1 Preliminaries

We recall some of the standard properties of pure extensions which are frequently used in the paper. Some of these are trivial consequences of the definitions. For the rest, appropriate references are mentioned.

For a commutative ring $A$, let $\text{Spec}\,A$ and Max$~A$ denote the sets of prime ideals and maximal ideals of $A$ respectively.

Let $A\subseteq B$ be commutative rings. If $A$ is a direct summand of $B$ (as an $A$-module), it follows from the definition that $A$ is a pure subring of $B$. This fact is implicitly used in this paper. Later, we will see two important instances when this happens:

1. (a) if $A$ is normal, and $B$ is a finite module over $A$; or

2. (b) if $A$ is a complete local ring, and $A\subseteq B$ is pure.

By an algebraic local ring we mean a localization of an affine algebra over a field $k$ at a maximal ideal.

A ring homomorphism $f:A~\rightarrow ~B$ is called cyclically pure if, for any ideal $I\subseteq B$, $A\cap IB=I$, or, equivalently, if any ideal in $A$ is contracted from some ideal in $B$. Note that analogues of Lemma 1.1 hold for cyclically pure extensions.

The following important result from [Reference Hochster6, Theorem 1.7] is often used in this paper. We state a special case which is sufficient for our purposes.

If $A$ is reduced, or Gorenstein, or a local Cohen–Macaulay ring of dimension ${\geqslant}2$, then ${\it\varphi}$ is a pure embedding if and only if it is cyclically pure.

The following result is proved in [Reference Hochster9].

Lemma 1.6 immediately implies that if $A$ is a normal affine subalgebra of $k[x_{1},x_{2},\ldots ,x_{n}]$, such that the extension is integral, then this is a pure extension.

The following criterion of purity can be found in [Reference Hochster and Roberts8, Corollary 5.3].

The following result and its proof were shown to us by C. Huneke to whom we are very thankful.

2 Some new observations

We use the following terminologies in this paper.

Let $f:~X\longrightarrow Y$ be a dominant morphism of affine algebraic varieties. Then we have the following.

1. (1) The morphism $f$ is called $\mathbf{pure}$ if the induced morphism $f^{\ast }:{\mathcal{O}}_{Y}\longrightarrow {\mathcal{O}}_{X}$ is pure.

2. (2) A point $x\in X$ is called $\mathbf{pure}$ if ${\mathcal{O}}_{Y,f(x)}\longrightarrow {\mathcal{O}}_{X,x}$ is pure. We call $f$$\mathbf{strongly~pure}$ if each $x\in X$ is pure.

3. (3) A point $y\in f(X)$ is called $\mathbf{completely~pure}$ if each point in $f^{-1}(y)$ is pure.

4. (4) A point $y\in Y$ is called $\mathbf{partially~pure}$ if some point in $f^{-1}(y)$ is pure.

5. (5) A point $y\in f(X)$ is called $\mathbf{pure}$ if ${\mathcal{O}}_{Y,y}\longrightarrow {\mathcal{O}}_{X}\otimes {\mathcal{O}}_{Y,y}$ is pure.

Note. An algebraic variety is always assumed to be irreducible unless otherwise specified.

Since given any $y\in f(X)$ and some $x\in f^{-1}(y)$, we have a natural factorization

it is easy to see that any completely pure point is partially pure, which, in turn, is pure.

Let $A{\hookrightarrow}B$ be a pure extension. Suppose that there is an element $x$ of $B$ satisfying a relation

$$\begin{eqnarray}a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}=0,\quad \text{with}~a_{i}\in A,\quad \forall i.\end{eqnarray}$$

Then, as an easy corollary of Lemma 1.5, it follows that $a_{0}$ is contained in the ideal generated by $a_{1},a_{2},\ldots ,a_{n}$ in $A$. A partial converse of this fact holds under special conditions.

Lemma 2.1. Let A be a Dedekind domain. Then, for any polynomial

$$\begin{eqnarray}f(x):=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n}\in A[x]\quad \text{of positive degree,}\end{eqnarray}$$

$A\longrightarrow A[x]/(f)$ is pure if $a_{0}$ is contained in the ideal generated by $a_{1},~a_{2},\ldots ,a_{n}$ in $A$.

Proof. In view of the local criteria for purity as in Lemma 1.3, after localizing the extension at a maximal ideal of $A$, we may assume that $A$ is a discrete valuation ring. First, we observe that, if B is a (cyclically) pure A-algebra, for any A-algebra C, $B\oplus C$ is also a (cyclically) pure A-algebra.

Now, coming to the proof, if $a_{0}=0$, there exists $m>0$ such that $x^{m}|f$ but $x^{m+1}$ does not divide $f$. Since $A[x]$ is a unique factorization domain, and $\gcd (x^{m},f/x^{m})=1$, we have

$$\begin{eqnarray}A~{\hookrightarrow}\frac{A[x]}{(f)}~{\hookrightarrow}\frac{A[x]}{(x^{m})}\oplus \frac{A[x]}{(f/x^{m})}.\end{eqnarray}$$

Now, $A[x]/(x^{m})$, being a free module over A, is pure. Hence, by the above observation,

$$\begin{eqnarray}A{\hookrightarrow}\frac{A[x]}{(x^{m})}\oplus \frac{A[x]}{(f/x^{m})}\end{eqnarray}$$

is also pure, and consequently so is $A{\hookrightarrow}A[x]/(f)$.

Otherwise, if $a_{0}\neq 0$, then $f$ can be factored as $f=ag$, where $a:=\gcd (a_{i})_{i=0}^{n}$, and $g$ is a primitive polynomial. Therefore, we have

$$\begin{eqnarray}\frac{A[x]}{(f)}=\frac{A[x]}{(ag)}~{\hookrightarrow}~\frac{A}{(a)}[x]\oplus \frac{A[x]}{(g)}.\end{eqnarray}$$

However, then, by [Reference Nagata14], $A[x]/(g)$ is a faithfully flat A-algebra. Therefore, the above observation, together with Lemma 1.7, implies that

$$\begin{eqnarray}A~{\hookrightarrow}~\frac{A}{(a)}[x]\oplus \frac{A[x]}{(g)}\end{eqnarray}$$

is pure, and consequently so is $A{\hookrightarrow}A[x]/(f)$.◻

The following lemma is in the same spirit, which shows that for Dedekind domains, the question of purity boils down to the question of surjectivity of the associated morphism of schemes.

Since $\mathfrak{p}$-primary ideals in $A$ are in one-to-one correspondence with the $\mathfrak{p}A_{\mathfrak{p}}$-primary ideals of $A_{\mathfrak{p}}$, it is sufficient to prove that any $\mathfrak{p}A_{\mathfrak{p}}$-primary ideal is contracted from some ideal in $B_{\mathfrak{p}}$. This is a special case of the following lemma.

Proof. Let $a$ be any nonzero, nonunit element in $A$ with a prime factorization (up to a unit)

$$\begin{eqnarray}a=\mathop{\prod }_{i=1}^{r}p_{i}^{e_{i}},\quad \text{with distinct primes}~p_{i}~\text{and positive integers}~e_{i}.\end{eqnarray}$$

If possible, let $a^{\ast }\in (a)B\cap A-(a)$. Then $a^{\ast }$ can be written as

$$\begin{eqnarray}a^{\ast }={\it\beta}\mathop{\prod }_{i=1}^{r}p_{i}^{f_{i}},\end{eqnarray}$$

where the $f_{i}$ are nonnegative integers, and ${\it\beta}\in A-\{0\}$ is relatively prime to $p_{i}$, for each $i$. Since $a^{\ast }\notin (a)$, there exists some $i$ for which $f_{i}<e_{i}$. Again, $a^{\ast }\in (a)B$ implies that there exists some $b\in B$ such that $a^{\ast }=ab$, implying that ${\it\beta}\prod _{j\neq i}p_{j}^{f_{j}}\in (p_{i})$, leading to a contradiction.◻

We have the following corollaries of Lemma 2.2 (using Lemma 1.5).

