# on the proof of Theorem 4.3.2 in Bruns & Herzog Cohen-Macaulay Rings" (Gotzmann's regularity theorem)

The theorem and the first part of its proof is shown below:

In particular, the authors conclude (2 lines below equation (2)) that

$(i): P_R(n) = {n + a_1 \choose a_1}+\cdots+ {n+a_r -(r-1) \choose a_r} + c$.

Question 1: Is there an immediate way to see how they arrived at this equation?

Question 2: It is not clear at first sight that $c$ is even an integer. Or it is? Anyway, the authors consider it obvious that $c$ is an integer and they proceed to show that it is non-negative via a rather complicated argument.

In trying to understand equation $(i)$ i came up with the following argument, which not only shows that $c$ is an integer, but that it is also non-negative:

My argument: There exists a unique expression

$(ii): P_R(n) = {n + a_1 \choose a_1}+{n + a_2-1 \choose a_2}+\cdots+ {n+a_s -(s-1) \choose a_s}$

with $a_1 \ge a_2 \ge \cdots \ge a_s \ge 0$. As in the argument in the proof, suppose that $P_R(n) - P_R(n-1)$ has a unique expression

$P_R(n) - P_R(n-1) = {n + b_1 \choose b_1}+\cdots+ {n+b_r -(r-1) \choose b_r}$

with $b_1 \ge b_2 \ge \cdots \ge b_r \ge 0$. Then using Pascal's triangle we must have that

${n + b_1 \choose b_1}+\cdots+ {n+b_r -(r-1) \choose b_r} = {n + a_1-1 \choose a_1-1}+\cdots+ {n+a_s-1 -(s-1) \choose a_s-1}$.

By the uniqueness of the expression on the left and the fact that $b_i \ge 0$ we can identify $a_i = b_i + 1$ for $i=1,\dots,r$ and we must also have that

${n + a_{r+1}-1-r \choose a_{r+1}-1}+\cdots+ {n+a_s-1 -(s-1) \choose a_s-1}=0$.

The only way this last equation can be satisfied is if $a_{r+1}=\cdots=a_s=0$. Substituting this in $(ii)$ we get precisely equation $(i)$, where $c:=s-r \ge 0$.

Question 3: Any comments regarding my argument?

• At this moment I can't see any flaw in your argument. (The only thing I'd change is that $a_{r+1}=⋯=a_s=0$ follows from the uniqueness, no need to write that sum equal to $0$.) Commented Dec 21, 2014 at 18:37
• @user26857: So basically we are using the uniqueness of the expression in exercise 4.2.17, right? Strange however that the exercise itself does not mention anything about uniqueness. Commented Dec 21, 2014 at 19:26
• I'd say so. Anyway, the uniqueness is out of any doubt. Commented Dec 21, 2014 at 19:34
• @user26857: How do you see that the uniqueness is out of any doubt? Commented Dec 21, 2014 at 19:38
• That equality for any $n\in\mathbb Z$ says that $P_R(X)$ is a sum of "combinatorial" polynomials and this is unique (by degree arguments). Commented Dec 21, 2014 at 19:50

It seems they used $P_R(k) - P_R(k-1) = {k + b_1 \choose b_1}+\cdots+ {k+b_r -(r-1) \choose b_r}$ by summing these up for $k=1,\dots,n$ and using the well known combinatorial identity $\sum_{k=0}^n{k + b_1 \choose b_1}={n + b_1 + 1\choose b_1 + 1}$. Then $c$ is nothing but $P_R(0)-1$ which is obviously an integer.