Consider the good old Pascal’s triangle:


Except for the first row, take the alternating sum of the entries. So for the second row we have $ 1 – 1 = 0$. For the third row we have $ 1 – 3 + 3 – 1 = 0$. For the fourth row we have $ 1 -4 + 6 – 4 + 1 = 0$, etc. So it seems that we have the following for $ n > 0$:

$ \sum_{i=0}^n (-1)^n\binom{n}{i} = 0$

One can use the binomial theorem to prove this. In the process of reviewing some material on regular sequences, I came up with a slightly different proof that could be the most pretentious and ridiculous proof of this fact. (However, even more ridiculous proofs using heavy machinery would be welcome in the comments.) The reader may wish to consult the previous post describing the Koszul resolution before reading onwards.

Before I introduce the proof, here is a standard definition: Given any $ R$-module $ A$, and a resolution $ 0\to F_n\to \dots\to F_0\to A$ of $ A$ by free modules, the Euler characteristic of $ A$ is the alternating sum $ \chi(A) = \sum_i (-1)^i r(F_i)$ where $ r(F_i)$ is the rank of the free module $ F_i$, which is well-defined because we are assuming $ R$ to be commutative. It is a fact that the Euler characteristic of a module with a finite free resolution is well-defined; that is, it does not depend on the chosen finite-free resolution.

The Proof

Let $ I$ be an ideal in a commutative ring $ R$ generated by a regular sequence of length $ n$, such as $ (x_1,\dots,x_n)$ in the polynomial ring $ F[x_1,\dots,x_n]$ where $ F$ is a commutative ring. Since the $ k$th term of the Koszul resolution is the wedge product $ \wedge^k N$ which has rank $ \tbinom{n}{k}$, the Euler characteristic of $R/I$ is

$ \chi(R/I) = \sum_{i=0}^n (-1)^i\binom{n}{i}$

It is known that if the annihilator of a module with a finite free resolution is nonzero, then the Euler characteristic is actually zero. For an ideal generated by a regular sequence, the first element of the sequence is of course a nonzerodivisor, and hence the above sum is zero.

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