Understanding the bit of the analysis map

Let $R$ be a non - absolutely no commutative ring. Confirm that the excellent $(x)$ of $R[x]$ is prime if and also just if $R$ is an indispensable domain name.

I'm working with the left - to - appropriate instructions now. I recognize that $R[x]/(x)$ is an indispensable domain name given that $(x)$ is prime. So I intend to deal with $r\in R$ and also make use of the analysis map $e_r:R[x]\to R$ offered by $f(x)\mapsto f(r)$, a surjective ring homomorphism, and also use the first isomorphism theory. Yet I'm having problem with the bit of $e_r$.

First off, is it also real that $\ker{(e_r)}=(x)$? If so, any kind of tips you can go down would certainly be wonderful. Many thanks!

0
2019-12-02 02:53:12
Source Share
Answers: 2

No, the kernel of $e_r$ contains all polynomials that are absolutely no at $r$ : bear in mind, $\mathrm{ker}(e_r) = \{ f(x)\in R[x]\mid e_r(f(x)) = f(r) = 0\}$. It will certainly have $(x-r)$ (thinking your ring has an identification), yet as a whole rings it might be all type of points ( modify : as Bill Dubuque mentions, the Factor Theorem will certainly get on hand in any kind of ring with $1$, yet in rings without $1$ unusual points might take place). Yet that does not actually matter below, due to the fact that you do not intend to consider an approximate $r$ in $R$. You intend to select a certain $r$ that offers you the kernel you desire.

And also for you to get $(x)$ as the kernel, you need to consider $e_0$, analysis at $0$. Trivially, $(x)$ is had in the kernel ; revealing that every component of the kernel remains in $(x)$ needs to be uncomplicated.

0
2019-12-03 04:19:57
Source

You get on the appropriate track by using : $\rm\ A/I\ $ is a domain name iff $\rm\ I\ $ is prime. For $\rm\ A = R[x],\ \ I = (x)\ $ the trouble lowers to revealing that $\rm R[x]/(x)\cong R\:.\:$ As you theorize, this is conveniently attained by showing an analysis map with bit $\rm\ (x)\:,\:$ i.e. an analysis which maps $\rm\ x\to \ \ldots$?

Probably you currently recognize the presence (and also individuality) of such analysis maps for polynomial rings. Actually this universal mapping property identifies polynomial rings. It can be used to offer glossy theoretical evidence of outcomes concerning polynomial rings. For a straightforward instance see this proof that $\rm\:x\:$ is not invertible in $\rm\: R[x]\:.$ The relevance of such global characterizations will certainly come to be more clear when you research global algebra and also category theory.

0
2019-12-03 04:19:27
Source