## 10.158 Formal smoothness of fields

In this section we show that field extensions are formally smooth if and only if they are separable. However, we first prove finitely generated field extensions are separable algebraic if and only if they are formally unramified.

Lemma 10.158.1. Let $k \subset K$ be a finitely generated field extension. The following are equivalent

$K$ is a finite separable field extension of $k$,

$\Omega _{K/k} = 0$,

$K$ is formally unramified over $k$,

$K$ is unramified over $k$,

$K$ is formally étale over $k$,

$K$ is étale over $k$.

**Proof.**
The equivalence of (2) and (3) is Lemma 10.148.2. By Lemma 10.143.4 we see that (1) is equivalent to (6). Property (6) implies (5) and (4) which both in turn imply (3) (Lemmas 10.150.2, 10.151.3, and 10.151.2). Thus it suffices to show that (2) implies (1). Choose a finitely generated $k$-subalgebra $A \subset K$ such that $K$ is the fraction field of the domain $A$. Set $S = A \setminus \{ 0\} $. Since $0 = \Omega _{K/k} = S^{-1}\Omega _{A/k}$ (Lemma 10.131.8) and since $\Omega _{A/k}$ is finitely generated (Lemma 10.131.16), we can replace $A$ by a localization $A_ f$ to reduce to the case that $\Omega _{A/k} = 0$ (details omitted). Then $A$ is unramified over $k$, hence $K/k$ is finite separable for example by Lemma 10.151.5 applied with $\mathfrak q = (0)$.
$\square$

Lemma 10.158.2. Let $k$ be a perfect field of characteristic $p > 0$. Let $K/k$ be an extension. Let $a \in K$. Then $\text{d}a = 0$ in $\Omega _{K/k}$ if and only if $a$ is a $p$th power.

**Proof.**
By Lemma 10.131.5 we see that there exists a subfield $k \subset L \subset K$ such that $k \subset L$ is a finitely generated field extension and such that $\text{d}a$ is zero in $\Omega _{L/k}$. Hence we may assume that $K$ is a finitely generated field extension of $k$.

Choose a transcendence basis $x_1, \ldots , x_ r \in K$ such that $K$ is finite separable over $k(x_1, \ldots , x_ r)$. This is possible by the definitions, see Definitions 10.45.1 and 10.42.1. We remark that the result holds for the purely transcendental subfield $k(x_1, \ldots , x_ r) \subset K$. Namely,

\[ \Omega _{k(x_1, \ldots , x_ r)/k} = \bigoplus \nolimits _{i = 1}^ r k(x_1, \ldots , x_ r) \text{d}x_ i \]

and any rational function all of whose partial derivatives are zero is a $p$th power. Moreover, we also have

\[ \Omega _{K/k} = \bigoplus \nolimits _{i = 1}^ r K\text{d}x_ i \]

since $k(x_1, \ldots , x_ r) \subset K$ is finite separable (computation omitted). Suppose $a \in K$ is an element such that $\text{d}a = 0$ in the module of differentials. By our choice of $x_ i$ we see that the minimal polynomial $P(T) \in k(x_1, \ldots , x_ r)[T]$ of $a$ is separable. Write

\[ P(T) = T^ d + \sum \nolimits _{i = 1}^ d a_ i T^{d - i} \]

and hence

\[ 0 = \text{d}P(a) = \sum \nolimits _{i = 1}^ d a^{d - i}\text{d}a_ i \]

in $\Omega _{K/k}$. By the description of $\Omega _{K/k}$ above and the fact that $P$ was the minimal polynomial of $a$, we see that this implies $\text{d}a_ i = 0$. Hence $a_ i = b_ i^ p$ for each $i$. Therefore by Fields, Lemma 9.28.2 we see that $a$ is a $p$th power.
$\square$

Lemma 10.158.3. Let $k$ be a field of characteristic $p > 0$. Let $a_1, \ldots , a_ n \in k$ be elements such that $\text{d}a_1, \ldots , \text{d}a_ n$ are linearly independent in $\Omega _{k/\mathbf{F}_ p}$. Then the field extension $k(a_1^{1/p}, \ldots , a_ n^{1/p})$ has degree $p^ n$ over $k$.

