In the book Homology of Local Rings of Gulliksen and Levin, it is mentioned that if we set $R=k[x_1,…,x_n]/\mathfrak{m}^d$ where $\mathfrak{m}$ is the homogeneous maximal ideal and consider the Koszul complex $K(R;x_1,…,x_n)$, "it is easy to see that"
$$Z_i(K)=\mathfrak{m}^{d-1}K_i+B_i(K)$$
for all $1\leq i\leq n$. I don't see how.
I can write down the differential $\partial_1,\partial_2$ and $\partial_n$ explicitly, so I can confirm the above statement for $i=1$ and $i=n$. In general I don't exactly know what the differentials look like.
In general all the matrices for the differentials have coefficients in $\mathfrak{m}$, hence one direction is clear. That's all I know
Is there a fast way to see this? Or is the only way to predict the differentials and compute $Z_i(K)$ and $B_i(K)$ explicitly?
Best Answer
The differential. Write $R = \mathbb k[y_1,\ldots,y_n]$ for your ring, so that any $d$-fold product of $y$s is zero, and let me write $\mathcal K(R,\mathbf y)$ for the Koszul complex of $R$. This complex is generated in degree $p$ by a sum $$\mathcal K_p(R,\mathbf y) = \bigoplus_{I\subseteq [p]} R\otimes \mathbb ke_I$$ and the differential sends $$ f\otimes e_I \longmapsto \sum_{i\in I} (-1)^i f\cdot y_i\otimes e_{I\smallsetminus i}.$$ That is, the differential is the same as the usual Koszul differential you know.
Cycles and boundaries. Now, you want to check that in this complex, we have that $Z = B + \mathfrak m^{d-1}K$.
One inclusion is easier than the other: of course $B\subseteq Z$, and because the differential above always multiplies by $y$s, then whenever you pick something in $\mathfrak m^{d-1}K$, you will land in $\mathfrak m^dK=0$. So $B+\mathfrak{m}^{d-1}K\subseteq Z$ is definitely true.
For the converse, suppose that you have an element $z\in Z$, that is, $dz=0$. You would like to write it as $dz' + \alpha$ where $\alpha\in \mathfrak{m}^{d-1}K$. We proceed by induction on $d$, noting that the case when $d=1$ the ring $R$ is just $\mathbb k$ and in this case the statement is obvious (all differentials are zero).
For the inductive step, consider the quotient map $R \longrightarrow R'$ where in $R'$ we quotient by $\mathfrak m^{d-1}$, which induces a map on Koszul complexes $\mathcal K(R,\mathbf y) \longrightarrow \mathcal K(R',\mathbf y')$. Given a cycle $z$ in the domain, its class $[z]$ is a cycle in the codomain, so $[z] = [dz_1] + [z_2]$ by induction, where $[z_2]\in \mathfrak m^{d-2}K$. This equality in the quotient complex simply means that $z = dz_1 + z_2 + z_3$ where $z_3\in \mathfrak m^{d-1}K$, so it suffices we address the $z_2$ term.
Now $z$ is a cycle and so are $dz_1$ and $z_3$, so $dz_2=0$. Thus, we are reduced to show the following:
Proof. Since $\mathfrak m^d=0$, we can assume that when we write down $f$, we only use elements in $R$ that are of the form $\lambda + \langle\mu,y\rangle$, that is, either constant or linear in the $y$s. But since multiplying $ \mathfrak m^{d-2}$ by a purely linear term lands in $ \mathfrak m^{d-1}$, we can simply assume that each $f_J$ is a linear combination of homogeneous elements of degree $d-2$ i.e. $f_J = \sum_{|\alpha|=d-2} \lambda_{\alpha,J} x^{\alpha} \otimes e_J$ where $x^\alpha = x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ and $|\alpha | = \sum_i \alpha_i$.
In this case, we see that $d(x^\alpha\otimes e_I) = \sum_{i\in I} (-1)^i x^{\alpha+e_i}\otimes e_I$ and are reduced to showing that
Proof. This follows from Euler's formula that if $h$ is homogeneous of degree $\lambda$ then $\langle x,\nabla h\rangle = \lambda h$, which gives a contracting homotopy for the 'Koszul differential' $x\cdot h$.