Lemma 10.46.1. Let $\varphi : R \to S$ be a surjective map with locally nilpotent kernel. Then $\varphi $ induces a homeomorphism of spectra and isomorphisms on residue fields. For any ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ is surjective with locally nilpotent kernel.

## 10.46 Universal homeomorphisms

Let $k'/k$ be an algebraic purely inseparable field extension. Then for any $k$-algebra $R$ the ring map $R \to k' \otimes _ k R$ induces a homeomorphism of spectra. The reason for this is the slightly more general Lemma 10.46.7 below.

**Proof.**
By Lemma 10.17.7 the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism onto the closed subset $V(\mathop{\mathrm{Ker}}(\varphi ))$. Of course $V(\mathop{\mathrm{Ker}}(\varphi )) = \mathop{\mathrm{Spec}}(R)$ because every prime ideal of $R$ contains every nilpotent element of $R$. This also implies the statement on residue fields. By right exactness of tensor product we see that $\mathop{\mathrm{Ker}}(\varphi )R'$ is the kernel of the surjective map $R' \to R' \otimes _ R S$. Hence the final statement by Lemma 10.32.3.
$\square$

Lemma 10.46.2. Let $k'/k$ be a field extension. The following are equivalent

for each $x \in k'$ there exists an $n > 0$ such that $x^ n \in k$, and

$k' = k$ or $k$ and $k'$ have characteristic $p > 0$ and either $k'/k$ is a purely inseparable extension or $k$ and $k'$ are algebraic extensions of $\mathbf{F}_ p$.

**Proof.**
Observe that each of the possibilities listed in (2) satisfies (1). Thus we assume $k'/k$ satisfies (1) and we prove that we are in one of the cases of (2). Discarding the case $k = k'$ we may assume $k' \not= k$. It is clear that $k'/k$ is algebraic. Hence we may assume that $k'/k$ is a nontrivial finite extension. Let $k'/k'_{sep}/k$ be the separable subextension found in Fields, Lemma 9.14.6. We have to show that $k = k'_{sep}$ or that $k$ is an algebraic over $\mathbf{F}_ p$. Thus we may assume that $k'/k$ is a nontrivial finite separable extension and we have to show $k$ is algebraic over $\mathbf{F}_ p$.

Pick $x \in k'$, $x \not\in k$. Pick $n, m > 0$ such that $x^ n \in k$ and $(x + 1)^ m \in k$. Let $\overline{k}$ be an algebraic closure of $k$. We can choose embeddings $\sigma , \tau : k' \to \overline{k}$ with $\sigma (x) \not= \tau (x)$. This follows from the discussion in Fields, Section 9.12 (more precisely, after replacing $k'$ by the $k$-extension generated by $x$ it follows from Fields, Lemma 9.12.8). Then we see that $\sigma (x) = \zeta \tau (x)$ for some $n$th root of unity $\zeta $ in $\overline{k}$. Similarly, we see that $\sigma (x + 1) = \zeta ' \tau (x + 1)$ for some $m$th root of unity $\zeta ' \in \overline{k}$. Since $\sigma (x + 1) \not= \tau (x + 1)$ we see $\zeta ' \not= 1$. Then

implies that

hence $\zeta ' \not= \zeta $ and

Hence every element of $k'$ which is not in $k$ is algebraic over the prime subfield. Since $k'$ is generated over the prime subfield by the elements of $k'$ which are not in $k$, we conclude that $k'$ (and hence $k$) is algebraic over the prime subfield.

Finally, if the characteristic of $k$ is $0$, the above leads to a contradiction as follows (we encourage the reader to find their own proof). For every rational number $y$ we similarly get a root of unity $\zeta _ y$ such that $\sigma (x + y) = \zeta _ y\tau (x + y)$. Then we find

and by our formula for $\tau (x)$ above we conclude $\zeta _ y \in \mathbf{Q}(\zeta , \zeta ')$. Since the number field $\mathbf{Q}(\zeta , \zeta ')$ contains only a finite number of roots of unity we find two distinct rational numbers $y, y'$ with $\zeta _ y = \zeta _{y'}$. Then we conclude that

which implies $\zeta _ y = 1$ a contradiction. $\square$

Lemma 10.46.3. Let $\varphi : R \to S$ be a ring map. If

for any $x \in S$ there exists $n > 0$ such that $x^ n$ is in the image of $\varphi $, and

$\mathop{\mathrm{Ker}}(\varphi )$ is locally nilpotent,

then $\varphi $ induces a homeomorphism on spectra and induces residue field extensions satisfying the equivalent conditions of Lemma 10.46.2.

