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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