**Proof.**
By induction on $n$. If $n = 1$ the result is Lemma 10.158.2. For the induction step, suppose that $k(a_1^{1/p}, \ldots , a_{n - 1}^{1/p})$ has degree $p^{n - 1}$ over $k$. We have to show that $a_ n$ does not map to a $p$th power in $k(a_1^{1/p}, \ldots , a_{n - 1}^{1/p})$. If it does then we can write

\begin{align*} a_ n & = \left(\sum \nolimits _{I = (i_1, \ldots , i_{n - 1}),\ 0 \leq i_ j \leq p - 1} \lambda _ I a_1^{i_1/p} \ldots a_{n - 1}^{i_{n - 1}/p}\right)^ p \\ & = \sum \nolimits _{I = (i_1, \ldots , i_{n - 1}),\ 0 \leq i_ j \leq p - 1} \lambda _ I^ p a_1^{i_1} \ldots a_{n - 1}^{i_{n - 1}} \end{align*}

Applying $\text{d}$ we see that $\text{d}a_ n$ is linearly dependent on $\text{d}a_ i$, $i < n$. This is a contradiction.
$\square$

Lemma 10.158.4. Let $k$ be a field of characteristic $p > 0$. The following are equivalent:

the field extension $K/k$ is separable (see Definition 10.42.1), and

the map $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective.

**Proof.**
Write $K$ as a directed colimit $K = \mathop{\mathrm{colim}}\nolimits _ i K_ i$ of finitely generated field extensions $k \subset K_ i$. By definition $K$ is separable if and only if each $K_ i$ is separable over $k$, and by Lemma 10.131.5 we see that $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective if and only if each $K_ i \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K_ i/\mathbf{F}_ p}$ is injective. Hence we may assume that $K/k$ is a finitely generated field extension.

Assume $k \subset K$ is a finitely generated field extension which is separable. Choose $x_1, \ldots , x_{r + 1} \in K$ as in Lemma 10.42.3. In this case there exists an irreducible polynomial $G(X_1, \ldots , X_{r + 1}) \in k[X_1, \ldots , X_{r + 1}]$ such that $G(x_1, \ldots , x_{r + 1}) = 0$ and such that $\partial G/\partial X_{r + 1}$ is not identically zero. Moreover $K$ is the field of fractions of the domain. $S = K[X_1, \ldots , X_{r + 1}]/(G)$. Write

\[ G = \sum a_ I X^ I, \quad X^ I = X_1^{i_1}\ldots X_{r + 1}^{i_{r + 1}}. \]

Using the presentation of $S$ above we see that

\[ \Omega _{S/\mathbf{F}_ p} = \frac{ S \otimes _ k \Omega _ k \oplus \bigoplus \nolimits _{i = 1, \ldots , r + 1} S\text{d}X_ i }{ \langle \sum X^ I \text{d}a_ I + \sum \partial G/\partial X_ i \text{d}X_ i \rangle } \]

Since $\Omega _{K/\mathbf{F}_ p}$ is the localization of the $S$-module $\Omega _{S/\mathbf{F}_ p}$ (see Lemma 10.131.8) we conclude that

\[ \Omega _{K/\mathbf{F}_ p} = \frac{ K \otimes _ k \Omega _ k \oplus \bigoplus \nolimits _{i = 1, \ldots , r + 1} K\text{d}X_ i }{ \langle \sum X^ I \text{d}a_ I + \sum \partial G/\partial X_ i \text{d}X_ i \rangle } \]

Now, since the polynomial $\partial G/\partial X_{r + 1}$ is not identically zero we conclude that the map $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{S/\mathbf{F}_ p}$ is injective as desired.

