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
\[ f_ j(a_1, \ldots , a_ n) = 0 \text{ in } S, \text{ for } j = 1, \ldots , m \]
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
\[ g_ j = f_ j(c_1 + \sum x_{1, l} d_ l , \ldots , c_ n + \sum x_{n, l} d_{n, l}) \in R[x_{i, l}] \]
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
\[ A \to B \to R^\wedge \]
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
\[ \xymatrix{ B \ar[rd] \ar@{..>}[r] & R' \ar@{..>}[d] \\ R \ar[r] \ar[u] & \kappa } \]
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
\[ R \subset R_{\mathfrak m'} \subset R^ h \subset R^\wedge , \]
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
\[ g_ j = f_ j(c_1 + \sum x_{1, l} d_ l , \ldots , c_ n + \sum x_{n, l} d_{n, l}) \in R[x_{i, l}] \]
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
\[ A \to B \to R^\wedge \]
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
\[ \xymatrix{ B \ar[rd] \ar@{..>}[r] & R' \ar@{..>}[d] \\ R \ar[r] \ar[u] & \kappa } \]
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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