# Confusion about a ring homomorphism theorem in textbook Analysis I by Amann/Escher

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!

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):

where the homomorphism (8.22) is

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

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!

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).
• 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! – LAD Apr 20 '19 at 14:58