Lemma 29.46.1. Let $A \to B$ be a ring map such that the induced morphism of schemes $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, resp. a universal homeomorphism inducing isomorphisms on residue fields, resp. universally closed, resp. universally closed and universally injective. Then for any $A$-subalgebra $B' \subset B$ the same thing is true for $f' : \mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$.

## 29.46 Universal homeomorphisms of affine schemes

In this section we characterize universal homeomorphisms of affine schemes.

**Proof.**
If $f$ is universally closed, then $B$ is integral over $A$ by Lemma 29.44.7. Hence $B'$ is integral over $A$ and $f'$ is universally closed (by the same lemma). This proves the case where $f$ is universally closed.

Continuing, we see that $B$ is integral over $B'$ (Algebra, Lemma 10.36.15) which implies $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is surjective (Algebra, Lemma 10.36.17). Thus if $A \to B$ induces purely inseparable extensions of residue fields, then the same is true for $A \to B'$. This proves the case where $f$ is universally closed and universally injective, see Lemma 29.10.2.

The case where $f$ is a universal homeomorphism follows from the remarks above, Lemma 29.45.5, and the obvious observation that if $f$ is surjective, then so is $f'$.

If $A \to B$ induces isomorphisms on residue fields, then so does $A \to B'$ (see argument in second paragraph). In this way we see that the lemma holds in the remaining case. $\square$

Lemma 29.46.2. Let $A$ be a ring. Let $B = \mathop{\mathrm{colim}}\nolimits B_\lambda $ be a filtered colimit of $A$-algebras. If each $f_\lambda : \mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, resp. a universal homeomorphism inducing isomorphisms on residue fields, resp. universally closed, resp. universally closed and universally injective, then the same thing is true for $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$.

**Proof.**
If $f_\lambda $ is universally closed, then $B_\lambda $ is integral over $A$ by Lemma 29.44.7. Hence $B$ is integral over $A$ and $f$ is universally closed (by the same lemma). This proves the case where each $f_\lambda $ is universally closed.

For a prime $\mathfrak q \subset B$ lying over $\mathfrak p \subset A$ denote $\mathfrak q_\lambda \subset B_\lambda $ the inverse image. Then $\kappa (\mathfrak q) = \mathop{\mathrm{colim}}\nolimits \kappa (\mathfrak q_\lambda )$. Thus if $A \to B_\lambda $ induces purely inseparable extensions of residue fields, then the same is true for $A \to B$. This proves the case where $f_\lambda $ is universally closed and universally injective, see Lemma 29.10.2.

The case where $f$ is a universal homeomorphism follows from the remarks above and Lemma 29.45.5 combined with the fact that prime ideals in $B$ are the same thing as compatible sequences of prime ideals in all of the $B_\lambda $.

If $A \to B_\lambda $ induces isomorphisms on residue fields, then so does $A \to B$ (see argument in second paragraph). In this way we see that the lemma holds in the remaining case. $\square$

Lemma 29.46.3. Let $A \subset B$ be a ring extension. Let $S \subset A$ be a multiplicative subset. Let $n \geq 1$ and $b_ i \in B$ for $1 \leq i \leq n$. If the set

is nonempty, then so is

**Proof.**
Omitted. Hint: clear denominators.
$\square$

Lemma 29.46.4. Let $A \subset B$ be a ring extension. If there exists $b \in B$, $b \not\in A$ and an integer $n \geq 2$ with $b^ n \in A$ and $b^{n + 1} \in A$, then there exists a $b' \in B$, $b' \not\in A$ with $(b')^2 \in A$ and $(b')^3 \in A$.

**Proof.**
Let $b$ and $n$ be as in the lemma. Then all sufficiently large powers of $b$ are in $A$. Namely, $(b^ n)^ k(b^{n + 1})^ i = b^{(k + i)n + i}$ which implies any power $b^ m$ with $m \geq n^2$ is in $A$. Hence if $i \geq 1$ is the largest integer such that $b^ i \not\in A$, then $(b^ i)^2 \in A$ and $(b^ i)^3 \in A$.
$\square$

Lemma 29.46.5. Let $A \subset B$ be a ring extension such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields. If $A \not= B$, then there exists a $b \in B$, $b \not\in A$ with $b^2 \in A$ and $b^3 \in A$.

