Suppose $R$ and $S$ are integral domains such that $R \subset S$. Suppose there exists a map (ring homomorphism) $\phi:S\rightarrow R$ such that $\phi$ is the identity on $R$.
I am trying to show that if $S$ is normal, then $R$ is normal as well, but I am struggling to see why this is the case.
Any hint/help will be appreciated. Thanks in advance.
Note: An integral domain is normal if it is its own integral closure in its field of fractions.
Since $Q(R)\subseteq Q(S)$, and $S$ is integrally closed, an element $a\in Q(R)$ which is integral over $R$ belongs to $S$.
Let $p\in R[X]$ be a monic polynomial of minimal degree such that $p(a)=0$. Then $\phi(p(a))=p(\phi(a))=0$. We can write $p(X)=(X-\phi(a))q(X)$ with $q\in R[X]$ monic. From $p(a)=0$ we get $a=\phi(a)\in R$ or $q(a)=0$. But the later contradicts the minimality of $\deg p$.