Universal property of the polynomial ring

Builds on Polynomial ring Ring homomorphism Evaluation homomorphism

Statement

Let $R$ and $S$ be commutative rings, let $\varphi \colon R \to S$ be a ring homomorphism and let $s \in S$ . Then there is exactly one ring homomorphism $\Phi \colon R[x] \to S$ such that $\Phi(r) = \varphi(r)$ for all $r \in R$ and $\Phi(x) = s$ . It is given by $\Phi(a_0 + a_1 x + \dots + a_n x^n) = \varphi(a_0) + \varphi(a_1) s + \dots + \varphi(a_n) s^n$ .

Remarks

With $S = R$ , $\varphi$ the identity and $s = c$ , the map $\Phi$ is the evaluation homomorphism at $c$ . In the language of algebras: for each commutative $R$ -algebra $A$ and each $a \in A$ , there is exactly one $R$ -algebra homomorphism $R[x] \to A$ that sends $x$ to $a$ . This property characterizes $R[x]$ up to a unique isomorphism.

Reading path

To understand this page from the ground up, read these 7 pages in order. The foundations come first.

Show the path as a graph

Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

no evidence level

Group concept
Ring definition
Polynomial ring definition
Ring homomorphism definition
Polynomial evaluation definition
Evaluation homomorphism definition
Universal property of the polynomial ring statement

Tag	`0DQMDSVSXJ`
URI	`m2://0DQMDSVSXJ`
Kind	statement, theorem
Writing	W0 stub
MSC	13F20
Status	This copy of the site has no build data. Run `m2 status 0DQMDSVSXJ`.
Uses	Polynomial ring, Ring homomorphism, Evaluation homomorphism

Universal property of the polynomial ring

Statement

Remarks

Reading path

Further reading

What this page needs for W1

What links here