**Proof.**
Recall that $A \subset B$ is integral (Lemma 29.44.7). By Lemma 29.46.1 we may assume that $B$ is generated by a single element over $A$. Hence $B$ is finite over $A$ (Algebra, Lemma 10.36.5). Hence the support of $B/A$ as an $A$-module is closed and not empty (Algebra, Lemmas 10.40.5 and 10.40.2). Let $\mathfrak p \subset A$ be a minimal prime of the support. After replacing $A \subset B$ by $A_\mathfrak p \subset B_\mathfrak p$ (permissible by Lemma 29.46.3) we may assume that $(A, \mathfrak m)$ is a local ring, that $B$ is finite over $A$, and that $B/A$ has support $\{ \mathfrak m\} $ as an $A$-module. Since $B/A$ is a finite module, we see that $I = \text{Ann}_ A(B/A)$ satisfies $\mathfrak m = \sqrt{I}$ (Algebra, Lemma 10.40.5). Let $\mathfrak m' \subset B$ be the unique prime ideal lying over $\mathfrak m$. Because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a homeomorphism, we find that $\mathfrak m' = \sqrt{IB}$. For $f \in \mathfrak m'$ pick $n \geq 1$ such that $f^ n \in IB$. Then also $f^{n + 1} \in IB$. Since $IB \subset A$ by our choice of $I$ we conclude that $f^ n, f^{n + 1} \in A$. Using Lemma 29.46.4 we conclude our lemma is true if $\mathfrak m' \not\subset A$. However, if $\mathfrak m' \subset A$, then $\mathfrak m' = \mathfrak m$ and we conclude that $A = B$ as the residue fields are isomorphic as well by assumption. This contradiction finishes the proof.
$\square$

Lemma 29.46.6. Let $A \subset B$ be a ring extension such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism. If $A \not= B$, then either there exists a $b \in B$, $b \not\in A$ with $b^2 \in A$ and $b^3 \in A$ or there exists a prime number $p$ and a $b \in B$, $b \not\in A$ with $pb \in A$ and $b^ p \in A$.

**Proof.**
The argument is almost exactly the same as in the proof of Lemma 29.46.5 but we write everything out to make sure it works.

Recall that $A \subset B$ is integral (Lemma 29.44.7). By Lemma 29.46.1 we may assume that $B$ is generated by a single element over $A$. Hence $B$ is finite over $A$ (Algebra, Lemma 10.36.5). Hence the support of $B/A$ as an $A$-module is closed and not empty (Algebra, Lemmas 10.40.5 and 10.40.2). Let $\mathfrak p \subset A$ be a minimal prime of the support. After replacing $A \subset B$ by $A_\mathfrak p \subset B_\mathfrak p$ (permissible by Lemma 29.46.3) we may assume that $(A, \mathfrak m)$ is a local ring, that $B$ is finite over $A$, and that $B/A$ has support $\{ \mathfrak m\} $ as an $A$-module. Since $B/A$ is a finite module, we see that $I = \text{Ann}_ A(B/A)$ satisfies $\mathfrak m = \sqrt{I}$ (Algebra, Lemma 10.40.5). Let $\mathfrak m' \subset B$ be the unique prime ideal lying over $\mathfrak m$. Because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a homeomorphism, we find that $\mathfrak m' = \sqrt{IB}$. For $f \in \mathfrak m'$ pick $n \geq 1$ such that $f^ n \in IB$. Then also $f^{n + 1} \in IB$. Since $IB \subset A$ by our choice of $I$ we conclude that $f^ n, f^{n + 1} \in A$. Using Lemma 29.46.4 we conclude our lemma is true if $\mathfrak m' \not\subset A$. If $\mathfrak m' \subset A$, then $\mathfrak m' = \mathfrak m$. Since $A \not= B$ we conclude the map $\kappa = A/\mathfrak m \to B/\mathfrak m' = \kappa '$ of residue fields cannot be an isomorphism. By Lemma 29.10.2 we conclude that the characteristic of $\kappa $ is a prime number $p$ and that the extension $\kappa '/\kappa $ is purely inseparable. Pick $b \in B$ whose image in $\kappa '$ is an element not contained in $\kappa $ but whose $p$th power is in $\kappa $. Then $b \not\in A$, $b^ p \in A$, and $pb \in A$ (because $pb \in \mathfrak m' = \mathfrak m \subset A$) as desired. $\square$

Proposition 29.46.7. Let $A \subset B$ be a ring extension. The following are equivalent

$\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields, and

every finite subset $E \subset B$ is contained in an extension

\[ A[b_1, \ldots , b_ n] \subset B \]such that $b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$ for $i = 1, \ldots , n$.

