Theorem 16.13.1. Let $R$ be a Noetherian local ring. Let $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$. Suppose that $(a_1, \ldots , a_ n) \in (R^\wedge )^ n$ is a solution in $R^\wedge $. If $R$ is a henselian G-ring, then for every integer $N$ there exists a solution $(b_1, \ldots , b_ n) \in R^ n$ in $R$ such that $a_ i - b_ i \in \mathfrak m^ NR^\wedge $.

## 16.13 The approximation property for G-rings

Let $R$ be a Noetherian local ring. In this case $R$ is a G-ring if and only if the ring map $R \to R^\wedge $ is regular, see More on Algebra, Lemma 15.50.7. In this case it is true that the henselization $R^ h$ and the strict henselization $R^{sh}$ of $R$ are G-rings, see More on Algebra, Lemma 15.50.8. Moreover, any algebra essentially of finite type over a field, over a complete local ring, over $\mathbf{Z}$, or over a characteristic zero Dedekind ring is a G-ring, see More on Algebra, Proposition 15.50.12. This gives an ample supply of rings to which the result below applies.

Let $R$ be a ring. Let $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$. Let $S$ be an $R$-algebra. In this situation we say a vector $(a_1, \ldots , a_ n) \in S^ n$ is a *solution in $S$* if and only if

Of course an important question in algebraic geometry is to see when systems of polynomial equations have solutions. The following theorem tells us that having solutions in the completion of a local Noetherian ring is often enough to show there exist solutions in the henselization of the ring.

**Proof.**
Let $c_ i \in R$ be an element such that $a_ i - c_ i \in \mathfrak m^ N$. Choose generators $\mathfrak m^ N = (d_1, \ldots , d_ M)$. Write $a_ i = c_ i + \sum a_{i, l} d_ l$. Consider the polynomial ring $R[x_{i, l}]$ and the elements

The system of equations $g_ j = 0$ has the solution $(a_{i, l})$. Suppose that we can show that $g_ j$ as a solution $(b_{i, l})$ in $R$. Then it follows that $b_ i = c_ i + \sum b_{i, l}d_ l$ is a solution of $f_ j = 0$ which is congruent to $a_ i$ modulo $\mathfrak m^ N$. Thus it suffices to show that solvability over $R^\wedge $ implies solvability over $R$.

Let $A \subset R^\wedge $ be the $R$-subalgebra generated by $a_1, \ldots , a_ n$. Since we've assumed $R$ is a G-ring, i.e., that $R \to R^\wedge $ is regular, we see that there exists a factorization

with $B$ smooth over $R$, see Theorem 16.12.1. Denote $\kappa = R/\mathfrak m$ the residue field. It is also the residue field of $R^\wedge $, so we get a commutative diagram

Since the vertical arrow is smooth, More on Algebra, Lemma 15.9.14 implies that there exists an étale ring map $R \to R'$ which induces an isomorphism $R/\mathfrak m \to R'/\mathfrak mR'$ and an $R$-algebra map $B \to R'$ making the diagram above commute. Since $R$ is henselian we see that $R \to R'$ has a section, see Algebra, Lemma 10.153.3. Let $b_ i \in R$ be the image of $a_ i$ under the ring maps $A \to B \to R' \to R$. Since all of these maps are $R$-algebra maps, we see that $(b_1, \ldots , b_ n)$ is a solution in $R$. $\square$

Given a Noetherian local ring $(R, \mathfrak m)$, an étale ring map $R \to R'$, and a maximal ideal $\mathfrak m' \subset R'$ lying over $\mathfrak m$ with $\kappa (\mathfrak m) = \kappa (\mathfrak m')$, then we have inclusions

by Algebra, Lemma 10.155.5 and More on Algebra, Lemma 15.45.3.

Theorem 16.13.2. Let $R$ be a Noetherian local ring. Let $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$. Suppose that $(a_1, \ldots , a_ n) \in (R^\wedge )^ n$ is a solution. If $R$ is a G-ring, then for every integer $N$ there exist

an étale ring map $R \to R'$,

a maximal ideal $\mathfrak m' \subset R'$ lying over $\mathfrak m$

a solution $(b_1, \ldots , b_ n) \in (R')^ n$ in $R'$

such that $\kappa (\mathfrak m) = \kappa (\mathfrak m')$ and $a_ i - b_ i \in (\mathfrak m')^ NR^\wedge $.