This also follows easily since such a morphism is faithfully flat (and Lemma 1.7).

The following corollary extends a cyclically pure version of Lemma 2.1 to the formal power series case.

Corollary 2.3.3. Let $A$ be a Dedekind domain, and let $f\in A[[x]]$ be a nonconstant formal power series. Then

$$\begin{eqnarray}A{\hookrightarrow}\frac{A[[x]]}{(f)}\end{eqnarray}$$

is cyclically pure if $f_{0}$ lies in the ideal generated by $f_{1},f_{2},\ldots$ , where $f(x):=f_{0}+f_{1}x+f_{2}x^{2}+\cdots \,$.

The following lemma is a special case of Lemma 1.1(1). We write it down in this particular form for the sake of future reference.

Proof. Since $A$ is normal, it suffices to show that ${\it\xi}$ is in every localization of $A$ at a height $1$ prime ideal $\mathfrak{p}$. Suppose that this is not true for some $\mathfrak{p}$. Then ${\it\xi}^{-1}$ must lie in the maximal ideal $\mathfrak{p}A_{\mathfrak{p}}$, that is, ${\it\xi}=1/a$ for some $a\in \mathfrak{p}A_{\mathfrak{p}}$. Since ${\it\xi}$ is integral over $B$, ${\it\xi}$ satisfies an integral equation ${\it\xi}^{n}+b_{1}{\it\xi}^{n-1}+\cdots +b_{n}=0$, with $b_{i}\in B$. By replacing ${\it\xi}$ by $1/a$, and clearing the denominator in the equation, we obtain

$$\begin{eqnarray}1+b_{1}a+b_{2}a^{2}+\cdots +b_{n}a^{n}=0.\end{eqnarray}$$

Therefore, $a$ is a unit in $B$. However, by the given condition, there exists $\mathfrak{q}\in \text{Spec}~B$ such that $\mathfrak{p}=\mathfrak{q}\cap A$, which implies that $a\in \mathfrak{q}$, leading to a contradiction.◻

Lemma 2.7. Let ${\it\varphi}:X\longrightarrow Y$ be a dominant morphism of affine varieties, with dim $X=m$ and dim $Y=n$. Assume that $Y$ is normal. Let $x\in X$ be such that any irreducible component of $~{\it\varphi}^{-1}({\it\varphi}(x))$, which contains $x$, has dimension equal to $m-n$. Then the induced local morphism

$$\begin{eqnarray}{\it\varphi}_{x}^{\ast }:{\mathcal{O}}_{Y,{\it\varphi}(x)}\longrightarrow {\mathcal{O}}_{X,x}\end{eqnarray}$$

is pure.

We add here the following remark.

Let $G$ be a reductive algebraic group acting on an affine regular $\mathbb{C}$-domain $A$. Then by Hochster–Roberts [Reference Hochster and Roberts7], the ring $A^{G}$ of $G$-invariants is a pure subalgebra of $A$. In a similar context, we can ask whether the ring of invariants of a $G_{a}$-action on a polynomial ring $\mathbb{C}[x_{1},\ldots ,x_{n}]$ is a pure subring of $\mathbb{C}[x_{1},\ldots ,x_{n}]$. The answer is positive if $n\leqslant 3$ but negative for $n\geqslant 4$. There are three crucial points to prove in the case $n\leqslant 3$. Namely, we have the following.

1. (1) The ring of invariants is regular.

2. (2) The quotient morphism

   $$\begin{eqnarray}q:\text{Spec}\,\mathbb{C}[x_{1},\ldots ,x_{n}]\rightarrow \text{Spec}\,\mathbb{C}[x_{1},\ldots ,x_{n}]^{G_{a}}\end{eqnarray}$$
   is surjective.

3. (3) The fibers of the quotient morphism $q$ are all of dimension one.

In fact, with all three conditions satisfied, $q$ is a faithfully flat morphism, and hence pure by Lemma 1.7. In the study of pure extensions, the surjectivity of the morphism $\text{Spec}\,B\rightarrow \text{Spec}\,A$ is always a key point. Bonnet [Reference Bonnet3] and later Kaliman [Reference Kaliman10] proved via topological arguments that if $n=3$, the quotient morphism $q$ is surjective. This is a fairly significant result in understanding the subalgebras of a polynomial ring, and it is desirable to give a proof which may elucidate the algebro-geometric background.

Proof. (1) If $n\leqslant 3$, we show that $B$ is faithfully flat over $A$, whence $A{\hookrightarrow}B$ is a pure embedding by Lemma 1.7. We only consider the case $n=3$. It is known that $A$ is a polynomial ring in two variables, and $q$ is equidimensional of dimension one (cf. [Reference Miyanishi12]). Hence, $B$ is flat over $A$. On the other hand, the surjectivity of $q$ is proved by [Reference Bonnet3] and [Reference Kaliman10]. Hence, $B$ is faithfully flat over $A$.

(2) The ring $A$ is generated over $\mathbb{C}$ by the following four elements:

$$\begin{eqnarray}\displaystyle X=x_{4},Y=x_{2}x_{4}-\frac{1}{2}x_{3}^{2},Z=x_{1}x_{4}^{2}-x_{2}x_{3}x_{4}+\frac{1}{3}x_{3}^{3}, & & \displaystyle \nonumber\\ \displaystyle W=9x_{1}^{2}x_{4}^{2}+(8x_{2}^{2}-18x_{1}x_{3})x_{2}x_{4}+6x_{1}x_{3}^{3}-3x_{2}^{2}x_{3}^{2}. & & \displaystyle \nonumber\end{eqnarray}$$

Hence, $\text{Spec}\,A$ is a hypersurface $X^{2}W=(2Y)^{3}+(3Z)^{2}$ in $\mathbb{A}^{4}$. We show that the set $\{(0,0,0,W)|W\neq 0\}$ is not in the image of $q$. In fact, for a closed point of $\text{Spec}\,A$ with $X=x_{4}=0$, we have $Y=-\frac{1}{2}x_{3}^{2}$ and $Z=\frac{1}{3}x_{3}^{3}$. If $x_{3}$ moves, the point $(-\frac{1}{2}x_{3}^{2},\frac{1}{3}x_{3}^{3})$ covers the curve $(2Y)^{3}+(3Z)^{2}=0$ in the $(Y,Z)$-plane. If $x_{3}\neq 0$, we can find $x_{1},x_{2}$ so that $W=6x_{1}x_{3}^{3}-3x_{2}^{2}x_{3}^{2}$ takes an arbitrary value. If $x_{3}=0$, then $W=0$. Hence, $\{(0,0,0,W)|W\neq 0\}$ is not in the image of $q$.

As for the flatness of $B$ over $A$, it suffices to show that $q$ is not equidimensional. It is clear that the general fibers of $q$ have dimension one. However, over the point $(0,0,0,0)$, the fiber is given by $\{x_{3}=x_{4}=0\}$, which is a plane. Therefore, we have shown that $A{\hookrightarrow}B$ is not a pure embedding.

A more straightforward way to see that $A{\hookrightarrow}B$ is not a pure embedding is to use that $IB\cap A\neq I$, for $I:=XA+YA$. In fact, $IB$ contains $X,Y,Z,W$, but $Z,W\not \in I$.

