Mathematics Asked by LAD on September 9, 2020
I’m doing Exercise 5 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.
If a ring $R$ is not commutative, an element $c$ is called central in $R$ if $c r=r c$ for every $r in R$. Construct from $R$ a ring $R[x]$ containing $R$ as a subring and a central element $x$ such that to any ring $R^{prime}$ with a central element $c$ and containing $R$ as a subring there is a unique morphism $E_{c}: R[x] rightarrow R^{prime}$ with $E_{c}(r)=r$ for all $r in R$ and $E_{c}(x)=c$.
Could you please verify if my attempt is fine or contains logical mistakes? Thank you so much for your help!
Let $R[x]$ be the set of all sequences $p:mathbb N to R$ such that ${n in mathbb N mid p_n neq 0}$ is finite. For $p,q in R[x]$, we define addition and multiplication by $$begin{aligned} (p + q)_n &= p_n+q_n \ (p q)_n &= sum_{k=0}^{n} p_{k} q_{n-k}end{aligned}$$
It’s then straightforward to verify that $R[x]$ is a ring. Define $x in R[x]$ by $x_n = 1$ if $n = 1$ and $x_n = 0$ otherwise. Then $x$ is a central element of $R[x]$. Furthermore, $x^k := underbrace{xcdots x}_{k text{ times}}$ is such that $(x^k)_n = 1$ if $n = k$ and $(x^k)_n = 0$ otherwise. We identify $a in R$ with $p in R[x]$ for which $p_n = a$ if $n=0$ and $p_n = 0$ otherwise. It follows that $p in R[x]$ can be represented uniquely as $p= sum_n p_n x^n$.
Consider the map $E_c:R[x] to R’, sum_n p_n x^n mapsto sum_n p_n c^n$. By construction, this is the unique morphism satisfying the required conditions.
This looks good to me! You didn't prove all of your claims (and as a result you didn't explicitly use all of the assumptions – e.g. you never mentioned the fact that $c$ is central), but it's clear you understand what's going on so I'm sure you could. In any case, this is absolutely the correct construction.
Correct answer by diracdeltafunk on September 9, 2020
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP