m\in\mathbb{N}}A^{\otimes m}\right)\otimes\left(\bigoplus\limits_{n\in\mathbb{N}}A^{\otimes n}\right)$
which is isomorphic to
$\bigoplus\limits_{(m,n)\in\mathbb{N}\times\mathbb{N}}\left(A^{\otimes m}\otimes A^{\otimes n}\right)$
which is isomorphic to
$\bigoplus\limits_{(m,n)\in\mathbb{N}\times\mathbb{N}}A^{\otimes(m+n)}$
Let’s change how we index these. Here we’re direct summing up over all pairs of natural numbers, but the abelian group we’re summing only depends on $m+n$ . So let’s first index by the sum of the natural numbers.
$\bigoplus\limits_{k\in\mathbb{N}}\bigoplus\limits_{m+n=k}A^{\otimes k}$

Okay, now we’ve got a single infinite direct sum over $k$ , and each term is direct sum of $k+1$ copies of $A^{\otimes k}$ . For each of these finite direct sums we can just add up the elements of $A^{\otimes k}$ from each copy. This gives a linear function $(k+1)A^{\otimes k}\rightarrow A^{\otimes k}$ . We can apply the right one to each direct summand to get a linear function
$R(A)\otimes R(A)\cong\bigoplus\limits_{k\in\mathbb{N}}\bigoplus\limits_{m+n=k}A^{\otimes k}\rightarrow\bigoplus\limits_{k\in\mathbb{N}}A^{\otimes k}=R(A)$
This is our ring structure.

As usual, there’s a universal property floating around. Any linear function $f$ from an abelian group $A$ to a ring $R$ extends uniquely to a ring homomorphism $\bar{f}$ from $R(A)$ to $R$ . Just define $\bar{f}(a_1\otimes...\otimes a_n)=f(a_1)...f(a_n)$ and extend linearly to find how $\bar{f}$ acts on a given direct summand of $R(A)$ . This justifies calling $R(A)$ the free ring on the abelian group $A$ .

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April 19, 2007 - Posted by John Armstrong | Ring theory

1 Comment »

[…] is exactly the free algebra on a vector space, and it’s just like we built the free ring on an abelian group. If we perform the construction on the dual space we get an algebra of […]

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