We note that Bonnet [Reference Bonnet3] constructed an example of $G_{a}$-action on $\mathbb{A}^{4}$ whose quotient morphism is not surjective.◻

3 Purity and surjectivity

In this section we study the relationship between the notions of purity and surjectivity of the induced morphism $\text{Spec}\,S~\rightarrow ~\text{Spec}\,R$ for an inclusion of rings $R~{\hookrightarrow}~S$. Of particular interest will be the cases when $R,S$ are analytic local rings, or complete local rings.

We now answer the questions raised in the introduction.

Let $R\subseteq S$ be algebraic local rings.

By Lemma 1.7, faithful flatness implies purity. Therefore, in answering Question 3, we have a generalization of the well-known result that flat maps are open.

We believe that this is a new result about pure morphisms. The proof of this result is quite nontrivial.

As a first step in relating purity to surjectivity we record the following easy consequence of Chevalley’s theorem, as in [Reference Matsumura11, Theorem 6], which tells us that the surjectivity of the $\text{Spec}\,$ map, associated to a local morphism, actually translates into the local surjectivity in Zariski-topology.

Theorem 3.1. Let $A\subseteq B$ be Noetherian rings of finite krull dimension, such that $A$ is an integral domain, and $B$ is a finitely generated algebra over $A$. Suppose that, for some $\mathfrak{p}\in \text{Spec}~B$, the induced morphism of schemes

$$\begin{eqnarray}\text{Spec}~B_{\mathfrak{p}}\longrightarrow \text{Spec}~A_{\mathfrak{p^{c}}}\end{eqnarray}$$

is surjective. Then there exists an open set $V\subseteq \text{Spec}~A$, such that $\mathfrak{p^{c}}\in V\subseteq ^{a}i(\text{Spec}~B)$, where $^{a}i:\text{Spec}\,B~\rightarrow ~\text{Spec}\,~A$ is induced by the natural inclusion $i:A\longrightarrow B$.

Corollary 3.1.1. Let ${\it\phi}:X\longrightarrow Y$ be a morphism of affine algebraic sets, with $Y$ being irreducible. If $x\in X$ is such that

$$\begin{eqnarray}\displaystyle \text{Spec}\,{\mathcal{O}}_{X,x}\longrightarrow \text{Spec}\,{\mathcal{O}}_{Y,{\it\phi}(x)} & & \displaystyle \nonumber\end{eqnarray}$$

is surjective, then there exists an open set $V\subseteq Y$, satisfying ${\it\phi}(x)\in V\subseteq {\it\phi}(X)$. In particular, the result is true when $x\in X$ is pure.

The next result shows that purity carries over to completion.

Proof. If ${\it\psi}:\hat{R}_{\mathfrak{m}}~\rightarrow ~{\hat{S}}_{\mathfrak{n}}$ is the induced natural map, then we get the following commutative diagram:

Given any ideal $\hat{I}$ of $\hat{R}_{\mathfrak{m}}$, by Krull’s intersection theorem, we have

$$\begin{eqnarray}\hat{I} =\mathop{\bigcap }_{n=1}^{\infty }(\hat{I} +\hat{\mathfrak{m}}^{n}).\end{eqnarray}$$

Since the ideal $\hat{I} +\hat{\mathfrak{m}}^{n}$ is $\hat{\mathfrak{m}}$-primary for each $n$, it is sufficient to show that any $\hat{\mathfrak{m}}$-primary ideal is contracted from some ideal in ${\hat{S}}_{\mathfrak{n}}$. If $\hat{\mathfrak{q}}$ is any $\hat{\mathfrak{m}}$-primary ideal in $\hat{R}_{\mathfrak{m}}$, then there exists an $\mathfrak{m}$-primary ideal $\mathfrak{q}$ in $R_{\mathfrak{m}}$ such that $\hat{\mathfrak{q}}=\mathfrak{q}\hat{R}_{\mathfrak{m}}$. Now, $R_{\mathfrak{m}}{\hookrightarrow}S_{\mathfrak{n}}$ being pure, there exists an ideal $\mathfrak{p}\in S_{\mathfrak{n}}$ satisfying $\mathfrak{p}\cap R_{\mathfrak{m}}=\mathfrak{q}$. Since completion is faithfully flat and hence pure by Lemma 1.7, there exists an ideal $\hat{\mathfrak{p}}\in \hat{S_{\mathfrak{n}}}$ such that $\hat{\mathfrak{p}}\cap S_{\mathfrak{n}}=\mathfrak{p}$. The diagram being commutative, ${\it\psi}^{-1}(\hat{\mathfrak{p}})\cap R_{\mathfrak{m}}=\mathfrak{q}$. Therefore, ${\it\psi}^{-1}(\hat{\mathfrak{p}})$ is also $\hat{\mathfrak{m}}$-primary, and consequently an extended ideal. Hence, ${\it\psi}^{-1}(\hat{\mathfrak{p}})=\hat{\mathfrak{q}}$, as the extended ideals are in one-to-one correspondence with the contracted ideals. Therefore,${\it\psi}:\hat{R}_{\mathfrak{m}}~\rightarrow ~{\hat{S}}_{\mathfrak{n}}$ is cyclically pure.

As a corollary, we get an affirmative answer to Question 1. This also follows from [Reference Hochster and Roberts7, Proposition 6.11], but we have given a direct elementary proof. ◻

Next, for Question 2, we begin by recalling a simple fact in commutative algebra.

Lemma 3.3. If $~\mathfrak{p}$ is a prime ideal in a Noetherian ring $R$, such that $\text{Spec}\,~(R/\mathfrak{p})$ is infinite, then

$$\begin{eqnarray}\mathfrak{p}~=~\mathop{\bigcap }_{\mathfrak{q}\in V(\mathfrak{p})\setminus \{\mathfrak{p}\}}\mathfrak{q}.\end{eqnarray}$$

Now, let $R$ be an algebraic local domain of dimension ${>}1$. Then, by a standard result in commutative algebra, $\hat{R}$ is a reduced ring. Therefore, in this case Lemma 3.3 has the following corollaries.

Proof. By [Reference Abhyankar1, Chapter 3, (10.14)], we have an injection $\hat{R}~\rightarrow ~{\hat{S}}$.

Consider first the case when $\mathfrak{m}S$ is primary for $\mathfrak{n}$. Then it is easy to see that ${\hat{S}}$ is integral over $\hat{R}$. In this case the result is well known.

So we assume that $\mathfrak{m}S$ is not primary for $\mathfrak{n}$. Let $d:=$ dim$~R$. We further assume that $d>1$, since the result is trivial for $d=1$.

We claim that it suffices to prove that given any prime ideal $\hat{\mathfrak{p}}\subset \hat{R}$ such that height $\hat{\mathfrak{p}}=d-1$, there is a prime ideal $\hat{\mathfrak{q}}\subset {\hat{S}}$ such that $\hat{\mathfrak{p}}=\hat{\mathfrak{q}}\cap \hat{R}$.

For, assuming this, if $\hat{\mathfrak{p}}\in \text{Spec}\,\hat{R}$ is of height less than $d-1$, then, by Corollary 3.3.2, there exists an ideal $I$ in ${\hat{S}}$ contracting to $\hat{\mathfrak{p}}$. However, any maximal element of the set $\{I{\vartriangleleft}{\hat{S}}\mid I\cap \hat{R}=\hat{\mathfrak{p}}\}$ is a prime ideal in $S$, and hence $\hat{\mathfrak{p}}$ must be in the image of $\text{Spec}\,{\hat{S}}$.

