## 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.

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.

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

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

2. $k' = k$ 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

$\zeta ' (\tau (x) + 1) = \zeta ' \tau (x + 1) = \sigma (x + 1) = \sigma (x) + 1 = \zeta \tau (x) + 1$

implies that

$\tau (x) (\zeta ' - \zeta ) = 1 - \zeta '$

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

$\tau (x) = (1 - \zeta ')/(\zeta ' - \zeta )$

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

$\zeta \tau (x) + y = \zeta _ y(\tau (x) + y)$

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

$y - y' = \sigma (x + y) - \sigma (x + y') = \zeta _ y(\tau (x + y)) - \zeta _{y'}\tau (x + y') = \zeta _ y(y - y')$

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

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

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

2. $\mathop{\mathrm{Ker}}(\varphi )$ 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

$x^ n \in \mathfrak q \Rightarrow y \in \mathfrak p \Rightarrow x^ n \in \mathfrak q'$

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

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

2. $\mathop{\mathrm{Ker}}(\varphi )$ 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

$(x + y)^{p^ a} = \sum \nolimits _{i, j \geq 0, i + j = p^ a} {p^ a \choose i, j} x^ iy^ j$

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

$p^{nr + mr'} \text{ divides } {p^ a \choose i, j}$

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

${p^ a \choose i, j} = \frac{p^ a}{i} {p^ a - 1 \choose i - 1, j}$

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

1. $k'$ 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

2. $k = k'$ 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

1. $S$ 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

2. $\mathop{\mathrm{Ker}}(\varphi )$ 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

1. $\varphi$ induces an injective map of spectra,

2. $\varphi$ 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

$\xymatrix{ \mathop{\mathrm{Spec}}(S') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(S) \ar[d] \\ \mathop{\mathrm{Spec}}(R') \ar[r] & \mathop{\mathrm{Spec}}(R) }$

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 10.17.8 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

$\kappa (\mathfrak p) \to F \to \kappa (\mathfrak r)$

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

$\kappa (\mathfrak p') \to \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} F \to \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak r)$

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

$\kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} F = \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p) \otimes _ R S = \kappa (\mathfrak p') \otimes _ R S = \kappa (\mathfrak p') \otimes _{R'} R' \otimes _ R S$

we conclude by the discussion in Remark 10.17.8. $\square$

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

1. $\varphi$ is integral,

2. $\varphi$ induces an injective map of spectra,

3. $\varphi$ 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

1. $\varphi$ is integral,

2. $\varphi$ induces an bijective map of spectra,

3. $\varphi$ 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

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

2. $S$ 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 10.17.9. 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

$(T - 1 \otimes x)^ n = ((T - 1 \otimes x)^{p^ k})^ m = (T^{p^ k} - 1 \otimes x^{p^ k})^ m$

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 10.17.8 we conclude that $\varphi$ is bijective on spectra.

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

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