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 15.44.3. 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 15.44.13.

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 $K$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A1W. Beware of the difference between the letter 'O' and the digit '0'.