**Proof.**
Assume (1). By transfinite induction we construct for each ordinal $\alpha $ an $A$-subalgebra $B_\alpha \subset B$ as follows. Set $B_0 = A$. If $\alpha $ is a limit ordinal, then we set $B_\alpha = \mathop{\mathrm{colim}}\nolimits _{\beta < \alpha } B_\beta $. If $\alpha = \beta + 1$, then either $B_\beta = B$ in which case we set $B_\alpha = B_\beta $ or $B_\beta \not= B$, in which case we apply Lemma 29.46.5 to choose a $b_\alpha \in B$, $b_\alpha \not\in B_\beta $ with $b_\alpha ^2, b_\alpha ^3 \in B_\beta $ and we set $B_\alpha = B_\beta [b_\alpha ] \subset B$. Clearly, $B = \mathop{\mathrm{colim}}\nolimits B_\alpha $ (in fact $B = B_\alpha $ for some ordinal $\alpha $ as one sees by looking at cardinalities). We will prove, by transfinite induction, that (2) holds for $A \to B_\alpha $ for every ordinal $\alpha $. It is clear for $\alpha = 0$. Assume the statement holds for every $\beta < \alpha $ and let $E \subset B_\alpha $ be a finite subset. If $\alpha $ is a limit ordinal, then $B_\alpha = \bigcup _{\beta < \alpha } B_\beta $ and we see that $E \subset B_\beta $ for some $\beta < \alpha $ which proves the result in this case. If $\alpha = \beta + 1$, then $B_\alpha = B_\beta [b_\alpha ]$. Thus any $e \in E$ can be written as a polynomial $e = \sum d_{e, i}b_\alpha ^ i$ with $d_{e, i} \in B_\beta $. Let $D \subset B_\beta $ be the set $D = \{ d_{e, i}\} \cup \{ b_\alpha ^2, b_\alpha ^3\} $. By induction assumption there exists an $A$-subalgebra $A[b_1, \ldots , b_ n] \subset B_\beta $ as in the statement of the lemma containing $D$. Then $A[b_1, \ldots , b_ n, b_\alpha ] \subset B_\alpha $ is an $A$-subalgebra of $B_\alpha $ as in the statement of the lemma containing $E$.

Assume (2). Write $B = \mathop{\mathrm{colim}}\nolimits B_\lambda $ as the colimit of its finite $A$-subalgebras. By Lemma 29.46.2 it suffices to show that $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields. Compositions of universally closed morphisms are universally closed and the same thing for morphisms which induce isomorphisms on residue fields. Thus it suffices to show that if $A \subset B$ and $B$ is generated by a single element $b$ with $b^2, b^3 \in A$, then (1) holds. Such an extension is integral and hence $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally closed and surjective (Lemma 29.44.7 and Algebra, Lemma 10.36.17). Note that $(b^2)^3 = (b^3)^2$ in $A$. For any ring map $\varphi : A \to K$ to a field $K$ we see that there exists a $\lambda \in K$ with $\varphi (b^2) = \lambda ^2$ and $\varphi (b^3) = \lambda ^3$. Namely, $\lambda = 0$ if $\varphi (b^2) = 0$ and $\lambda = \varphi (b^3)/\varphi (b^2)$ if not. Thus $B \otimes _ A K$ is a quotient of $K[x]/(x^2 - \lambda ^2, x^3 - \lambda ^3)$. This ring has exactly one prime with residue field $K$. This implies that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is bijective and induces isomorphisms on residue fields. Combined with universal closedness this shows (1) is true, see Lemmas 29.45.5 and 29.10.2. $\square$

Proposition 29.46.8. Let $A \subset B$ be a ring extension. The following are equivalent

$\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, and

every finite subset $E \subset B$ is contained in an extension

\[ A[b_1, \ldots , b_ n] \subset B \]such that for $i = 1, \ldots , n$ we have

$b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or

there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.

