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Let $\phi:R\to R'$ be a ring homomorphism. Let $N'$ be either an ideal of either $\phi[R]$ or of $R'$. Show that $\phi^{-1}[N']$ is an ideal of $R$.

My attempt so far:

$N'$ is an ideal of $R'$ or $\phi[R]$, so $aN'\subseteq N',\forall a\in R'$ and $N'b\subseteq N',\forall b\in R'$. Since $aN'\subseteq N'$, $\phi^{-1}(aN')\subseteq\phi^{-1}[N']$. But, $\phi$ being a homomorphism does not necessarily imply that $\phi^{-1}$ is a homomorphism. How do I show that it is, in this case? Or what other direction should I try to prove this?

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    $\begingroup$ If $N'$ is an ideal of $\phi(R)$, then you don't necessarily have $aN' \subseteq N'$ for $a \in R'$. Example: $R = \mathbb{Z}, R' = \mathbb{Z}[X]$, and $\phi$ is the inclusion map. Then $2\mathbb{Z}$ is an ideal of $\phi(R) = \mathbb{Z}$, but is clearly not closed under arbitrary multiplication by elements of $\mathbb{Z}[X]$. $\endgroup$ – D_S Jan 24 '16 at 22:08
  • $\begingroup$ Isn't $R'$ a subset of $\phi[R]$? $\endgroup$ – Matt G Jan 24 '16 at 22:10
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    $\begingroup$ Generally not, but $\phi(R)$ is always a subset of $R'$. $\endgroup$ – D_S Jan 24 '16 at 22:11
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$\phi^{-1}$ as it is meant here is not a homomorphism; it is not even a function. By $\phi^{-1} (N')$, what is meant is the set consisting of those $r \in R$ for which $\phi(r) \in N'$.

Let's show that $\phi^{-1}(N')$ is closed under arbitrary multiplication. Let $r \in \phi^{-1}(N')$, and let $s$ be any element of $R$. We need to show that $sr$ is an element of $\phi^{-1}(N')$. By definition, this means that $\phi(sr) \in N'$.

Since $\phi$ is a ring homomorphism, $\phi(sr) = \phi(s) \phi(r)$, with $\phi(r) \in N'$. Since $N'$ is either an ideal of $\phi(R)$ or of $R'$, $N'$ is closed under multiplication by elements in $\phi(R)$. In particular, $\phi(s) \in \phi(R)$, and so $\phi(s)\phi(r) \in N'$.

Similarly, you can show that $\phi^{-1}(N')$ is closed under subtraction, and then you're done.

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