Sufficiency

(After EGA IV₂. Theorem 5.8.6.)

Suppose A satisfies S₂ and R₁. Then A in particular satisfies S₁ and R₀; hence, it is reduced. If ${\displaystyle {\mathfrak {p}}_{i},\,1\leq i\leq r}$ are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields ${\displaystyle \kappa ({\mathfrak {p}}_{i})=Q(A/{\mathfrak {p}}_{i})}$: see total ring of fractions of a reduced ring. That means we can write ${\displaystyle 1=e_{1}+\dots +e_{r}}$ where ${\displaystyle e_{i}}$ are idempotents in ${\displaystyle \kappa ({\mathfrak {p}}_{i})}$ and such that ${\displaystyle e_{i}e_{j}=0,\,i\neq j}$. Now, if A is integrally closed in K, then each ${\displaystyle e_{i}}$ is integral over A and so is in A; consequently, A is a direct product of integrally closed domains Ae_i's and we are done. Thus, it is enough to show that A is integrally closed in K.

For this end, suppose

${\displaystyle (f/g)^{n}+a_{1}(f/g)^{n-1}+\dots +a_{n}=0}$

where all f, g, a_i's are in A and g is moreover a non-zerodivisor. We want to show:

${\displaystyle f\in gA}$.

Now, the condition S₂ says that ${\displaystyle gA}$ is unmixed of height one; i.e., each associated primes ${\displaystyle {\mathfrak {p}}}$ of ${\displaystyle A/gA}$ has height one. This is because if ${\displaystyle {\mathfrak {p}}}$ has height greater than one, then ${\displaystyle {\mathfrak {p}}}$ would contain a non zero divisor in ${\displaystyle A/gA}$. However, ${\displaystyle {\mathfrak {p}}}$ is associated to the zero ideal in ${\displaystyle A/gA}$ so it can only contain zero divisors, see here. By the condition R₁, the localization ${\displaystyle A_{\mathfrak {p}}}$ is integrally closed and so ${\displaystyle \phi (f)\in \phi (g)A_{\mathfrak {p}}}$, where ${\displaystyle \phi :A\to A_{\mathfrak {p}}}$ is the localization map, since the integral equation persists after localization. If ${\displaystyle gA=\cap _{i}{\mathfrak {q}}_{i}}$ is the primary decomposition, then, for any i, the radical of ${\displaystyle {\mathfrak {q}}_{i}}$ is an associated prime ${\displaystyle {\mathfrak {p}}}$ of ${\displaystyle A/gA}$ and so ${\displaystyle f\in \phi ^{-1}({\mathfrak {q}}_{i}A_{\mathfrak {p}})={\mathfrak {q}}_{i}}$; the equality here is because ${\displaystyle {\mathfrak {q}}_{i}}$ is a ${\displaystyle {\mathfrak {p}}}$-primary ideal. Hence, the assertion holds.

Necessity

Suppose A is a normal ring. For S₂, let ${\displaystyle {\mathfrak {p}}}$ be an associated prime of ${\displaystyle A/fA}$ for a non-zerodivisor f; we need to show it has height one. Replacing A by a localization, we can assume A is a local ring with maximal ideal ${\displaystyle {\mathfrak {p}}}$. By definition, there is an element g in A such that ${\displaystyle {\mathfrak {p}}=\{x\in A|xg\equiv 0{\text{ mod }}fA\}}$ and ${\displaystyle g\not \in fA}$. Put y = g/f in the total ring of fractions. If ${\displaystyle y{\mathfrak {p}}\subset {\mathfrak {p}}}$, then ${\displaystyle {\mathfrak {p}}}$ is a faithful ${\displaystyle A[y]}$-module and is a finitely generated A-module; consequently, ${\displaystyle y}$ is integral over A and thus in A, a contradiction. Hence, ${\displaystyle y{\mathfrak {p}}=A}$ or ${\displaystyle {\mathfrak {p}}=f/gA}$, which implies ${\displaystyle {\mathfrak {p}}}$ has height one (Krull's principal ideal theorem).

For R₁, we argue in the same way: let ${\displaystyle {\mathfrak {p}}}$ be a prime ideal of height one. Localizing at ${\displaystyle {\mathfrak {p}}}$ we assume ${\displaystyle {\mathfrak {p}}}$ is a maximal ideal and the similar argument as above shows that ${\displaystyle {\mathfrak {p}}}$ is in fact principal. Thus, A is a regular local ring. ${\displaystyle \square }$