Let $\mathfrak{m}S=Q_{1}\cap Q_{2}\cap \cdots \cap Q_{r+1}$, where $Q_{i}$ are primary for prime ideals $\mathfrak{p_{i}}$ strictly smaller than $\mathfrak{n}$ for $1\leqslant i\leqslant r$, and $Q_{r+1}$ (if it occurs in a minimal primary decomposition) is primary for $\mathfrak{n}$. Suppose that $\hat{\mathfrak{p}}$ is not of the form $\hat{\mathfrak{q}}\cap \hat{R}$ for any prime ideal of $\hat{\mathfrak{q}}$ in $S$. Then $\hat{\mathfrak{p}}{\hat{S}}=Q_{1}^{\prime }\cap \cdots \cap Q_{s}^{\prime }\cap Q_{s+1}^{\prime }$, where $Q_{j}^{\prime }$ is primary for a prime ideal $\mathfrak{p}_{j}^{\prime }$ which contracts to some $\mathfrak{p}_{j_{i}}$ for $1\leqslant j\leqslant s,~j_{i}\leqslant r$, and $Q_{s+1}^{\prime }$ is primary for $\hat{\mathfrak{n}}$. Clearly, $s\geqslant 1$.

We write $S=k[x_{1},\ldots ,x_{n}]_{(x_{1},\ldots ,x_{n})},R=k[f_{1},\ldots ,f_{e}]_{(f_{1},\ldots ,f_{e})}$. It follows that $f_{l}\in \mathfrak{p}_{i}$ for each $l$, and $1\leqslant i\leqslant r+1$, where $\mathfrak{p}_{r+1}=\mathfrak{n}$. Let $\hat{\mathfrak{p}}=({\it\varphi}_{1},\ldots ,{\it\varphi}_{m})\subset k[[f_{1},\ldots ,f_{e}]]$. Then ${\it\varphi}_{j}\in \mathfrak{p}_{{\it\beta}}^{\prime }$ for each $j$, and ${\it\beta}\leqslant s+1$. Each $\mathfrak{p}_{{\it\beta}}^{\prime }$ contains some $\mathfrak{p}_{{\it\beta}_{i}}$. Hence, each $f_{{\it\alpha}}$ is contained in each $\mathfrak{p}_{{\it\beta}}^{\prime }$ for ${\it\beta}\leqslant s+1$. Thus, ${\it\varphi}_{j}\in Q_{{\it\beta}}^{\prime }$ for each $j$, and ${\it\beta}\leqslant s+1$.

There exists $N>0$ such that ${\mathfrak{p}_{{\it\beta}}^{\prime }}^{N}\subset Q_{{\it\beta}}^{\prime }$ for each ${\it\beta}$. Hence, we can write

$$\begin{eqnarray}{\it\varphi}_{i}={\it\varphi}_{j0}+{\it\varphi}_{j1},\end{eqnarray}$$

where each ${\it\varphi}_{j0}$ is a polynomial in $f_{1},f_{2},\ldots ,f_{e}$, and ${\it\varphi}_{j1}\in \hat{\mathfrak{n}}\cdot Q_{{\it\beta}}^{\prime }$. It follows that the ideal $({\it\varphi}_{jo},\ldots ,{\it\varphi}_{m0})$ generates $\cap _{1}^{s+1}Q_{{\it\beta}}^{\prime }$ mod $\hat{\mathfrak{n}}\cdot (\cap _{1}^{s+1}Q_{{\it\beta}}^{\prime })$. By Nakayama’s lemma, this implies that $({\it\varphi}_{10},\ldots ,{\it\varphi}_{m0})=\hat{\mathfrak{p}}{\hat{S}}$.

By [Reference Artin2, pages 32–33], we can assume that $({\it\varphi}_{10},\ldots ,{\it\varphi}_{m0})\hat{R}$ is of height $d-1$. Let $I=({\it\varphi}_{10},\ldots ,{\it\varphi}_{m0})\subset R$.

Claim.

$IS\cap R\subset \sqrt{I}$.

To see this, we write the primary decomposition of $I$ as $I=\cap J_{i}$, where the $J_{i}$ are primary for distinct prime ideals $\widetilde{J_{i}}$. Since $\text{Spec}\,S~\rightarrow ~\text{Spec}\,R$ is surjective, we have $\widetilde{J_{i}}S\cap R=\widetilde{J_{i}}$. It follows that $IS\cap R\subset \cap \widetilde{J_{i}}=\sqrt{I}$.

Now, $(IS){\hat{S}}\cap S=IS$, since ${\hat{S}}$ is a faithfully flat extension of $S$. It follows that $I{\hat{S}}\cap \hat{R}=(\hat{\mathfrak{p}}{\hat{S}})\cap \hat{R}$ is an ideal, which contracts to an ideal contained in $\sqrt{I}$. Hence, $\hat{\mathfrak{p}}{\hat{S}}$ cannot contract to an ideal which is primary for $\hat{\mathfrak{n}}$.

This contradiction shows that $\hat{\mathfrak{p}}$ is contracted from ${\hat{S}}$. This completes the proof of Theorem 3.4.◻

The next result gives an affirmative answer to Question 3 for morphisms of complex analytic varieties. This generalizes the well-known result that a flat morphism is open.

We need the following well-known result from complex analysis.

We use this as follows.

Let $g(z,w)$ be holomorphic in variables $z_{1},z_{2},\ldots ,z_{r}$ and one variable $w$ in an open neighborhood $U$ of $(0,0)$ in $\mathbb{C}^{r+1}$. Suppose that $g=b_{0}(z)+b_{1}(z)w+\cdots +b_{n}(z)w^{n}+\cdots \,,$ where $n$ is the smallest positive integer such that $b_{n}$ is a unit in the convergent power series ring $\mathbb{C}\{z\}$. Assume for simplicity that $b_{n}=1+c_{n}$, where $c_{n}(z)$ is a nonunit.

To see this, suppose $|w|={\it\delta}<1/2$. Since $c_{n}+b_{n+1}w+b_{n+2}w^{2}+\cdots \,$ vanishes at the origin in $\mathbb{C}^{r+1}$, by continuity we can assume that ${\it\delta}_{1},{\it\delta}$ are so small that $|z_{i}|<{\it\delta}_{1}$ in $N_{1}$, $|w|<{\it\delta}$ in $U_{2}$, and $|c_{n}+b_{n+1}w+\cdots |<1/2$ in $N_{1}\times U_{2}$. Since $b_{0},b_{1},\ldots ,b_{n-1}$ are nonunits, we can find an open neighborhood $N_{1}^{\prime }$ of $0\in C^{r}$ such that $|(b_{0}+b_{1}w+\cdots +b_{n-1}w^{n-1})/w^{n}|<1/2$ for all $(z)\in N_{1}^{\prime }$ and $|w|={\it\delta}$. Taking $U_{1}=N_{1}\cap N_{1}^{\prime }$, we get the required result.

Now by Rouché’s theorem, for any fixed $(z)\in U_{1}$, the numbers of zeroes of $b_{0}+b_{1}w+\cdots +w^{n}$ and $g$ in $U_{2}$ are the same. Clearly, by shrinking $U_{1}$ if necessary, for any $(z)\in U_{1}$ the equation $b_{0}+b_{1}w+\cdots +w^{n}=0$ has $n$ zeroes in $U_{2}$. Hence, this is also true for $g$ inside $U_{2}$. This proves that the image of any neighborhood of $(0,0)$ in $\{g=0\}$ under the map $(z,w)~\rightarrow ~(z)$ contains a neighborhood of $0$ in $\mathbb{C}^{r}$.

In the application of the above ideas in the proof of Theorem 3.5 when dim $V>1$, we have $b_{n}=c_{0}+c_{n}$, where $c_{0}\in \mathbb{C}$, with $1/2\leqslant |c_{0}|\leqslant 1$, and $c_{n}$ is a nonunit. Then the rest of the above argument can be changed to get an upper bound on $|w|$ which is a root of $g=0$.

