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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.

  1. The tensor algebra $\bigotimes R$ is generated by pure tensors of the form $$ r_1 \otimes \cdots \otimes r_n = (r_1 \cdots r_n) \cdot 1_R \otimes \cdots \otimes 1_R. $$
  2. 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?

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Other that simplifying into $\bigwedge R=R\oplus R$, you are completely right.

You can also go for a different approach, by defining the degree$-n$ terms as follows. Namely by $\bigwedge^nM=M^{\otimes n}/N$, where $N$ is the submodule generated by the dublicate terms. Then we have $\bigwedge M=\bigoplus_n \bigwedge^n M$.

From this definition, it is obvious that we always have $\bigwedge^0 M=R$ and $\bigwedge^1 M=M$. It is also fairly easy to see that $\bigwedge^n R=0$ for $n\geqslant 2$, as $r_1\otimes r_2\otimes ...\otimes r_n=r_1...r_n 1\otimes ...\otimes 1=0$.

So as a result, we get $\bigwedge R=\bigwedge^0 R\oplus\bigwedge^1 R=R\oplus R$.

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  • $\begingroup$ Thank you for the confirmation. While typing up the question, I got the feeling that I can identify $1_R$ with an indeterminant $X$. If I do this, can I say that the tensor algebra of $R$ is in fact a polynomial ring? In this case then, the exterior algebra quotients out the ideal $X^2$? $\endgroup$ – Boni Lindsley Jul 12 '16 at 18:04
  • $\begingroup$ This is indeed an identification that works. $\endgroup$ – Mar Jul 12 '16 at 18:17

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