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I was reading the book (Combinatorial Commutative Algebra) of Ezra Miller, and I got stuck when they give the example of the Koszul complex (which, in the book, is given by taking the reduced chain complex of $\Delta^n$, labeling the rows and columns appropiately and renumbering the homological degrees so that the empty set sits in homological degree $0$):

They say that, for $S=k[x,y,z]$, the Koszul complex is given by $$K_{\bullet}:0\leftarrow S\stackrel{\begin{bmatrix}1 &1&1\end{bmatrix}}{\leftarrow}S^3\stackrel{\begin{bmatrix}0&1&1\\ 1&0&-1\\-1&-1&0\end{bmatrix}}{\leftarrow}S^3\stackrel{\begin{bmatrix}1\\1\\-1\end{bmatrix}}{\leftarrow}S\stackrel{}{\leftarrow}0.$$ After that, they say that this is a free resolution of $k=S/ \langle x,y,z\rangle$. So, what we need for it to be true, is that the map $S^3\to S$ in $K_\bullet$ must have as image $\langle x,y,z\rangle$.

But everything I know tells me that map is surjective, so I am confused about how can it be a free resolution. Perhaps I am misunderstanding something.

LeviathanTheEsper
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1 Answers1

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The book defines the action of a monomial matrix using the labels.

In a complex given by monomial matrices $(\lambda_{qp})$ with labels $x^{a_p},x^{a_q}$ in the columns and rows, the scalar entry $\lambda_{qp}$ indicates that the basis vector of $S(−a_p)$ should map to an element that has coefficient $\lambda_{qp}$ on the monomial that is $x^{a_p-a_q}$ times the basis vector of $S(−a_q)$.

In short, the vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ in $S^3$ are sent to $x,y,z$ in $S$ respectively. It's not the same chain complex, since $(f,g,h)\mapsto xf+yg+zh$. Thus, the image is contained in $\langle x,y,z\rangle$, and is in fact, the same. So the cokernel is $S/\langle x,y,z\rangle$.

LeviathanTheEsper
  • 3,163