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I'd really like to understand the proof that Paul Monsky wrote about the finiteness of the de Rham cohomology of algebraic varieties. I'd like it very much because it seems to explain in concrete terms some of Dwork's deformation theory and it may help me to proceed to concrete computations.

MR301017 — Monsky, P. “Finiteness of de Rham cohomology”. Amer. J. Math. 94 (1972), 237–245.

Unfortunately, I'm stuck quite early in the proof, lemma 2.1 in fact. If I may, since the paragraph I'm stuck with is short and self-contained, let me show it to you.

(the last words are missing, they say: “the lemma follows”.)

I think I'm OK with every single word of this extract, however, I don't understand why a proof follows...

In fact, I don't even understand why this lemma can conceivably be true... Indeed, its says that two Koszul homology are isomorphic, but the first one is built with $n$ operators whereas the second is built with $n+1$ operators, so that the $H_{n+1}$ of the first is always zero but the $H_{n+1}$ of the second may be non zero, a priori.

The two questions I'd like to ask you are:

  • Why am I wrong about the $H_{n+1}$?
  • Why does “the lemma follow”?

Thank you very much!

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As for your issue with the $n+1$-st term, observe that it is $0$ for both, since $H_{n+1}$ of the second complex is the kernel of $\frac{\partial}{\partial T}+ f$ which is injective (use $f$ is not a zero-divisor).

The "lemma follows" since Monsky has reduced the homology of the first complex to that of the complex $K_\cdot(A[T];E_1+L_{Tf_1},\dots,E_n+L_{Tf_n},\frac{\partial}{\partial T}+ L_f)$ (I assume he uses the notation $L_x$ to mean the operator "multiplication by $x$"). Now, the big exact sequence and the considerations right after it show that the homology of my last complex is the one of the first complex in his statement, by shifting: $H_1$ is the cokernel of the last operator $\frac{\partial}{\partial T}+ L_f$ which is $A_f/A$ and each other $H_i$ is the $\text{ker}/\text{coker}$ of $E_i+L_{Tf_i}$ on $A[T]$ which coincides with $E_i$ on $A_f/A$.

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    $\begingroup$ Thanks ! So know I understand the lemma, and I can write a proof of it with the explicit map of complexes from the second (in the lemma) to the first, given by $\alpha + \beta\wedge e_{n+1} \mapsto \alpha$, and checking everything. However, I hardly understand why you and Monsky can curcumvent this computation. Is that just trivial for someone used to Koszul homology ? I'm not sure to understand you shifting argument. $\endgroup$
    – Lierre
    Sep 1, 2012 at 13:04
  • $\begingroup$ The "shifting argument" simply means that if you have a complex $C_\dot$, then the complex $\cdots\to C_0\stackrel{0}{\to}C_0$ has the same homology. And this is what happens in Monsky's paper, by observing $A_f/A=A[T]/(\partial_T+L_f)A[T]$. Tha arrows are simply $x\wedge(e_1\wedge\dots\wedge e_s)\mapsto \bar{x}\wedge(e_1\wedge\dots\wedge e_s)$ where "bar" is reduction $\pmod{(\partial_T+L_f)A[T]}$. $\endgroup$
    – Filippo Alberto Edoardo
    Sep 3, 2012 at 0:35

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