**Proof.**
Assume (1) and (2). Let $\mathfrak q, \mathfrak q'$ be primes of $S$ lying over the same prime ideal $\mathfrak p$ of $R$. Suppose $x \in S$ with $x \in \mathfrak q$, $x \not\in \mathfrak q'$. Then $x^ n \in \mathfrak q$ and $x^ n \not\in \mathfrak q'$ for all $n > 0$. If $x^ n = \varphi (y)$ with $y \in R$ for some $n > 0$ then

which is a contradiction. Hence there does not exist an $x$ as above and we conclude that $\mathfrak q = \mathfrak q'$, i.e., the map on spectra is injective. By assumption (2) the kernel $I = \mathop{\mathrm{Ker}}(\varphi )$ is contained in every prime, hence $\mathop{\mathrm{Spec}}(R) = \mathop{\mathrm{Spec}}(R/I)$ as topological spaces. As the induced map $R/I \to S$ is integral by assumption (1) Lemma 10.36.17 shows that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R/I)$ is surjective. Combining the above we see that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is bijective. If $x \in S$ is arbitrary, and we pick $y \in R$ such that $\varphi (y) = x^ n$ for some $n > 0$, then we see that the open $D(x) \subset \mathop{\mathrm{Spec}}(S)$ corresponds to the open $D(y) \subset \mathop{\mathrm{Spec}}(R)$ via the bijection above. Hence we see that the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism.

To see the statement on residue fields, let $\mathfrak q \subset S$ be a prime lying over a prime ideal $\mathfrak p \subset R$. Let $x \in \kappa (\mathfrak q)$. If we think of $\kappa (\mathfrak q)$ as the residue field of the local ring $S_\mathfrak q$, then we see that $x$ is the image of some $y/z \in S_\mathfrak q$ with $y \in S$, $z \in S$, $z \not\in \mathfrak q$. Choose $n, m > 0$ such that $y^ n, z^ m$ are in the image of $\varphi $. Then $x^{nm}$ is the residue of $(y/z)^{nm} = (y^ n)^ m/(z^ m)^ n$ which is in the image of $R_\mathfrak p \to S_\mathfrak q$. Hence $x^{nm}$ is in the image of $\kappa (\mathfrak p) \to \kappa (\mathfrak q)$. $\square$

Lemma 10.46.4. Let $\varphi : R \to S$ be a ring map. Assume

$S$ is generated as an $R$-algebra by elements $x$ such that $x^2, x^3 \in \varphi (R)$, and

$\mathop{\mathrm{Ker}}(\varphi )$ is locally nilpotent,

Then $\varphi $ induces isomorphisms on residue fields and a homeomorphism of spectra. For any ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ also satisfies (a) and (b).

**Proof.**
Assume (a) and (b). The map on spectra is closed as $S$ is integral over $R$, see Lemmas 10.41.6 and 10.36.22. The image is dense by Lemma 10.30.6. Thus $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective. If $\mathfrak q \subset S$ is a prime lying over $\mathfrak p \subset R$ then the field extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is generated by elements $\alpha \in \kappa (\mathfrak q)$ whose square and cube are in $\kappa (\mathfrak p)$. Thus clearly $\alpha \in \kappa (\mathfrak p)$ and we find that $\kappa (\mathfrak q) = \kappa (\mathfrak p)$. If $\mathfrak q, \mathfrak q'$ were two distinct primes lying over $\mathfrak p$, then at least one of the generators $x$ of $S$ as in (a) would have distinct images in $\kappa (\mathfrak q) = \kappa (\mathfrak p)$ and $\kappa (\mathfrak q') = \kappa (\mathfrak p)$. This would contradict the fact that both $x^2$ and $x^3$ do have the same image. This proves that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is injective hence a homeomorphism (by what was already shown).

