Show that there is a unique morphism $E_{c}: R[x] rightarrow R^{prime}$ with $E_{c}(r)=r$ for all $r in R$ and $E_{c}(x)=c$

Mathematics Asked by LAD on September 9, 2020

I’m doing Exercise 5 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

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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.

One Answer

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

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