Example 16.13.3. Let $(R, \mathfrak m)$ be a Noetherian local ring with henselization $R^ h$. The map on completions $R^\wedge \to (R^ h)^\wedge$ is an isomorphism, see More on Algebra, Lemma 15.45.3. Since also $R^ h$ is Noetherian (ibid.) we may think of $R^ h$ as a subring of its completion (because the completion is faithfully flat). In this way we see that we may identify $R^ h$ with a subring of $R^\wedge$.

Let us try to understand which elements of $R^\wedge$ are in $R^ h$. For simplicity we assume $R$ is a domain with fraction field $K$. Clearly, every element $f$ of $R^ h$ is algebraic over $R$, in the sense that there exists an equation of the form $a_ n f^ n + \ldots + a_1 f + a_0 = 0$ for some $a_ i \in R$ with $n > 0$ and $a_ n \not= 0$.

Conversely, assume that $f \in R^\wedge$, $n \in \mathbf{N}$, and $a_0, \ldots , a_ n \in R$ with $a_ n \not= 0$ such that $a_ n f^ n + \ldots + a_1 f + a_0 = 0$. If $R$ is a G-ring, then, for every $N > 0$ there exists an element $g \in R^ h$ with $a_ n g^ n + \ldots + a_1 g + a_0 = 0$ and $f - g \in \mathfrak m^ N R^\wedge$, see Theorem 16.13.2. We'd like to conclude that $f = g$ when $N \gg 0$. If this is not true, then we find infinitely many roots $g$ of $P(T)$ in $R^ h$. This is impossible because (1) $R^ h \subset R^ h \otimes _ R K$ and (2) $R^ h \otimes _ R K$ is a finite product of field extensions of $K$. Namely, $R \to K$ is injective and $R \to R^ h$ is flat, hence $R^ h \to R^ h \otimes _ R K$ is injective and (2) follows from More on Algebra, Lemma 15.45.13.

Conclusion: If $R$ is a Noetherian local domain with fraction field $K$ and a G-ring, then $R^ h \subset R^\wedge$ is the set of all elements which are algebraic over $K$.

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