Lemma 10.115.2 (051N)—The Stacks project

Lemma 10.115.2. Let $R$ be a ring. Let $g \in R[x_1, \ldots , x_ n]$ be an element which is nonconstant, i.e., $g \not\in R$. For $e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_ n = 1$ the polynomial

\[ g(x_1 + x_ n^{e_1}, x_2 + x_ n^{e_2}, \ldots , x_{n - 1} + x_ n^{e_{n - 1}}, x_ n) = ax_ n^ d + \text{lower order terms in }x_ n \]

where $d > 0$ and $a \in R$ is one of the nonzero coefficients of $g$.

Proof. Write $g = \sum _{\nu \in N} a_\nu x^\nu $ with $a_\nu \in R$ not zero. Here $N$ is a finite set of multi-indices as in Lemma 10.115.1 and $x^\nu = x_1^{\nu _1} \ldots x_ n^{\nu _ n}$. Note that the leading term in

\[ (x_1 + x_ n^{e_1})^{\nu _1} \ldots (x_{n-1} + x_ n^{e_{n-1}})^{\nu _{n-1}} x_ n^{\nu _ n} \quad \text{is}\quad x_ n^{e_1\nu _1 + \ldots + e_{n-1}\nu _{n-1} + \nu _ n}. \]

Hence the lemma follows from Lemma 10.115.1 which guarantees that there is exactly one nonzero term $a_\nu x^\nu $ of $g$ which gives rise to the leading term of $g(x_1 + x_ n^{e_1}, x_2 + x_ n^{e_2}, \ldots , x_{n - 1} + x_ n^{e_{n - 1}}, x_ n)$, i.e., $a = a_\nu $ for the unique $\nu \in N$ such that $e \cdot \nu $ is maximal. $\square$

Comments (2)

Comment #2530 by Giulia Battiston on May 02, 2017 at 07:38

There is a small typo in the formula of the proof: one sees "~~is~~" instead of " is ", which makes it hardly readable.

Comment #2566 by Johan on May 25, 2017 at 18:40

Thanks Giulia. Fixed here.

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