4
$\begingroup$

I am studying by myself and I needed help for few question which I am confused how give proof of that. Let $\varphi : J \to K$ be a ring epimorphism with $\varphi(1) = 1$, where $J$ and $K$ are commutative rings with $1$. Prove the following or give a choice of $J$, $K$, and $\varphi$ where the claim fails.

  1. If $J$ is an integral domain, then $\varphi(J)$ is an integral domain.
  2. If $(k) \unlhd K$ is a principal ideal, i. e., generated by a single element, then the preimage $\varphi^{-1}((k))$ is a principal ideal in $J$.

Help me to understand to how to solve this question?

$\endgroup$
  • $\begingroup$ For your second question, what is $S$? Also, do you mean $\varphi^{-1}[(s)]$? $\endgroup$ – Hayden Mar 31 '14 at 17:37
  • $\begingroup$ Also, what is $R$? You keep on introducing new objects without indication of what they are $\endgroup$ – Hayden Mar 31 '14 at 17:39
  • $\begingroup$ I have updated my question. $\endgroup$ – user1413 Mar 31 '14 at 17:41
  • $\begingroup$ When you work in the category of (commutative) unital rings you don't have to mention $\varphi(1) = 1$, that is part of the definition. Is $\varphi$ really just an epimorphism in the categorical sense or do you mean a surjective homomorphism? $\endgroup$ – Dune Mar 31 '14 at 17:48
  • $\begingroup$ Yes I do mean Surjective Homomorphism!! $\endgroup$ – user1413 Mar 31 '14 at 17:58
3
$\begingroup$

I will assume that $\varphi: J\rightarrow K$ because of the mixed notation.

For the first question, suppose $J=\mathbb{Z}$ and $K=\mathbb{Z}/m\mathbb{Z}$, where $m$ is not prime, and $\varphi: J\rightarrow K$ is the canonical quotient transformation (which is surjective and thus epic). $\mathbb{Z}/m\mathbb{Z}$ is not an integral domain; if $m=ab$, then $ab\equiv 0 \mod m$ but $a,b$ are non-zero modulo $m$.

For the second question, take $J=\mathbb{F}[x,y]$ and $K=\mathbb{F}$. Then take $\varphi:\mathbb{F}[x,y]\rightarrow \mathbb{F}$ such that $\varphi$ is constant on constant polynomials and sends $x$ and $y$ to 0. The kernel of this is the (non-principal ideal) $(x,y)$. Thus, we take $k=0$, so that $(0)=\{0\}=\varphi(\ker \varphi)$ and thus $\varphi^{-1}[(0)]=\ker\varphi$ is not a principal ideal.

$\endgroup$
  • $\begingroup$ So therefore $\varphi(J)$ is not integral domain. right? $\endgroup$ – user1413 Mar 31 '14 at 17:54
  • $\begingroup$ Yep, it isn't required to be an integral domain $\endgroup$ – Hayden Mar 31 '14 at 18:01
  • $\begingroup$ Oh ok got it. So the proof that I thought was right. Appreciate all for your help. $\endgroup$ – user1413 Mar 31 '14 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.