Assume $k \subset K$ is a finitely generated field extension and that $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective. (This part of the proof is the same as the argument proving Lemma 10.44.1.) Let $x_1, \ldots , x_ r$ be a transcendence basis of $K$ over $k$ such that the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset K$ is minimal. If $K$ is separable over $k(x_1, \ldots , x_ r)$ then we win. Assume this is not the case to get a contradiction. Then there exists an element $\alpha \in K$ which is not separable over $k(x_1, \ldots , x_ r)$. Let $P(T) \in k(x_1, \ldots , x_ r)[T]$ be its minimal polynomial. Because $\alpha $ is not separable actually $P$ is a polynomial in $T^ p$. Clear denominators to get an irreducible polynomial

\[ G(X_1, \ldots , X_ r, T) = \sum a_{I, i} X^ I T^ i \in k[X_1, \ldots , X_ r, T] \]

such that $G(x_1, \ldots , x_ r, \alpha ) = 0$ in $L$. Note that this means $k[X_1, \ldots , X_ r, T]/(G) \subset L$. We may assume that for some pair $(I_0, i_0)$ the coefficient $a_{I_0, i_0} = 1$. We claim that $\text{d}G/\text{d}X_ i$ is not identically zero for at least one $i$. Namely, if this is not the case, then $G$ is actually a polynomial in $X_1^ p, \ldots , X_ r^ p, T^ p$. Then this means that

\[ \sum \nolimits _{(I, i) \not= (I_0, i_0)} x^ I\alpha ^ i \text{d}a_{I, i} \]

is zero in $\Omega _{K/\mathbf{F}_ p}$. Note that there is no $k$-linear relation among the elements

\[ \{ x^ I\alpha ^ i \mid a_{I, i} \not= 0 \text{ and } (I, i) \not= (I_0, i_0)\} \]

of $K$. Hence the assumption that $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective this implies that $\text{d}a_{I, i} = 0$ in $\Omega _{k/\mathbf{F}_ p}$ for all $(I, i)$. By Lemma 10.158.2 we see that each $a_{I, i}$ is a $p$th power, which implies that $G$ is a $p$th power contradicting the irreducibility of $G$. Thus, after renumbering, we may assume that $\text{d}G/\text{d}X_1$ is not zero. Then we see that $x_1$ is separably algebraic over $k(x_2, \ldots , x_ r, \alpha )$, and that $x_2, \ldots , x_ r, \alpha $ is a transcendence basis of $L$ over $k$. This means that the degree of inseparability of the finite extension $k(x_2, \ldots , x_ r, \alpha ) \subset L$ is less than the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$, which is a contradiction.
$\square$

Lemma 10.158.5. Let $k \subset K$ be an extension of fields. If $K$ is formally smooth over $k$, then $K$ is a separable extension of $k$.

**Proof.**
Assume $K$ is formally smooth over $k$. By Lemma 10.138.9 we see that $K \otimes _ k \Omega _{k/\mathbf{Z}} \to \Omega _{K/\mathbf{Z}}$ is injective. Hence $K$ is separable over $k$ by Lemma 10.158.4.
$\square$

Lemma 10.158.6. Let $k \subset K$ be an extension of fields. Then $K$ is formally smooth over $k$ if and only if $H_1(L_{K/k}) = 0$.

**Proof.**
This follows from Proposition 10.138.8 and the fact that a vector spaces is free (hence projective).
$\square$

Lemma 10.158.7. Let $k \subset K$ be an extension of fields.

If $K$ is purely transcendental over $k$, then $K$ is formally smooth over $k$.

If $K$ is separable algebraic over $k$, then $K$ is formally smooth over $k$.

If $K$ is separable over $k$, then $K$ is formally smooth over $k$.

**Proof.**
For (1) write $K = k(x_ j; j \in J)$. Suppose that $A$ is a $k$-algebra, and $I \subset A$ is an ideal of square zero. Let $\varphi : K \to A/I$ be a $k$-algebra map. Let $a_ j \in A$ be an element such that $a_ j \mod I = \varphi (x_ j)$. Then it is easy to see that there is a unique $k$-algebra map $K \to A$ which maps $x_ j$ to $a_ j$ and which reduces to $\varphi $ mod $I$. Hence $k \subset K$ is formally smooth.

