Let $\varphi : R \rightarrow S$ be a ring homomorphism and $T$ be a subring of $R$. Let $r \in R$ be integral over $T$.
I would like to show that the image $\varphi(r)$ is integral over the subring $\varphi(T)$ of $S$.
My attempt: Let $r \in R$. Since $r$ is integral over $T$, $r$ is a root of a monic polynomial with coefficients in $T$, i.e. $r^n + c_{n-1}r^{n-1} + \cdots + c_0 = 0$, where $c_{n-1}, \ldots, c_0 \in T$. My idea is to apply $\varphi$ to this equation, but I am unsure if this approach is correct :
$\varphi(r^n + c_{n-1}r^{n-1} + \cdots + c_0) = 0$
$\Leftrightarrow \varphi(r^n) + \varphi(c_{n-1})\varphi(r^{n-1}) + \cdots + \varphi(c_0) = 0$
$\Leftrightarrow (\varphi(r))^n + \varphi(c_{n-1})(\varphi(r))^{n-1} + \cdots + \varphi(c_0) = 0$
where $\varphi(c_{n-1}), \ldots, \varphi(c_0) \in \varphi(T) \subseteq S$. Hence, $\varphi(r)$ is integral over $\varphi(T)$.
Moreover, how can I find a counterexample that the converse is not true, i.e. if $\varphi(r)$ is integral over $\varphi(T)$, then $r$ is not necessarily integral over $T$ ?
Thanks for your help.
Take $R=T[X]$, $S=T[X]/(p)$, where $p$ is any monic, non-constant, polynomial in $R$, $\varphi: R\to S$ the canonical map. Then $\varphi (R)=S$, the restriction of $\varphi$ to $T$ is injective, and any element of $S$ is integral over $T$. Yet $X$, for instance, is not integral over $T$.