We implicitly use the following observation. If $I\subseteq R$ is an ideal such that one of the conditions of Lemma 1.5 is satisfied for the inclusion $R/I\subseteq S/IS$, then this is also a pure embedding.

We assume that $p,q$ are the origins in $\mathbb{C}^{r},\mathbb{C}^{r+m}$ respectively.

Let $R=\mathbb{C}\{z_{1},z_{2},\ldots ,z_{r}\}/I,S=\mathbb{C}\{z_{1},\ldots ,z_{r},w_{1},\ldots ,w_{m}\}/J$. Here, $r+m$ may be larger than the embedding dimension of $S$. Let $J=(f_{1},\ldots ,f_{l})$.

Write $f_{i}=a_{i0}+a_{i1}(z)w^{A_{1}}+a_{i2}(z)w^{A_{2}}+\cdots \,$, where each $A_{j}$ is a multi-index $(j_{1},\ldots ,j_{m})$, and $w^{A_{j}}=w_{1}^{j_{1}}\cdots w_{m}^{j_{m}}$, etc. For any multi-index $A_{j}$, we denote $|A_{j}|=\sum _{a}j_{a}$.

Let $U$ be the set of points in $\mathbb{C}^{r+m}$ such that $|z_{i}|\leqslant {\it\delta}_{0},|w|\leqslant {\it\delta}_{0}<1/2$ for a suitable ${\it\delta}_{0}$.

Case 1. Suppose that, for some $f_{i}$, one of the coefficients $a_{i{\it\alpha}}$ is a unit in $R$. (This includes the case when $W$ is a product of $V$ and another variety.)

By making a unitary (linear) change of coordinates involving only $w_{1},\ldots ,w_{m}$ (so that Euclidean distances are preserved), we can assume that the monomial corresponding to $a_{i{\it\alpha}}$ is $w_{m}^{j_{m}}$ for some $j_{m}$. By the Weierstrass preparation theorem, we can write $f_{i}=u_{i}g_{i}$, where $u_{i}$ is a unit in $\mathbb{C}\{z_{1},\ldots ,w_{m}\}$, and $g_{i}=w_{m}^{j_{m}}+b_{m-1}w_{m}^{j_{m}-1}+\cdots +b_{0}$. The coefficients $b_{0},b_{1},\ldots ,b_{m-1}$ are nonunits in $\{z_{1},\ldots ,z_{r}\}$. It follows that $S$ is integral over a subring $S_{1}$ such that $R\subseteq S_{1}\subseteq S$ and $S_{1}:=\mathbb{C}\{z_{1},\ldots ,w_{1},\ldots ,w_{m-1}\}/J_{1}$. Let $W_{1}$ be the germ in $\mathbb{C}^{r+m-1}$ corresponding to $J_{1}$. Then the image of any neighborhood of $(0,0)\in W$ under the map $(z,w)~\rightarrow ~(z,w_{1},\ldots ,w_{m-1})$ contains a neighborhood of the origin in $W_{1}$. Any such point in $W$ is contained in $\{f_{i}=0\}$.

Now we are in a position to use Rouché’s theorem. By the discussion above, we can find open neighborhoods $U_{1},U_{2}$ of the origins in $\mathbb{C}^{r+m-1},\mathbb{C}$ respectively such that over any point in $W_{1}\cap U_{1}$ there are exactly $j_{m}$ points in $(U_{1}\times U_{2})\cap \{f_{i}=0\}$, one of which lies in $W$.

Since $R~\rightarrow ~S$ is pure, so is $R~\rightarrow ~S_{1}$. By induction, the result is true for $W_{1}~\rightarrow ~V$. Hence, it is also true for $W~\rightarrow ~V$.

To complete the induction for Case 1, we must prove the result for $m=1$.

However, the same argument as above shows that in this case the morphism ${\it\pi}$ satisfies the assertion of the theorem.

Case 2. Now we can assume that for each $i$ every $a_{i{\it\alpha}}$ is a nonunit in $R$.

Case 2.1. Suppose first that dim $R=1$ and $R=\mathbb{C}\{z\}$. The proof of this case is the crucial point of the proof of the general case.

Since ${\it\pi}$ is nonconstant, using the well-known result in $1$-variable complex analysis that a nonconstant holomorphic map is open, we can easily deduce the result. However, we must prove a uniform statement for all $1$-dimensional sections of $V$ obtained by intersecting with linear subspaces when dim $V>1$ and their scheme-theoretic inverse images in $W$. We consider a neighborhood $U$ of $p$ in $V$ defined by points $p^{\prime }$ with $|z_{k}|\leqslant {\it\delta}$ for a suitable ${\it\delta}<1$, which we will choose later, and $k$ varies over the set $\{1,2,\ldots ,r\}$.

Now, each $a_{i{\it\alpha}}$ is a power series in one variable $z$ vanishing at the origin. For any element $f\in J$ let $z^{e}$ be the largest power of $z$ which divides all $a_{{\it\alpha}}$, where $f=\sum _{{\it\alpha}}a_{{\it\alpha}}A^{{\it\alpha}}$. Let $J^{\prime }$ be the ideal generated by all $f/z^{e}$ as $f$ varies over elements in $J$. Then $W$ is locally the union of $\{z=0\}$ and the variety $W^{\prime }$ defined by $J^{\prime }$.

For simplicity, assume that $J^{\prime }$ is generated by $f_{1}^{\prime },\ldots ,f_{l}^{\prime }$, and $f_{i}^{\prime }={\rm\Sigma}a_{i{\it\alpha}}^{\prime }w^{{\it\alpha}}$, where $f_{i}^{\prime }=f_{i}/z^{e_{i}}$. No $a_{i0}^{\prime }$ can be a unit in $R$, for otherwise $J^{\prime }$ is the unit ideal and $W^{\prime }={\it\phi}$. This implies that ${\it\pi}$ is constant near $q$. This contradicts the hypothesis that $R~\rightarrow ~S$ is pure (hence injective). However, by construction for each $i$, some $a_{i{\it\alpha}}^{\prime }$ with $|{\it\alpha}|>0$ is a unit. Arguing as above, we can reduce to the case $m=1$, in which case we get the required result.

Since we need to prove this result for all $1$-dimensional sections of $V$ when dim $V>1$, we must prove a quantitative statement involving an upper bound on $|a_{ij}|$.

Again, let $f:=f_{1}$, and let $f=a_{0}+a_{1}w+\cdots \,$.

Let $min_{i}~ord_{z}(a_{i})=e$. Let $a_{i}^{\prime }:=a_{i}/z^{e}$. Our equation is of the form $a_{0}^{\prime }(z)+a_{1}^{\prime }(z)+\cdots +a_{n}^{\prime }(z)w^{n}+a_{n+1}^{\prime }(z)w^{n+1}+\cdots =0$, such that all the $a_{i}^{\prime }$ are nonunits for $i<n$, and $a_{n}^{\prime }$ is a unit. By the application of Rouché’s theorem discussed earlier we deduce the result in this case.

This completes the proof of Case 2.1 when dim $R=1,m=1$ and $V$ is smooth.

Case 2.2. Now we consider the case when dim $V=1,~m=1$, and $V$ is a singular curve.

The normalization of ${\mathcal{O}}_{V,p}$ in its quotient field is isomorphic to $\mathbb{C}\{z\}$. Let $V^{\prime }~\rightarrow ~V$ be the normalization. Let $W^{\prime }$ be the fiber product $W\times _{V}V^{\prime }$. Both the morphisms $V^{\prime }~\rightarrow ~V,~W^{\prime }~\rightarrow ~W$ are finite, and $W^{\prime }~\rightarrow ~V^{\prime }$ is pure by Lemma 1.4. By taking small open neighborhoods of points in $V^{\prime }$ lying over $p$ we are reduced to the previous case.