Since $\varphi $ induces a homeomorphism on spectra, it is in particular surjective on spectra which is a property preserved under any base change, see Lemma 10.30.3. Therefore for any $R \to R'$ the kernel of the ring map $R' \to R' \otimes _ R S$ consists of nilpotent elements, see Lemma 10.30.6, in other words (b) holds for $R' \to R' \otimes _ R S$. It is clear that (a) is preserved under base change. $\square$

Lemma 10.46.5. Let $p$ be a prime number. Let $n, m > 0$ be two integers. There exists an integer $a$ such that $(x + y)^{p^ a}, p^ a(x + y) \in \mathbf{Z}[x^{p^ n}, p^ nx, y^{p^ m}, p^ my]$.

**Proof.**
This is clear for $p^ a(x + y)$ as soon as $a \geq n, m$. In fact, pick $a \gg n, m$. Write

For every $i, j \geq 0$ with $i + j = p^ a$ write $i = q p^ n + r$ with $r \in \{ 0, \ldots , p^ n - 1\} $ and $j = q' p^ m + r'$ with $r' \in \{ 0, \ldots , p^ m - 1\} $. The condition $(x + y)^{p^ a} \in \mathbf{Z}[x^{p^ n}, p^ nx, y^{p^ m}, p^ my]$ holds if

If $r = r' = 0$ then the divisibility holds. If $r \not= 0$, then we write

Since $r \not= 0$ the rational number $p^ a/i$ has $p$-adic valuation at least $a - (n - 1)$ (because $i$ is not divisible by $p^ n$). Thus ${p^ a \choose i, j}$ is divisible by $p^{a - n + 1}$ in this case. Similarly, we see that if $r' \not= 0$, then ${p^ a \choose i, j}$ is divisible by $p^{a - m + 1}$. Picking $a = np^ n + mp^ m + n + m$ will work. $\square$

Lemma 10.46.6. Let $k'/k$ be a field extension. Let $p$ be a prime number. The following are equivalent

$k'$ is generated as a field extension of $k$ by elements $x$ such that there exists an $n > 0$ with $x^{p^ n} \in k$ and $p^ nx \in k$, and

$k = k'$ or the characteristic of $k$ and $k'$ is $p$ and $k'/k$ is purely inseparable.

**Proof.**
Let $x \in k'$. If there exists an $n > 0$ with $x^{p^ n} \in k$ and $p^ nx \in k$ and if the characteristic is not $p$, then $x \in k$. If the characteristic is $p$, then we find $x^{p^ n} \in k$ and hence $x$ is purely inseparable over $k$.
$\square$

Lemma 10.46.7. Let $\varphi : R \to S$ be a ring map. Let $p$ be a prime number. Assume

$S$ is generated as an $R$-algebra by elements $x$ such that there exists an $n > 0$ with $x^{p^ n} \in \varphi (R)$ and $p^ nx \in \varphi (R)$, and

$\mathop{\mathrm{Ker}}(\varphi )$ is locally nilpotent,

Then $\varphi $ induces a homeomorphism of spectra and induces residue field extensions satisfying the equivalent conditions of Lemma 10.46.6. For any ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ also satisfies (a) and (b).

**Proof.**
Assume (a) and (b). Note that (b) is equivalent to condition (2) of Lemma 10.46.3. Let $T \subset S$ be the set of elements $x \in S$ such that there exists an integer $n > 0$ such that $x^{p^ n} , p^ n x \in \varphi (R)$. We claim that $T = S$. This will prove that condition (1) of Lemma 10.46.3 holds and hence $\varphi $ induces a homeomorphism on spectra. By assumption (a) it suffices to show that $T \subset S$ is an $R$-sub algebra. If $x \in T$ and $y \in R$, then it is clear that $yx \in T$. Suppose $x, y \in T$ and $n, m > 0$ such that $x^{p^ n}, y^{p^ m}, p^ n x, p^ m y \in \varphi (R)$. Then $(xy)^{p^{n + m}}, p^{n + m}xy \in \varphi (R)$ hence $xy \in T$. We have $x + y \in T$ by Lemma 10.46.5 and the claim is proved.

Since $\varphi $ induces a homeomorphism on spectra, it is in particular surjective on spectra which is a property preserved under any base change, see Lemma 10.30.3. Therefore for any $R \to R'$ the kernel of the ring map $R' \to R' \otimes _ R S$ consists of nilpotent elements, see Lemma 10.30.6, in other words (b) holds for $R' \to R' \otimes _ R S$. It is clear that (a) is preserved under base change. Finally, the condition on residue fields follows from (a) as generators for $S$ as an $R$-algebra map to generators for the residue field extensions. $\square$

Lemma 10.46.8. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ induces an injective map of spectra,

$\varphi $ induces purely inseparable residue field extensions.

