# Prove: The pre-image of an ideal is an ideal.

Let $\phi : R \to S$ be a homomorphism. If $N$ is an ideal of $S$, then $\phi ^{-1} (N)$ is an ideal of $R$.

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Clearly $\phi^{-1}(N)$ is nonempty.

Suppose $x,y\in \phi^{-1}(N)$.

Then $\phi(x),\phi(y)\in N$ so $\phi(x)-\phi(y)=\phi(x-y)\in N$.

So $x-y\in\phi^{-1}(N)$.

If $a\in R, x\in \phi^{-1}(N)$.

Then $\phi(ax)=\phi(a)\phi(x) \in N$, since $\phi(a)\in S$ and $\phi(x)\in N$.

So $ax\in\phi^{-1}(N)$, and similarly for the reverse multiplication.

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it is the kernel of $R \rightarrow S \rightarrow S/N$

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