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I am reading Section 7: Groups and Homomorphisms, Chapter 1: Foundation, textbook Analysis I by Herbert Amann and Joachim Escher.

First of all, I am so sorry for posting many screenshots. Since the information is too complicated for me to summarize, I have no way but to do so. There is a proof of Remark 8.20(c) that I could not understand. I have been stuck at this proof for two weeks in spite of re-reading the proof many times. Please help me get over it!

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I got stuck at the below arguments in Remark 8.20(c).

Clearly, $p = \sum_{\alpha} p_\alpha X^\alpha$ can be written in the form $$\sum_{j=0}^n q_j X^j_m$$ for suitable $n \in \Bbb N$ and $q_j \in K[X_1,\cdots,X_{m-1}]$. This suggests a proof by induction on the number of indeterminates: For $m = 1$, the claim is true by Remark 8.19(d).

Here is the Remark 8.19(d):

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where the homomorphism (8.22) is

enter image description here


My questions:

  1. For $m=1$, $m-1=0$. What is $K[X_0]$? The index of $X$ starts from $1$ and the authors say $K[X_1,\cdots,X_{m-1}]$.

  2. What is $X^j_m$?

Previously, the authors define

enter image description here

In my understanding, $X^j_m = \begin{cases}1, &j=m \\ 0, &j\neq m \end{cases}$. As a result, $X^j_m \in K$.

  1. I can not see how Remark 8.19(d) helps to prove the case where $m=1$. Please elaborate on this points!

Thank you for your help!

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  1. The induction argument using $K[X_1,\dots,X_{m-1}]$ is only used for $m>1$. If you had to interpret it when $m=1$ you would probably say that $K[X_1,\dots,X_{m-1}]=K$, because the indexing set is empty in that case, but this is not necessary here, you can just ignore it when $m=1$ if it confuses you.

  2. You are mixing different notations (granted, you are not helped by the notations in the book which are incredibly confusing in my opinion). When they define $X_\beta^\alpha$ as $0$ or $1$, what they mean is that $X^\alpha$ is a function from $\mathbb{N}^m$ to $R$, and that its value on $\beta\in \mathbb{N}^m$ is $1$ when $\alpha=\beta$. On the other hand, $X_m^j$ is just the $j$th power of the element $X_m$ in the ring $R[X_1,\dots,X_m]$. If you want to be really formal, $X_m$ is the element $X^\alpha$ for $\alpha=(0,\dots,0,1)$, with the $1$ in the $m$th slot. So if you really want to see $X_m^j$ as a function $\mathbb{N}^m\to R$ (which I don't think is not such a good idea), it is the function which sends $\beta\in \mathbb{N}^m$ to $1$ if $\beta = (0,\dots,0,j)$, and to $0$ otherwise.

  3. I'm not sure I understand this question, remark 8.19(d) is exactly the case $m=1$, so I would think the link is quite clear. The morphism described in (8.22) is precisely the morphism described in (8.30) when $m=1$ (since $R[X]$ is $R[X_1,\dots,X_m]$ when $m=1$), so saying that it is injective (which is remark 8.19(d)) is the case $m=1$ of remark 8.20(d).

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  • $\begingroup$ Thank you so much for your detailed answer! As you explained, $X^j_m \in K[X_1,\dots,X_m]$. As the authors mention, $q_j \in K[X_1,\cdots,X_{m-1}]$. As a result, $q_j$ and $X^j_m$ belong to two different rings of polynomials. I could not understand how the product $q_j X^j_m$ makes sense. Please elaborate more on this point! $\endgroup$ – LAD Apr 20 '19 at 14:58

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