**Proof.**
We could deduce this theorem from Theorem 16.13.1 using that the henselization $R^ h$ is a G-ring by More on Algebra, Lemma 15.50.8 and writing $R^ h$ as a directed colimit of étale extension $R'$. Instead we prove this by redoing the proof of the previous theorem in this case.

Let $c_ i \in R$ be an element such that $a_ i - c_ i \in \mathfrak m^ N$. Choose generators $\mathfrak m^ N = (d_1, \ldots , d_ M)$. Write $a_ i = c_ i + \sum a_{i, l} d_ l$. Consider the polynomial ring $R[x_{i, l}]$ and the elements

The system of equations $g_ j = 0$ has the solution $(a_{i, l})$. Suppose that we can show that $g_ j$ as a solution $(b_{i, l})$ in $R'$ for some étale ring map $R \to R'$ endowed with a maximal ideal $\mathfrak m'$ such that $\kappa (\mathfrak m) = \kappa (\mathfrak m')$. Then it follows that $b_ i = c_ i + \sum b_{i, l}d_ l$ is a solution of $f_ j = 0$ which is congruent to $a_ i$ modulo $(\mathfrak m')^ N$. Thus it suffices to show that solvability over $R^\wedge $ implies solvability over some étale ring extension which induces a trivial residue field extension at some prime over $\mathfrak m$.

Let $A \subset R^\wedge $ be the $R$-subalgebra generated by $a_1, \ldots , a_ n$. Since we've assumed $R$ is a G-ring, i.e., that $R \to R^\wedge $ is regular, we see that there exists a factorization

with $B$ smooth over $R$, see Theorem 16.12.1. Denote $\kappa = R/\mathfrak m$ the residue field. It is also the residue field of $R^\wedge $, so we get a commutative diagram

Since the vertical arrow is smooth, More on Algebra, Lemma 15.9.14 implies that there exists an étale ring map $R \to R'$ which induces an isomorphism $R/\mathfrak m \to R'/\mathfrak mR'$ and an $R$-algebra map $B \to R'$ making the diagram above commute. Let $b_ i \in R'$ be the image of $a_ i$ under the ring maps $A \to B \to R'$. Since all of these maps are $R$-algebra maps, we see that $(b_1, \ldots , b_ n)$ is a solution in $R'$. $\square$

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 $K$.

Here is another variant of the main theorem of this section.

Lemma 16.13.4. Let $R$ be a Noetherian ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$. Suppose that $(a_1, \ldots , a_ n) \in ((R_\mathfrak p)^\wedge )^ n$ is a solution. If $R_\mathfrak p$ is a G-ring, then for every integer $N$ there exist

an étale ring map $R \to R'$,

a prime ideal $\mathfrak p' \subset R'$ lying over $\mathfrak p$

a solution $(b_1, \ldots , b_ n) \in (R')^ n$ in $R'$

such that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$ and $a_ i - b_ i \in (\mathfrak p')^ N(R'_{\mathfrak p'})^\wedge $.

**Proof.**
By Theorem 16.13.2 we can find a solution $(b'_1, \ldots , b'_ n)$ in some ring $R''$ étale over $R_\mathfrak p$ which comes with a prime ideal $\mathfrak p''$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak p'')$ and $a_ i - b'_ i \in (\mathfrak p'')^ N(R''_{\mathfrak p''})^\wedge $. We can write $R'' = R' \otimes _ R R_\mathfrak p$ for some étale $R$-algebra $R'$ (see Algebra, Lemma 10.143.3). After replacing $R'$ by a principal localization if necessary we may assume $(b'_1, \ldots , b'_ n)$ come from a solution $(b_1, \ldots , b_ n)$ in $R'$. Setting $\mathfrak p' = R' \cap \mathfrak p''$ we see that $R''_{\mathfrak p''} = R'_{\mathfrak p'}$ which finishes the proof.
$\square$

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