# 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 $E_{c}(x)=c$

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

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