If $R\subseteq S$ is a split extension of Noetherian domains and $S$ is normal, then is $R$ normal?

Question

Let $R\subseteq S$ be a ring extension of Noetherian domains. $S$ naturally has an $R$-module structure. Assume $R$ is a direct summand of $S$ as an $R$-module.

If $S$ is integrally closed in its own field of fractions, then is $R$ also integrally closed in its own field of fractions?

I am willing to assume that $S$ is module finite over $R$.

Answer 1

Let $\frac{r_1}{r_2}\in\operatorname{Frac}R\subseteq \operatorname{Frac}S$ be integral over $R$. Then $\frac{r_1}{r_2}$ is integral over $S$ which is integrally closed, so there is an $s\in S$ such that $$r_1=r_2s.$$ Now, if $R$ is a direct summand of $S$, i.e. $S\simeq R \oplus S'$, we may in a unique way write $s=r_3+s'$ with $r_3\in R, s'\notin R$. Thus, $$r_1 = r_2r_3 +r_2s',$$ but then $s'=0$ since $r_1$ has the unique decomposition $r_1+0$ in $R\oplus S'$. Hence $r_1=r_2r_3$ and $\frac{r_1}{r_2}=r_3\in R$, so $R$ is integrally closed.