Prove: The pre-image of an ideal is an ideal.
Solution 1:
it is the kernel of $R \rightarrow S \rightarrow S/N$
Solution 2:
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.