Case 3. Now assume that dim $V\geqslant 2$.

Case 3.1. To explain the main idea we first assume that $V$ is smooth of dimension $2$.

First, we prove two technical results which are useful in the arguments below.

Let $f=a_{0}(z_{1},z_{2})+a_{1}(z_{1},z_{2})w+\cdots +a_{n}(z_{1},z_{2})w^{n}+\cdots \,$ be a convergent power series in variables $z_{1},z_{2},w$, such that $a_{i}(z_{1},z_{2})$ is a nonunit in $R:=\mathbb{C}\{z_{1},z_{2}\}$ for every $i$. For any linear homogeneous polynomial $L(z_{1},z_{2})$, consider the ring $R_{0}:=\mathbb{C}\{z_{1},z_{2}\}/(L)=\mathbb{C}\{{\it\tau}\}$. Clearly, we can take $\overline{z_{1}}$ or $\overline{z_{2}}$ as ${\it\tau}$. Assume that $\overline{z_{1}}={\it\tau}$. Let $\overline{a_{i}}$ be the image of $a_{i}$ in $R_{0}$. Then $\overline{a_{i}}$ is a nonunit. Let $e_{i}>0$ be the ${\it\tau}$-order of $\overline{a_{i}}$. Consider the set of integers $e_{i}-\{min_{j}~e_{j}\}$. We assume that for any $L$ we have $e_{0}-\{min_{j}~e_{j}\}>0$. Let $n_{L}$ be the smallest integer $i$ (depending on $L$) such that $e_{i}=\{min_{j}~e_{j}\}$.

The next result concerns upper bounds on the coefficients of the terms occurring in the application of Rouché’s theorem as the linear homogeneous polynomials $L$ vary.

As above, let $g:=\overline{f}/\overline{z_{1}}^{e}=b_{0}+b_{1}w+\cdots \,$ be a nonunit in $\mathbb{C}\{z_{1},z_{2},w\}/(L)$, with $f\in J$, such that for some $n$ there is a term in $g$ of the form $b_{n}w^{n}$, where $b_{n}$ is a unit in $R$. To start with, by making a suitable unitary (linear) change of coordinates involving $z_{1},z_{2}$, we can assume that $b_{n}=c_{0}+c_{n}$, where $c_{0}$ is a constant, with $1/2\leqslant |c_{0}|\leqslant 1$, and $c_{n}$ is a nonunit. In order not to complicate the notation, we assume that $c_{0}=1$. We also assume that $n_{L}$ is the smallest such integer. Then $n_{L}>0$ for all $L$ such that $L$ does not divide $f$. For a fixed $f$, the set of integers $n_{L}$ is bounded above. Hence, for simplicity of exposition, we assume that $n_{L}$ is a constant $n\geqslant 1$.

We can assume that there are suitable ${\it\delta}_{1}>0,{\it\delta}_{2}>0$ such that $|c_{n}+b_{n+1}w+b_{n+2}w^{2}+\cdots |<1/2$ if $|z_{1}|\leqslant {\it\delta}_{1},|z_{2}|\leqslant {\it\delta}_{1},|w|\leqslant {\it\delta}_{2}$.

Since $b_{i}$ is a nonunit for $i<n$, we can assume that ${\it\delta}_{1}$ is such that $|(b_{0}+b_{1}w+\cdots +b_{n-1}w^{n-1})<|w|^{n}/2$ when $|z_{1}|\leqslant {\it\delta}_{1},~|w|={\it\delta}_{2}$. Now, it follows that $|b_{0}+b_{1}w+\cdots |<|b_{0}+b_{1}w+\cdots +b_{n-1}w^{n-1}+w^{n}|$ when $|w|={\it\delta}_{2}$. By Roché’s theorem, for any $|z_{1}|\leqslant {\it\delta}_{1},|z_{2}|\leqslant {\it\delta}_{1}$, the numbers of zeros of the equations $b_{0}+b_{1}w+\cdots =0$ and $b_{0}+b_{1}w+\cdots +b_{n-1}w^{n-1}+w^{n}=0$ inside $|w|<{\it\delta}_{2}$ are the same. As in the discussion above, this number can be assumed to be $n$.

We must show that the numbers ${\it\delta}_{1},{\it\delta}_{2}$ can be chosen independent of $L$.

This can be seen as follows.

Any line $L$ in $\mathbb{C}^{2}$ has the equation $z_{2}=cz_{1}$ for some constant $c$, or $z_{1}=0$. We give the argument assuming that $|c|\leqslant 1$, since if $|c|>1$ then we can take $\overline{z_{2}}={\it\tau}$.

We need to analyze the power series $f(z_{1},cz_{1},w)$ more carefully. If some monomial ${\it\gamma}z_{1}^{a}z_{2}^{b}$ occurs in $f$ with ${\it\gamma}\in \mathbb{C}$ a nonzero constant, then there is a term ${\it\gamma}c^{b}z_{1}^{a+b}$ in $f(z_{1},cz_{1},w)$. The maximum possible number of monomials of total degree $a+b$ in $z_{1},z_{2}$ is $a+b+1$. For terms $b_{i}$ with $i<n$, if $|z_{1}|\leqslant {\it\delta}_{1}<1/2$, then $(a+b+1)|c^{b}z_{1}^{a+b}|\leqslant |(a+b+1)z_{1}^{a+b}|\leqslant (a+b+1){\it\delta}^{a+b}$. Since $n$ is bounded, there are at most $n\cdot (a+b+1)$ terms containing $z_{1}^{a+b}$ for $i<n$. Collecting all the terms with the same monomial $z_{1}^{a+b}$, we get $n$ convergent power series of the form $\sum _{l}{\it\gamma}_{l}{\it\delta}^{l}$, which tend to $0$ as ${\it\delta}_{1}$ tends to $0$. Hence, ${\it\delta}_{1}$ and ${\it\delta}_{2}<1$ can be chosen small and independent of $L$. The argument for the series $c_{n}+b_{n+1}w+b_{n+2}w^{2}+\cdots \,$ is similar but more complicated. In this case we get a convergent power series in variables ${\it\delta}_{1},{\it\delta}_{2}$ of the form $\sum _{l}{\it\gamma}_{l}({\it\delta}_{1}){\it\delta}_{2}^{l}$, where ${\it\gamma}_{l}$ is a convergent power series in ${\it\delta}_{1}$. Hence, a similar argument shows that $|c_{n}+b_{n+1}w+\cdots |<1/2$ when $|z_{1}|\leqslant {\it\delta}_{1},~{\it\delta}_{2}$ are sufficiently small (independent of $L$).

Now, as in the proof when dim $V=1$, we prove that the image of any Euclidean neighborhood of $O$ in $W$ contains a Euclidean neighborhood of $O$ in $V$.

As before, let $f_{1},\ldots ,f_{l}$ be the equations defining $W$.

Let $J_{i}$ be the ideal generated by all the $a_{i{\it\alpha}}$ in ${\mathcal{O}}_{V,p}$. We consider a prime ideal $\mathfrak{p}$ which defines an irreducible curve $C$ in $V$ which is an irreducible component of a hyperplane section of $V$. It suffices to show that there is an ${\it\epsilon}>0$ such that for every such curve $C$ there is an ${\it\epsilon}$-neighborhood of $p$ in $C$ which is contained in the image of ${\it\pi}$.

For the application of Rouché’s theorem, we use $f:=f_{1}$ below, write$f=\sum a_{{\it\alpha}}w^{A_{{\it\alpha}}}$, and assume that $f|_{L}$ is not identically zero for every $L$.

