Let $\phi: R \to R'$ be a ring isomorphism and $I$ an ideal of $R$. Define $\phi(I)=\{\phi(i): i \in I\}$.
Show that $\frac RI \cong \frac {R'}{\phi(I)}$.

To use the first isomorphism theorem, I was trying to show that the kernel of $\pi \circ \phi$ was $I$, where $\pi: R'\to \frac {R'}{\phi(I)}$. It seems to me this follows from the definition of $I$, but my professor said I needed to use the injectivity of $\phi$ for one of the steps in one of the inclusions. I marked the step with an asterisk:

$I \supseteq \ker\pi \circ \phi$: $$ i \in \ker\pi \circ \phi\implies \overline{\phi(i)}=\overline 0 \implies \phi(i)\in \phi(I)\overset{\ast}{\implies} i \in I$$

So these are my questions:

  1. Why doesn't this just follow from the definition of $\phi(I)$?
  2. If we do need the injectivity here, there must be an example where $\phi$ is not injective and there's an element $\phi(i)$ in $\phi(I)$ where $i \notin I$, but I can't think of it. Can you give me an example of this?

At the point where you have $\phi(i)\phi(I)$, you deduce only $i\in \phi^{-1}(\phi(I))$, which is equal to $I$ if $\phi$ is injective.

To have another example, take the canonical morphism $\mathbf Z/4\mathbf Z\to \mathbf Z/2\mathbf Z$, and the ideal $I=0$ in $\mathbf Z/4\mathbf Z$. Then $\;\phi^{-1}(\phi(I))=2\mathbf Z/4\mathbf Z$.

  • $\begingroup$ In your example, the canonical morphism sends $0$ and $2$ to $0$, so $\phi^{-1}(\phi(I))=\phi^{-1}((0))=\{0, 2\} \cong 2\Bbb Z/4\Bbb Z$, right? $\endgroup$ – a girl Jul 30 '15 at 0:32
  • $\begingroup$ That's correct. $\endgroup$ – Bernard Jul 30 '15 at 0:33

For any polynomial $f\in \Bbb Z[X]$ define $\phi(f)=f(0)$. Here $\phi$ is a surjective morphism from $\Bbb Z[X]$ to $\Bbb Z$. Consider the ideal $I=\langle X+2\rangle\subset \Bbb Z[X]$. Clearly $2\in \phi(I)$ and $\phi(3X+2)=2$, but $3X+2\notin I$.

  • $\begingroup$ Thank you for this example. This is all very clear to me now :) $\endgroup$ – a girl Jul 30 '15 at 0:34

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.