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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?
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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 $\;\phi^{-1}(\phi(I))=2\mathbf Z/4\mathbf Z$.

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  • $\begingroup$ 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
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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 $3X+2\notin I$.

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

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