Why can't $\prod_{i=1}^d a_i^{e_i}$ be in the ideal $(a_1^{e_1 + 1}, \ldots, a_d^{e_d+1})$?

Question

Let $A$ be a regular local ring of dimension $d$ with maximal ideal $\mathfrak{m} = (a_1,\ldots,a_d)$. Let further $e_1,\ldots,e_d$ be natural numbers and set $$a := \prod_{i=1}^d a_i^{e_i}.$$ Now assume I want to prove something by contradiction and I am able to show that $$\tag{1} a \in (a_1^{e_1+1},\ldots,a_d^{e_d+1}) =: I,$$ is there an easy way to get an obvious contradiction from here on?

Note that regularity is necessary here: Let $R = K[X^2,XY,Y^2]_{(X^2,XY,Y^2)}$ with maximal ideal $(X^2,XY,Y^2)$, then $R$ is not regular. If $(e_1, e_2, e_3) = (1,2,1)$, then $y = X^2 \cdot (XY)^2 \cdot Y^2 = X^4 Y^4$, which is contained in the ideal $I = (X^4, X^3 Y^3, Y^3)$.

What I have done so far: In simple examples one can be convinced that the statement $(1)$ is a contradiction because the rings are sufficiently nice.

Example for nice situation: Let $R = K[X,Y,Z]_{(X,Y,Z)}$ with maximal ideal $(X,Y,Z)$ and assume I was able to show that e.g. $XY^3 \in (X^2, Y^4, Z).$ This is should be wrong but even in this simple example I am not able to deduct a clear contradiction. This is wrong as xavierm02 pointed out, but imho the general argument should use the regularity of $R$ more explicitly, maybe by using that $R$ is a unique factorization domain or that $a_1^{e_1+1},\ldots,a_d^{e_d+1}$ is a regular sequence.

Another nice situation is $d = 1$. In this case $R$ is a discrete valuation ring and the statement would reduce to $a^e \in (a^{e+1})$, i.e. $a^e = u \cdot a^{e+1}$ which is absurd.

Any help would be greatly appreciated.

Answer 1

There is probably a simpler proof, but here is one way to prove this. The Lemma and Theorem below are from Hochster's notes on tight closure, starting on p. 118.

Let $R$ be a commutative ring.

Definition. Let $x_1,x_2,\ldots,x_n \in R$ and let $M$ be an $R$-module. The sequence $x_1,x_2,\ldots,x_n$ is a possibly improper regular sequence on $M$ if $x_1$ is a nonzerodivisor on $M$ and if, for all $0 \le i \le n-1$, the element $x_{i+1}$ is a nonzerodivisor on $M/(x_1,x_2,\ldots,x_i)$. A possibly improper regular sequence $x_1,x_2,\ldots,x_n$ is a regular sequence on $M$ if $(x_1,x_2,\ldots,x_n)M \ne M$.

Lemma. Let $M$ be an $R$-module, and let $x_1,x_2,\ldots,x_n$ be a possibly improper regular sequence on $M$. If $u_1,u_2,\ldots,u_n \in M$ are elements such that $$ \sum_{j=1}^n x_ju_j = 0,\tag{$*$}\label{eq:lem} $$ then we have $u_j \in (x_1,x_2,\ldots,x_n)M$ for all $j$.

Proof. We induce on $n$. If $n = 1$, then \eqref{eq:lem} reads $x_1u_1 = 0$, which implies $u_1 = 0 \in (x_1,x_2,\ldots,x_n)M$ since $x_1$ is a nonzerodivisor on $M$ by definition of a possibly improper regular sequence. Now suppose $n > 1$. Then, by reducing \eqref{eq:lem} modulo $(x_1,x_2,\ldots,x_{n-1})$, we have $$x_nu_n \equiv 0 \mod (x_1,x_2,\ldots,x_{n-1})M.$$ By definition of a possibly improper regular sequence, we have $$u_n = \sum_{j=1}^{n-1}x_jv_j$$ for some $v_1,v_2,\ldots,v_{n-1} \in M$, hence $$\sum_{j=1}^{n-1} x_j(u_j + x_nv_j) = 0.$$ By the inductive hypothesis, this implies $u_j + x_nv_j \in (x_1,x_2,\ldots,x_{n-1})M$ for all $j$, and therefore $u_j \in (x_1,x_2,\ldots,x_n)M$ for all $j$. $\blacksquare$

Theorem. Let $x_1,x_2,\ldots,x_n$ be a regular sequence on a $R$-module $M$, and let $I$ be the ideal $(x_1,x_2,\ldots,x_n) \subset R$. Let $a_1,a_2,\ldots,a_n$ be nonnegative integers, and suppose $u,u_1,u_2,\ldots,u_n \in M$ are such that $$x_1^{a_1}x_2^{a_2} \cdots x_n^{a_n} u = \sum_{j=1}^n x_j^{a_j+1}u_j.\tag{$**$}\label{eq:thm}$$ Then, $u \in IM$.

Proof. We induce on the number of nonzero $a_j$. If $a_j = 0$ for all $j$, then we are done, since \eqref{eq:thm} reads $u = \sum_{j=1}^n x_ju_j$. Now suppose $a_i > 0$ for some $i$, and let $y = \prod_{j \ne i} x_j^{a_j}$. We can then rearrange \eqref{eq:thm} to read $$\sum_{j \ne i} x_j^{a_j+1} u_j - x_i^{a_i}(yu-x_iu_i) = 0.$$ Since taking powers of the $x_j$ still gives a regular sequence, the Lemma above implies that $$yu-x_iu_i \in x_i^{a_i}M + (x_j^{a_j+1})_{j \ne i}\,M,$$ and so $$yu = \biggl( \prod_{j \ne i} x_j^{a_j} \biggr) u \in x_i M + (x_j^{a_j+1})_{j \ne i}\,M.$$ By inductive hypothesis, we then have that $u \in IM$. $\blacksquare$

Corollary. Let $x_1,x_2,\ldots,x_n$ be a regular sequence on $R$, and let $a_1,a_2,\ldots,a_n$ be nonnegative integers. Then, we have $$x_1^{a_1}x_2^{a_2} \cdots x_n^{a_n} \notin (x_1^{a_1+1},x_2^{a_2+1},\ldots,x_n^{a_n+1}).$$

Proof. If the inclusion does hold, then the Theorem implies $1 \in (x_1,x_2,\ldots,x_n)$, contradicting the fact that $x_1,x_2,\ldots,x_n$ is a regular sequence. $\blacksquare$