D+F 7.3.24

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

(a) Prove that if $J$ is an ideal of $S$ then $\varphi^{-1}(J)$ is an ideal of $R$. In other words, the preimage of an ideal is an ideal.
$\textit{Proof}$ : Ideals are subrings that are closed under left and right multiplication by arbitrary ring elements. To show that $\varphi^{-1}(J)$ is a subring, we show that it is nonempty and closed under both ring operations. Since $J$ is an ideal in $S$, $$0\in{J} \Longrightarrow \varphi^{-1}(0)\in{\varphi^{-1}(J)} \Longrightarrow \varphi^{-1}(J) \neq \emptyset$$ Let $r_1$, $r_2 \in{\varphi^{-1}(J)}$. So, $\varphi(r_1)$, $\varphi(r_2) \in{J}$.
$J$ is an ideal, so $$\varphi(r_1)\varphi(r_2)=\varphi(r_1r_2)\in{J} \Longrightarrow r_1r_2\in{\varphi^{-1}(J)}$$ So, $\varphi^{-1}(J)$ is closed under multiplication making $\varphi^{-1}(J)$ a subring of $R$.
To show that $\varphi^{-1}(J)$ is an ideal, we show that it is closed under multiplication by all elements of $R$. Let $x\in{R}$. $$\varphi(xr)=\varphi(x)\varphi(r)$$ $J$ is an ideal, so, $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under left multiplication. Similarly, $$\varphi(rx)=\varphi(r)\varphi(x)$$ $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under right multiplication. $\therefore \varphi^{-1}(J)$ is an ideal of $R$ $\blacksquare$

Note: we could have skipped showing that $\varphi^{-1}(J)$ is closed under multiplication by its own elements, since this is taken care of in showing that $\varphi^{-1}(J)$ is closed with left and right multiplication by arbitrary elements in $R$.

(b) Prove that if $\varphi$ is surjective (onto) and $I$ is an ideal of $R$, then $\varphi(I)$ is an ideal of $S$. Give an example where this fails if $\varphi$ is not surjective. In other words, images of ideals are ideals under surjective ring homomorphisms.
$\textit{Proof}$ : Since $I$ is an ideal of $R$, $$0\in{I} \Longrightarrow \varphi(0)\in{\varphi(I)} \Longrightarrow \varphi(I) \neq \emptyset$$ Let $a,b\in{I}$. $I$ is an ideal, so $a+b\in{I}$. $$\varphi(a+b)=\varphi(a)+\varphi(b)\in{\varphi(I)}$$ So, $\varphi(I)$ is closed under addition.
Now, let $y\in{I}$ and $s\in{S}$ . Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and also $\varphi(x)=y$ with $x\in{I}$. $$xr\in{I} \Longrightarrow \varphi(xr)=\varphi(x)\varphi(r)\in{\varphi(I)}$$ $$rx\in{I} \Longrightarrow \varphi(rx)=\varphi(r)\varphi(x)\in{\varphi(I)}$$ So, $\varphi(I)$ is a left and right ideal, and thus an ideal of $S$.

If we relax the condition that $\varphi$ is surjective, consider the following example:
Let $\phi:\mathbb{Z} \rightarrow \mathbb{Q}$ be the identity map. $\mathbb{Z}$ is trivially an ideal of itself, but its image , which is still just $\mathbb{Z}$ is not an ideal of $\mathbb{Q}$ since $\frac{1}{2} \cdot 1 = \frac{1}{2} \not \in{\mathbb{Z}}$ i.e. $\mathbb{Z}$ is not closed under multiplication by elements in $\mathbb{Q}$.