Then for any ring map $R \to R'$ properties (1) and (2) are true for $R' \to R' \otimes _ R S$.

**Proof.**
Set $S' = R' \otimes _ R S$ so that we have a commutative diagram of continuous maps of spectra of rings

Let $\mathfrak p' \subset R'$ be a prime ideal lying over $\mathfrak p \subset R$. If there is no prime ideal of $S$ lying over $\mathfrak p$, then there is no prime ideal of $S'$ lying over $\mathfrak p'$. Otherwise, by Remark 10.18.5 there is a unique prime ideal $\mathfrak r$ of $F = S \otimes _ R \kappa (\mathfrak p)$ whose residue field is purely inseparable over $\kappa (\mathfrak p)$. Consider the ring maps

By Lemma 10.25.1 the ideal $\mathfrak r \subset F$ is locally nilpotent, hence we may apply Lemma 10.46.1 to the ring map $F \to \kappa (\mathfrak r)$. We may apply Lemma 10.46.7 to the ring map $\kappa (\mathfrak p) \to \kappa (\mathfrak r)$. Hence the composition and the second arrow in the maps

induces bijections on spectra and purely inseparable residue field extensions. This implies the same thing for the first map. Since

we conclude by the discussion in Remark 10.18.5. $\square$

Lemma 10.46.9. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ is integral,

$\varphi $ induces an injective map of spectra,

$\varphi $ induces purely inseparable residue field extensions.

Then $\varphi $ induces a homeomorphism from $\mathop{\mathrm{Spec}}(S)$ onto a closed subset of $\mathop{\mathrm{Spec}}(R)$ and for any ring map $R \to R'$ properties (1), (2), (3) are true for $R' \to R' \otimes _ R S$.

**Proof.**
The map on spectra is closed by Lemmas 10.41.6 and 10.36.22. The properties are preserved under base change by Lemmas 10.46.8 and 10.36.13.
$\square$

Lemma 10.46.10. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ is integral,

$\varphi $ induces an bijective map of spectra,

$\varphi $ induces purely inseparable residue field extensions.

Then $\varphi $ induces a homeomorphism on spectra and for any ring map $R \to R'$ properties (1), (2), (3) are true for $R' \to R' \otimes _ R S$.

Lemma 10.46.11. Let $\varphi : R \to S$ be a ring map such that

the kernel of $\varphi $ is locally nilpotent, and

$S$ is generated as an $R$-algebra by elements $x$ such that there exist $n > 0$ and a polynomial $P(T) \in R[T]$ whose image in $S[T]$ is $(T - x)^ n$.

Then $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism and $R \to S$ induces purely inseparable extensions of residue fields. Moreover, conditions (1) and (2) remain true on arbitrary base change.

**Proof.**
We may replace $R$ by $R/\mathop{\mathrm{Ker}}(\varphi )$, see Lemma 10.46.1. Assumption (2) implies $S$ is generated over $R$ by elements which are integral over $R$. Hence $R \subset S$ is integral (Lemma 10.36.7). In particular $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective and closed (Lemmas 10.36.17, 10.41.6, and 10.36.22).

Let $x \in S$ be one of the generators in (2), i.e., there exists an $n > 0$ be such that $(T - x)^ n \in R[T]$. Let $\mathfrak p \subset R$ be a prime. The $\kappa (\mathfrak p) \otimes _ R S$ ring is nonzero by the above and Lemma 10.18.6. If the characteristic of $\kappa (\mathfrak p)$ is zero then we see that $nx \in R$ implies $1 \otimes x$ is in the image of $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$. Hence $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$ is an isomorphism. If the characteristic of $\kappa (\mathfrak p)$ is $p > 0$, then write $n = p^ k m$ with $m$ prime to $p$. In $\kappa (\mathfrak p) \otimes _ R S[T]$ we have

and we see that $mx^{p^ k} \in R$. This implies that $1 \otimes x^{p^ k}$ is in the image of $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$. Hence Lemma 10.46.7 applies to $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$. In both cases we conclude that $\kappa (\mathfrak p) \otimes _ R S$ has a unique prime ideal with residue field purely inseparable over $\kappa (\mathfrak p)$. By Remark 10.18.5 we conclude that $\varphi $ is bijective on spectra.

The statement on base change is immediate. $\square$

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