**Proof.**
The proof is exactly the same as the proof of Proposition 29.46.7 except for the following changes:

Use Lemma 29.46.6 instead of Lemma 29.46.5 which means that for each successor ordinal $\alpha = \beta + 1$ we either have $b_\alpha ^2, b_\alpha ^3 \in B_\beta $ or we have a prime $p$ and $pb_\alpha , b_\alpha ^ p \in B_\beta $.

If $\alpha $ is a successor ordinal, then take $D = \{ d_{e, i}\} \cup \{ b_\alpha ^2, b_\alpha ^3\} $ or take $D = \{ d_{e, i}\} \cup \{ pb_\alpha , b_\alpha ^ p\} $ depending on which case $\alpha $ falls into.

In the proof of (2) $\Rightarrow $ (1) we also need to consider the case where $B$ is generated over $A$ by a single element $b$ with $pb, b^ p \in B$ for some prime number $p$. Here $A \subset B$ defines a universal homeomorphism for example by Algebra, Lemma 10.46.7.

This finishes the proof. $\square$

Lemma 29.46.9. Let $p$ be a prime number. Let $A \to B$ be a ring map which induces an isomorphism $A[1/p] \to B[1/p]$ (for example if $p$ is nilpotent in $A$). The following are equivalent

$\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, and

the kernel of $A \to B$ is a locally nilpotent ideal and for every $b \in B$ there exists a $p$-power $q$ with $qb$ and $b^ q$ in the image of $A \to B$.

**Proof.**
If (2) holds, then (1) holds by Algebra, Lemma 10.46.7. Assume (1). Then the kernel of $A \to B$ consists of nilpotent elements by Algebra, Lemma 10.30.6. Thus we may replace $A$ by the image of $A \to B$ and assume that $A \subset B$. By Algebra, Lemma 10.46.5 the set

is an $A$-subalgebra of $B$ (being closed under products is trivial). We have to show $B' = B$. If not, then according to Lemma 29.46.6 there exists a $b \in B$, $b \not\in B'$ with either $b^2, b^3 \in B'$ or there exists a prime number $\ell $ with $\ell b, b^\ell \in B'$. We will show both cases lead to a contradiction, thereby proving the lemma.

Since $A[1/p] = B[1/p]$ we can choose a $p$-power $q$ such that $qb \in A$.

If $b^2, b^3 \in B'$ then also $b^ q \in B'$. By definition of $B'$ we find that $(b^ q)^{q'} \in A$ for some $p$-power $q'$. Then $qq'b, b^{qq'} \in A$ whence $b \in B'$ which is a contradiction.

Assume now there exists a prime number $\ell $ with $\ell b, b^\ell \in B'$. If $\ell \not= p$ then $\ell b \in B'$ and $qb \in A \subset B'$ imply $b \in B'$ a contradiction. Thus $\ell = p$ and $b^ p \in B'$ and we get a contradiction exactly as before. $\square$

Lemma 29.46.10. Let $A$ be a ring. Let $x, y \in A$.

If $x^3 = y^2$ in $A$, then $A \to B = A[t]/(t^2 - x, t^3 - y)$ induces bijections on residue fields and a universal homeomorphism on spectra.

If there is a prime number $p$ such that $p^ px = y^ p$ in $A$, then $A \to B = A[t]/(t^ p - x, pt - y)$ induces a universal homeomorphism on spectra.

**Proof.**
We will use the criterion of Lemma 29.45.5 to check this. In both cases the ring map is integral. Thus it suffices to show that given a field $k$ and a ring map $\varphi : A \to k$ the $k$-algebra $B \otimes _ A k$ has a unique prime ideal whose residue field is equal to $k$ in case (1) and purely inseparable over $k$ in case (2). See Lemma 29.10.2.

In case (1) set $\lambda = 0$ if $\varphi (x) = 0$ and set $\lambda = \varphi (y)/\varphi (x)$ if not. Then $B = k[t]/(t^2 - \lambda ^2, t^3 - \lambda ^2)$. Thus the result is clear.

