## The ring homomorphic image of an ideal is an ideal

Let $\varphi : R \rightarrow S$ be a ring homomorphism.

1. Prove that if $J \subseteq S$ is an ideal, then $\varphi^\ast[J] \subseteq R$ is an ideal. Apply this to the special case when $R$ is a subring of $S$ and $\varphi$ is the inclusion map to deduce that if $J \subseteq S$ is an ideal and $R \subseteq S$ a subring, then $J \cap R \subseteq R$ is an ideal.
2. Prove that if $\varphi$ is surjective and $I \subseteq R$ an ideal then $\varphi[I] \subseteq S$ is an ideal. Give an example where this fails if $\varphi$ is not surjective.

1. Let $x,y \in \varphi^\ast[J]$. Now $0 \in J$ and $\varphi(0) = 0$, so that $0 \in \varphi^\ast[J]$. By hypothesis, $\varphi(x-y) = \varphi(x) - \varphi(y) \in J$, so that $\varphi^\ast[J]$ is closed under subtraction. Now let $r \in R$. We have $\varphi(rx) = \varphi(r)\varphi(x) \in J$ since $J$ is an ideal of $S$; thus $rx \in \varphi^\ast[J]$. Similarly, $xr \in \varphi^\ast[J]$. Thus $\varphi^\ast[J]$ is an ideal of $R$.

The final deduction is immediate since $\iota^\ast[J] = J \cap R$.

2. Note that $0 \in I$, so that $\varphi(0) \in \varphi[I]$. Now let $x,y \in \varphi[I]$. Now $x = \varphi(r)$ and $y = \varphi(s)$ for some $r,s \in I$. Since $x-y = \varphi(r)-\varphi(s)$ $= \varphi(r-s) \in \varphi[I]$, $\varphi[I]$ is closed under subtraction. Now let $s \in S$. Since $\varphi$ is surjective, there exists $t \in R$ with $\varphi(t) = s$. Now $sx = \varphi(t) \varphi(r)$ $= \varphi(tr) \in \varphi[I]$ since $I$ is an ideal; thus $sx \in \varphi[I]$. Similarly, $xs \in \varphi[I]$. Thus $\varphi[I]$ is an ideal of $S$.

For the counterexample, consider the inclusion $\iota : \mathbb{Z} \rightarrow \mathbb{Q}$. $\mathbb{Z}$ is an ideal of itself but not of $\mathbb{Q}$ since $2 \cdot \frac{1}{3} = \frac{2}{3}$ is not an integer.