By Lemma 1.3, the extension $R/\mathfrak{p}\subseteq S/\mathfrak{p}S$ is also pure.

Again, first consider the case when $R/\mathfrak{p}$ is smooth. As before, we can assume that $R/\mathfrak{p}=\mathbb{C}\{z\}$.

The functions $|a_{i{\it\alpha}}|_{C}$ have the same upper bound as $|a_{i{\it\alpha}}|$ on $V$. It is possible that for some such $C$ and an $i$ all $a_{i{\it\alpha}}\in \mathfrak{p}$. Such an $f_{i}$ can be ignored for the consideration that follows.

We now consider only those $C$ for which $a_{i{\it\alpha}}|_{C}$ are not all identically zero.

Now, $J_{i}\not \subset \mathfrak{p}$. As above, let $e_{i}$ be the $z$-order of $\overline{f_{i}}$.

Claim. For any $\mathfrak{p}$ such that $J_{i}\not \subset \mathfrak{p}$, the set of integers $e_{i}$ as $L$ varies is finite.

This follows using the primary decomposition of $J_{i}$, and the hypothesis that dim $V=2$.

Now, $f/z^{e_{i}}=a_{0}^{\prime }+a_{1}^{\prime }w+\cdots +a_{n}^{\prime }w^{n}+\cdots \,$, where $a_{n}^{\prime }$ is a unit, and no $a_{i}^{\prime }$ is a unit for $i<n$. The set of integers $e_{i},n_{L}$ is bounded above.

Hence, given any ${\it\eta}>0$, we can find a ${\it\delta}$ such that whenever $|z_{i}|\leqslant {\it\delta}$, we have $|a_{i}^{\prime }|<1/2$ for $i<n$, and this for all $L$. Similarly, $|c_{n}+a_{n+1}^{\prime }w+a_{n+2}^{\prime }w^{2}+\cdots |<1/2$ if $|z_{i}|\leqslant {\it\delta},~|w|\leqslant {\it\delta}$ for all $L$. Therefore, in the application of Rouché’s theorem there is a uniform upper bound on the absolute value of at least one $w$ corresponding to any $(z)\in U_{1}$.

Now the proof is similar to the case when dim $R=1$, and $R$ is smooth.

A similar proof to that above works when dim $R/\mathfrak{p}=1$, and $R/\mathfrak{p}$ is singular.

Case 3.2. Now, let dim $V>2$.

We again consider the ideals $J_{i}$ and irreducible components $C$ of $1$-dimensional sections of $V$ with corresponding local ring $R/\mathfrak{p}$. If some $f_{i}\in \mathfrak{p}$, with the notation above, then such an $f_{i}$ can be ignored for the consideration that follows. If there is a $\mathfrak{p}$ which contains all $J_{i}$, then every point in the curve $C$ is in the image of ${\it\pi}$.

Now, assume that not all $f_{i}\in \mathfrak{p}$. The proof is the same as in the two-dimensional case. (The proof of the claim is also not difficult.)

We leave the details to the reader.

This completes the proof of Theorem 3.5.

Theorem 3.5 has an interesting corollary. There exist convergent power series $f_{1},f_{2},\ldots ,f_{m}$ in variables $z_{1},\ldots ,z_{n}$ with $m>n$ such that there is no convergent power series relation between the $f_{i}$, but there is a formal power series relation between them [Reference Gabrièlov4]. This gives an inclusion $R:=\mathbb{C}\{f_{1},f_{2},\ldots ,f_{m}\}\subset S:=\mathbb{C}\{z_{1},z_{2},\ldots ,z_{n}\}$.

4 Some positive results in special cases

Coming back to the quasi-finiteness question raised in Section 1, we will see in Section 5 that even a pure morphism of affine planes need not be quasi-finite. However, the result is true for subalgebras generated by monomials.

Proof. First, suppose that $A{\hookrightarrow}k^{[n]}$ is a pure embedding. Then $A$ is also a normal domain. Let $A_{1},A_{2},\ldots ,A_{m}$ be the $n$-tuples of exponents corresponding to $f_{1},\ldots ,f_{m}$; namely, $A_{i}:=(a_{i1},a_{i2},\ldots ,a_{in})$ when $f_{i}=x_{1}^{a_{i1}}x_{2}^{a_{i2}}\cdots x_{n}^{a_{in}}$. Since $^{a}i$ is generically quasi-finite, the $m\times n$-matrix $A$, whose row vectors are $A_{1},\ldots ,A_{m}$, has rank $n$. After reindexing, if required, we may assume that $A_{1},\ldots ,A_{n}$ are $Q$-independent. Hence, they have maximal rank in the $Q$-vector space generated by $(1,0,\ldots ,0),(0,1,0,\ldots ,0),\ldots ,(0,0,\ldots ,1)$. Then, for some integer $r>0$, $r(1,0,0,\ldots ,0)$ lies in the abelian subgroup generated by $A_{1},A_{2},\ldots ,A_{n}$. After a change of indices, we can write

$$\begin{eqnarray}r(1,0,\ldots ,0)=b_{1}A_{1}+\cdots +b_{s}A_{s}-(b_{s+1}A_{s+1}+\cdots +b_{n}A_{n}),\end{eqnarray}$$

where each $b_{i}$ is nonnegative. This means that

$$\begin{eqnarray}f_{1}^{b_{1}}\cdots f_{s}^{b_{s}}=x_{1}^{r}f_{s+1}^{b_{s+1}}\cdots f_{n}^{b_{n}}.\end{eqnarray}$$

Since $A$ is a pure subring of the integral domain $k^{[n]}$, it follows from Lemma 1.1(2) that $x_{1}^{r}\in A$, and similarly for $x_{2},\ldots ,x_{n}$. Hence, the powers $x_{i}^{N}$ are in $A$ for some $N>0$.

The converse follows from Lemma 1.6. ◻

Generalizing Theorem 4.1, we can then ask whether the following conditions are equivalent.

We know that any finite morphism is quasi-finite and surjective. Since $A$ is normal, by Lemma 1.6, any surjective and quasi-finite morphism must be pure. Again, it is not difficult to see that any quasi-finite, graded morphism is actually finite. Therefore, the question that remains open is whether a polynomial ring is quasi-finite over a graded, pure subring of the same dimension with induced grading.

Let $A{\hookrightarrow}B$ be an inclusion of affine domains over $k$. In order to approach the question of quasi-finiteness raised in the introduction, we construct the unique subalgebra $C$ of $B$, containing $A$, and satisfying the following properties.

1. (1) The morphism $\text{Spec}\,C\rightarrow \text{Spec}\,A$, induced by the inclusion $A{\hookrightarrow}C$, is a quasi-finite surjective morphism.

2. (2) $C$ is an affine domain over $k$.

3. (3) If $C^{\prime }\subset B$ is an affine domain over $k$ such that $A{\hookrightarrow}C^{\prime }$, and the induced morphism $\text{Spec}\,C^{\prime }\rightarrow \text{Spec}\,A$ is quasi-finite and surjective, then $C^{\prime }\subseteq C$.

We call $C$ the quasi-finite closure of $A$ in $B$.