In case (2) if the characteristic of $k$ is $p$, then we obtain $\varphi (y) = 0$ and $B = k[t]/(t^ p - \varphi (x))$ which is a local Artinian $k$-algebra whose residue field is either $k$ or a degree $p$ purely inseparable extension of $k$. If the characteristic of $k$ is not $p$, then setting $\lambda = \varphi (y)/p$ we see $B = k[t]/(t - \lambda ) = k$ and we conclude as well. $\square$

Lemma 29.46.11. Let $A \to B$ be a ring map.

If $A \to B$ induces a universal homeomorphism on spectra, then $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of finitely presented $A$-algebras $B_ i$ such that $A \to B_ i$ induces a universal homeomorphism on spectra.

If $A \to B$ induces isomorphisms on residue fields and a universal homeomorphism on spectra, then $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of finitely presented $A$-algebras $B_ i$ such that $A \to B_ i$ induces isomorphisms on residue fields and a universal homeomorphism on spectra.

**Proof.**
Proof of (1). We will use the criterion of Algebra, Lemma 10.127.4. Let $A \to C$ be of finite presentation and let $\varphi : C \to B$ be an $A$-algebra map. Let $B' = \varphi (C) \subset B$ be the image. Then $A \to B'$ induces a universal homeomorphism on spectra by Lemma 29.46.1. By Algebra, Lemma 10.127.2 we can write $B' = \mathop{\mathrm{colim}}\nolimits _{i \in I} B_ i$ with $A \to B_ i$ of finite presentation and surjective transition maps. By Algebra, Lemma 10.127.3 we can choose an index $0 \in I$ and a factorization $C \to B_0 \to B'$ of the map $C \to B'$. We claim that $\mathop{\mathrm{Spec}}(B_ i) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism for $i$ sufficiently large. The claim finishes the proof of (1).

Proof of the claim. By Lemma 29.45.6 the ring map $A_{red} \to B'_{red}$ induces a universal homeomorphism on spectra. Thus $A_{red} \subset B'_{red}$ by Algebra, Lemma 10.30.6. Setting $A' = \mathop{\mathrm{Im}}(A \to B')$ we have surjections $A \to A' \to A_{red}$ inducing bijections $\mathop{\mathrm{Spec}}(A_{red}) = \mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(A)$. Thus $A' \subset B'$ induces a universal homeomorphism on spectra. By Proposition 29.46.8 and the fact that $B'$ is finite type over $A'$ we can find $n$ and $b'_1, \ldots , b'_ n \in B'$ such that $B' = A'[b'_1, \ldots , b'_ n]$ and such that for $j = 1, \ldots , n$ we have

$(b'_ j)^2, (b'_ j)^3 \in A'[b'_1, \ldots , b'_{j - 1}]$, or

there exists a prime number $p$ with $pb'_ j, (b'_ j)^ p \in A'[b'_1, \ldots , b'_{j - 1}]$.

Choose $b_1, \ldots , b_ n \in B_0$ lifting $b'_1, \ldots , b'_ n$. For $i \geq 0$ denote $b_{j, i}$ the image of $b_ j$ in $B_ i$. For large enough $i$ we will have for $j = 1, \ldots , n$

$b_{j, i}^2, b_{j, i}^3 \in A_ i[b_{1, i}, \ldots , b_{j - 1, i}]$, or

there exists a prime number $p$ with $pb_{j, i}, b_{j, i}^ p \in A_ i[b_{1, i}, \ldots , b_{j - 1, i}]$.

Here $A_ i \subset B_ i$ is the image of $A \to B_ i$. Observe that $A \to A_ i$ is a surjective ring map whose kernel is a locally nilpotent ideal. After increasing $i$ more if necessary, we may assume $B_ i$ is generated by $b_1, \ldots , b_ n$ over $A_ i$, in other words $B_ i = A_ i[b_1, \ldots , b_ n]$. By Algebra, Lemmas 10.46.7 and 10.46.4 we conclude that $A \to A_ i \to A_ i[b_1] \to \ldots \to A_ i[b_1, \ldots , b_ n] = B_ i$ induce universal homeomorphisms on spectra. This finishes the proof of the claim.

The proof of (2) is exactly the same. $\square$

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