In the book "Cohen-Macaulay rings" by Bruns and Herzog, the quick introduction of tensor algebra and exterior algebra left me a bit bewildered. After referring to the section on tensor algebra from The Stacks Project, I would like to check that my understanding is correct with what should be a basic example.
Let $R$ be a commutative unital ring. I would like to consider the tensor and exterior algebrae of this ring as a module over itself.
- The tensor algebra $\bigotimes R$ is generated by pure tensors of the form $$ r_1 \otimes \cdots \otimes r_n = (r_1 \cdots r_n) \cdot 1_R \otimes \cdots \otimes 1_R. $$
- The exterior algebra $\bigwedge M$ of an $R$-module $M$ is the tensor algebra $\bigotimes M$ in which terms with duplicate "factors" are identified with $0$. In the case of the ring $M = R$, because all pure tensors can be rewritten as above, all components of degree greater than $1$ is removed, leaving behind $$ \bigwedge R = R \oplus {\bigwedge}^1 R. $$
I am posting here because this is not what I had expected. When I was looking at this example, going by my intuition that duplicates are removed, I originally set out to show that the exterior algebra is in fact just $R$ itself. The extra "base ring" $R$ of degree $0$ seems to break my intuition of how it works.
Am I missing a step in how to simply the direct sum further?