With the notation as in Theorem 4.3, set $Y:=$ Spec $(B),X:=$ Spec $(A)$, and $f:Y\rightarrow X$ is the morphism induced by the inclusion $A{\hookrightarrow}B$. We assume that $A$ and $B$ are normal. Let $F$ be the closed set of $X$ consisting of the points $x$, such that $f^{-1}(x)$ contains an irreducible component of positive dimension. Let ${\rm\Delta}$ be the union of all irreducible components of codimension one in $f^{-1}(F)$, and let $Y^{\prime }:=Y\setminus {\rm\Delta}$. Then $Y^{\prime }$ is an affine scheme $\text{Spec}\,B^{\prime }$, where $B^{\prime }$ is the ideal quotient $(B:J):=\{f\in \mathbb{Q}(B)|fJ^{n}\subseteq B~\text{ for some }n\}$, where $J$ is the defining ideal of ${\rm\Delta}$. In fact, since $B$ is normal, $B^{\prime }$ is the intersection of $B_{\mathfrak{q}}$, where $\mathfrak{q}$ ranges over all prime ideals of height one of $B$ such that $V(\mathfrak{q})\not \subseteq {\rm\Delta}$. We prove the following result.

Let $S$ be the polynomial ring in $n$-variables over $k$, and let $G$ be a reductive algebraic subgroup of $GL(n,k)$, acting naturally on $S$. Then the ring of invariants $R:=S^{G}$ is a graded subring of $S$. Let $\mathfrak{m},\mathfrak{n}$ be the irrelevant maximal ideals of $R,S$ respectively.

The next result may be well known to the experts.

5 Examples, and more observations on pure extensions

To start with, let us consider the following two morphisms of affine domains, which manifest the subtlety of pure extensions.

Example 5.1.

$$\begin{eqnarray}{\it\phi}:k[x,y]{\hookrightarrow}\frac{k[x,y,z]}{(xz^{2}-yz-xy)}\end{eqnarray}$$

and

$$\begin{eqnarray}{\it\psi}:k[x,y]{\hookrightarrow}\frac{k[x,y,z]}{(xz^{2}-yz-xy^{2})}~.\end{eqnarray}$$

Then ${\it\phi}$ is not pure, but ${\it\psi}$ is pure.

Proof. To show that ${\it\phi}$ is not pure, look at the ideal $I\subseteq k[x,y]$, generated by $x^{4}$ and $y^{4}+3x^{2}y^{3}+x^{4}y^{2}$. Then $xy^{4}+2x^{3}y^{3}\notin I$, whereas $xy^{4}+2x^{3}y^{3}\in I(k[x,y,z]/(xz^{2}+yz+xy))\cap k[x,y]$ because of the relation

$$\begin{eqnarray}x^{4}z^{5}\in (y^{4}+3x^{2}y^{3}+x^{4}y^{2})z+(xy^{4}+2x^{3}y^{3}).\end{eqnarray}$$

The relation is obtained by successive substitution $yz+xy$ for $xz^{2}$. The purity of ${\it\psi}$, on the other hand, is a consequence of the following lemma.

Lemma 5.2. Let $f$ be a polynomial in $k[x,y]$, divisible by $y^{2}$. Then the map

$$\begin{eqnarray}{\rm\Psi}:k[x,y]\longrightarrow \frac{k[x,y,z]}{(xz^{2}+yz+f)}\end{eqnarray}$$

is pure.

Proof. First, note that the associated morphism of schemes is surjective, since it is surjective on the closed points. Therefore, by local criteria for purity as in Lemma 1.3, it is sufficient to check that the induced local morphism over the origin

$$\begin{eqnarray}{\rm\Psi}_{0}:k[x,y]_{(x,y)}\longrightarrow \frac{k[x,y,z]}{(xz^{2}+yz+f)}\otimes k[x,y]_{(x,y)}\end{eqnarray}$$

is pure, as, over any point other than the origin, the morphism is quasi-finite, and hence is pure by Lemma 1.6. Again, in view of the observation that any partially pure point is pure, it suffices if we can show that the induced map

$$\begin{eqnarray}{\rm\Psi}_{0}^{\prime }:k[x,y]_{(x,y)}\longrightarrow \Big(\frac{k[x,y,z]}{(xz^{2}+yz+f)}\Big)_{(x,y,z)}\end{eqnarray}$$

is pure. However, the polynomial $xz^{2}+yz+f$, as a polynomial in $z$, has roots in $k[[x,y]]$ since $y^{2}|f$. Therefore, the corresponding map of completions

$$\begin{eqnarray}\hat{{\rm\Psi}}_{0}^{^{\prime }}:k[[x,y]]\longrightarrow \frac{k[[x,y,z]]}{(xz^{2}+yz+f)}\end{eqnarray}$$

is pure. Consequently, in view of the commutative diagram,

Lemma 2.4 implies that ${\rm\Psi}_{0}^{\prime }$ is pure. Hence, ${\rm\Psi}$ is pure. This completes the proof of Lemma 5.2 and shows that ${\it\psi}$ is a pure morphism.◻

The notion of pure morphism generalizes the ring of invariants of a polynomial ring under some group actions. If a finite group $G$ acts on $A:=k[x_{1},~x_{2},\ldots ,~x_{n}]$, then $A$ is finite over the ring of invariants $A^{G}$. However, in the case of a pure extension of normal, affine domains of same dimension, the map may not even be quasi-finite.

We saw in Example 5.3 that the local morphism at the origin is pure in spite of the fiber over it containing a line. However, in this case, the fiber also contains an isolated point, i.e., a component of “correct” dimension. Therefore, it is natural to ask whether the purity is due to the presence of the “correct” dimensional component in the fiber. Therefore, we pose the following question.

Unfortunately, this question has a negative answer, as the following example shows.

Example 5.5. (The pure locus is not open in general)

Consider the affine morphism

$$\begin{eqnarray}\mathbb{{\it\theta}}:\mathbb{A}^{3}\longrightarrow \mathbb{A}^{2},\quad \text{given by}~(x,y,z)\mapsto (xz,yz).\end{eqnarray}$$

The fiber over $(0,0)\in \mathbb{A}^{2}$ is the union of the line $\{(x,y,z)\in \mathbb{A}^{3}~|~x=y=0\}$ and the plane $\{(x,y,z)\in \mathbb{A}^{3}~|~z=0\}$. Now ${\mathcal{O}}_{\mathbb{A}^{2},(0,0)}\longrightarrow {\mathcal{O}}_{\mathbb{A}^{3},(0,0,0)}$ is pure by Proposition 4.5, since $k[xz,yz]$ is the ring of invariants of $k[x,y,z]$ under the action of $k^{\ast }$, given by $({\it\lambda},(x,y,z))\mapsto ({\it\lambda}x,{\it\lambda}y,{\it\lambda}^{-1}z)$. However, it follows from the following Theorem 5.6 that there exist points in $\mathbb{A}^{3}$, arbitrarily close to the origin, such that $\text{Spec}~{\mathcal{O}}_{\mathbb{A}^{3},(x,y,z)}\longrightarrow \text{Spec}~{\mathcal{O}}_{\mathbb{A}^{2},{\it\theta}(x,y,z)}$ is not surjective, and consequently $\text{Spec}~{\mathcal{O}}_{\mathbb{A}^{3},(x,y,z)}\longrightarrow \text{Spec}~{\mathcal{O}}_{\mathbb{A}^{2},{\it\theta}(x,y,z)}$ is not pure.

Theorem 5.6 has the following two interesting corollaries.

Corollary 5.6.2. The nonequidimensional pure extension

$$\begin{eqnarray}k[xz,yz]{\hookrightarrow}k[x,y,z]\end{eqnarray}$$

shows that a pure morphism need not be strongly pure.

Note. If we choose the underlying field to be the field of complex numbers $\mathbb{C}$, then the above proof also tells us that the pure locus is not even open in Euclidean topology, as any nonempty Zariski-open set in a variety is dense in the Euclidean topology.

In Section 1, we saw that pure extensions are preserved under tensor products. However, as the following example shows, this fails to hold true for the composite of subalgebras.