Well, I've been learning Tensor algebra and related topics.

So, I wonder tensor algebra is related to somehow familiar with me, so called group ring.

$\mathbb{Z}$: a ring of integers.

Let G be a abelian group, i.e., G is a $\mathbb{Z}$-module.

Now we consider group ring of G over $\mathbb{Z}$, say $\mathbb{Z}$(G).

Since G is (bi)$\mathbb{Z}$-module (Recall $\mathbb{Z}$ is commutative ring), So we can consider tensor algebra of G, T(G).

I would say T(G) is isomorphic to $\mathbb{Z}$(G).

Is it correct? OR Is it related to each other somehow?

Thank you!


One thing to notice first off is that $\mathbb{Z}G$ is a commutative ring if $G$ is abelian, whereas $T(G)$ is not. In $\mathbb{Z}G$ we start with an abelian group operation (which we think of as multiplication) and we impose this "free" addition structure on things to make it a ring, which yields a commutative ring if $G$ is abelian. In $T(G)$ we start out with a group operation which we think of as addition, and impose this "free" (noncommutative) multiplication structure on things to make it a ring. One thing to note is that in the former construction, we could have started with a non-abelian group, whereas in constructing the tensor algebra, we need an abelian group.

From the above description, it should be clear that even if we made the tensor multiplication commutative (and so took the symmetric algebra) the two structures would not coincide. The multiplicative identity in $\mathbb{Z}G$ is $e$, the identity element of $G$, whereas the multiplicative identity of $T(G)$ is the integer $1$.

In the language of category theory, the tensor algebra functor $\mathsf{Ab} \to \mathsf{Ring}: A \mapsto T(A)$ is left adjoint to the functor $\mathsf{Ring} \to \mathsf{Ab}$ which forgets the multiplicative structure. In other words, the tensor algebra is the most general multiplicative structure we could impose on an abelian group. The group ring functor $\mathsf{Grp} \to \mathsf{Ring}$ (or $\mathsf{Ab} \to \mathsf{Comm}$, if you like) taking $G \mapsto \mathbb{Z}G$, is left adjoint to the functor $\mathsf{Ring} \to \mathsf{Grp}$ which takes a ring to its group of multiplicative units. This is reminiscent of a "forgetful" functor, but not quite, because it is not faithful.

So both structures are the "most general" version of an operation in some sense. But they're different operations, which can be confusing since we can write abelian groups additively or multiplicatively.

  • $\begingroup$ Thank you very much! Can you tell me that the meaning of 'the multiplicative identity of T(G) is the integer 1'? I can't understand why '1' is the multiplicative identity of T(G). $\endgroup$ – nicksohn Dec 8 '15 at 7:21
  • $\begingroup$ @nicksohn It is indeed a little confusing at first. In addition to all the tensor powers $G$, $G \otimes G$, $G \otimes G \otimes G$ which are summands in the tensor algebra (these are the degree 1, 2, and 3 terms), there is also a degree zero term which is a copy of $\mathbb{Z}$. Multiplication with these degree zero tensors works just like multiplication by a scalar. $\endgroup$ – Eric Auld Dec 8 '15 at 7:29
  • $\begingroup$ AH! I got it!! Thank you!! and one more question! Why the functor Ring ->Grp which is left adjoint to Group Ring functor is not faithful? is there specific a example? $\endgroup$ – nicksohn Dec 8 '15 at 7:32
  • $\begingroup$ I'm glad. :-) To see it's not faithful, the functor takes $\mathbb{Z}[X]$ to $\{\pm 1\}$ which are the only units. Now consider the ring maps from $\mathbb{Z}[X]$ to an arbitrary ring $R$. They are in bijection with the elements of $R$...we just have to decide where to send $X$. But all of these maps go to the same map under the functor. $\endgroup$ – Eric Auld Dec 8 '15 at 7:39

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