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I was reading about Koszul Complex and I am fairly new to the topic. So lemme start with a ring $R$, not necesarilly commutative and let $x\in R$ be central.

Denote $K(x)$ to be the chain complex $0\to R\to R \to 0$ and the map from $R$ to $R$ is just multiplying by $x$.

I was reading a proof which starts with:

Considering the following natural SES of chain complexes $$0\to R\to K(x)\to R[-1]\to 0$$

I was wondering what complex is denoted by $R$ alone?

I thought about a couple possibilities but none seem to give a SES of chain complexes. For instance, I thought about $\cdots\to R\to R\to \cdots$ where morphisms are just identity ones but that does not give a SES.

I couldn't seem to find any more background on this matter in the notes but if it is the preferred to post the entire proof, I will try to do so.

Thanks in advance!

UnsinkableSam
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To me it looks like the complex which has just an $R$ term in degree $0$. Hence with homological grading the degree $0$ SES looks like $$0\to R\to R\to 0\to 0$$ and degree $-1$ looks like $$0\to 0\to R\to R\to 0$$ which are both exact. At any other degree you just get a SES of zeroes.

Jackozee Hakkiuz
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