Showing a commutative ring with field subring has a unique ring homomorphism on polynomials

Question

Consider the following problem, given without solution in a german abstract algebra text book:

Let $R$ be a commutative ring with $1$ and $K \subseteq R$ a subring, such that $1 \in K$ and $K$ is a field.

Show that for each $r\in R$ there exists a unique ring homomorphism $\varphi: K \left[X\right] \rightarrow R$, such that $\varphi(a) = a$ for $a \in K$ and $\varphi(X) = r$.

$K\left[X\right]$ is the ring of all polynomials over $K$ with variable $X$.

I do not know at all about how to approach this. This is the beginning of a very introductory course on abstract algebra, and I feel like I'm missing the required tools here. How can this be proven?

Define, for $p\in K[X]$, $\phi(p)=p(a)$ and check the definition of homomorphism. — conditionalMethod, Oct 26 '19 at 17:08
@conditionalMethod I do not quite follow - since $a$ is not a fixed value here, how am I allowed to define $\phi$ like that without binding $a$? The only thing fixed is $r$ — badapple11, Oct 26 '19 at 17:22
@conditionalMethod Yea, that worked. The only thing left is to show that this homomorphism is unique, which I am now stuck at. Any hints for that? :) — badapple11, Oct 26 '19 at 17:56

score 3 · Answer 1 · answered Oct 26 '19 at 17:15

3

This is immediate from the universal property that defines the polynomial ring.

answered Oct 26 '19 at 17:15

Hagen von Eitzen

374,180

Showing a commutative ring with field subring has a unique ring homomorphism on polynomials

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