$R_1\subsetneq R_2$ integral domains, $R_1$ integrally closed. Is $R_2$ integrally closed? > Let $R_1\subset R_2$ be integral domains (i.e., commutative rings with unit without zero divisors). > > Suppose $R_1, R_2$ have the same fraction field $F$ and that $R_1$ is integrally closed. > > Can we say that $R_2$ is integrally closed? If $x\in F\setminus R_1$, we have $x$ transcendental over $R_1$, since it is integrally closed. Consequently the polynomial ring $R_1[x]$ is also integrally closed. So the answer to the question is yes when $R_2=R_1[x]$. But is it true in general? How can I see that?

Here is a counterexample: Let $S_{1},S_{2} \subset \mathbb{Z}^{\oplus 2}$ be the submonoids $S_{1} := \mathbb{N} \cdot (0,1) + \mathbb{N} \cdot (1,0)$ and $S_{2} := \\{\mathbb{N} \cdot (-1,1) + \mathbb{N} \cdot (1,0)\\} \setminus \\{(-1,1)\\}$ and let $R_{i} := \mathbb{Z}[S_{i}]$ be the monoid algebra. Then $R_{1} \simeq \mathbb{Z}[t_{1},t_{2}]$ is a polynomial ring over a normal ring, so it's normal; but $R_{2}$ contains $(t_{1}^{-1}t_{2})^{2}$ but not $t_{1}^{-1}t_{2}$ itself.