In case (2) we see that $k \subset K$ is a colimit of étale ring extensions. An étale ring map is formally étale (Lemma 10.150.2). Hence this case follows from Lemma 10.150.3 and the trivial observation that a formally étale ring map is formally smooth.

In case (3), write $K = \mathop{\mathrm{colim}}\nolimits K_ i$ as the filtered colimit of its finitely generated sub $k$-extensions. By Definition 10.42.1 each $K_ i$ is separable algebraic over a purely transcendental extension of $k$. Hence $K_ i/k$ is formally smooth by cases (1) and (2) and Lemma 10.138.3. Thus $H_1(L_{K_ i/k}) = 0$ by Lemma 10.158.6. Hence $H_1(L_{K/k}) = 0$ by Lemma 10.134.9. Hence $K/k$ is formally smooth by Lemma 10.158.6 again.
$\square$

slogan
Lemma 10.158.8. Let $k$ be a field.

If the characteristic of $k$ is zero, then any extension field of $k$ is formally smooth over $k$.

If the characteristic of $k$ is $p > 0$, then $k \subset K$ is formally smooth if and only if it is a separable field extension.

**Proof.**
Combine Lemmas 10.158.5 and 10.158.7.
$\square$

Here we put together all the different characterizations of separable field extensions.

Proposition 10.158.9. Let $k \subset K$ be a field extension. If the characteristic of $k$ is zero then

$K$ is separable over $k$,

$K$ is geometrically reduced over $k$,

$K$ is formally smooth over $k$,

$H_1(L_{K/k}) = 0$, and

the map $K \otimes _ k \Omega _{k/\mathbf{Z}} \to \Omega _{K/\mathbf{Z}}$ is injective.

If the characteristic of $k$ is $p > 0$, then the following are equivalent:

$K$ is separable over $k$,

the ring $K \otimes _ k k^{1/p}$ is reduced,

$K$ is geometrically reduced over $k$,

the map $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective,

$H_1(L_{K/k}) = 0$, and

$K$ is formally smooth over $k$.

**Proof.**
This is a combination of Lemmas 10.44.1, 10.158.8 10.158.5, and 10.158.4.
$\square$

Here is yet another characterization of finitely generated separable field extensions.

Lemma 10.158.10. Let $k \subset K$ be a finitely generated field extension. Then $K$ is separable over $k$ if and only if $K$ is the localization of a smooth $k$-algebra.

**Proof.**
Choose a finite type $k$-algebra $R$ which is a domain whose fraction field is $K$. Lemma 10.140.9 says that $k \to R$ is smooth at $(0)$ if and only if $K/k$ is separable. This proves the lemma.
$\square$

Lemma 10.158.11. Let $k \subset K$ be a field extension. Then $K$ is a filtered colimit of global complete intersection algebras over $k$. If $K/k$ is separable, then $K$ is a filtered colimit of smooth algebras over $k$.

**Proof.**
Suppose that $E \subset K$ is a finite subset. It suffices to show that there exists a $k$ subalgebra $A \subset K$ which contains $E$ and which is a global complete intersection (resp. smooth) over $k$. The separable/smooth case follows from Lemma 10.158.10. In general let $L \subset K$ be the subfield generated by $E$. Pick a transcendence basis $x_1, \ldots , x_ d \in L$ over $k$. The extension $k(x_1, \ldots , x_ d) \subset L$ is finite. Say $L = k(x_1, \ldots , x_ d)[y_1, \ldots , y_ r]$. Pick inductively polynomials $P_ i \in k(x_1, \ldots , x_ d)[Y_1, \ldots , Y_ r]$ such that $P_ i = P_ i(Y_1, \ldots , Y_ i)$ is monic in $Y_ i$ over $k(x_1, \ldots , x_ d)[Y_1, \ldots , Y_{i - 1}]$ and maps to the minimum polynomial of $y_ i$ in $k(x_1, \ldots , x_ d)[y_1, \ldots , y_{i - 1}][Y_ i]$. Then it is clear that $P_1, \ldots , P_ r$ is a regular sequence in $k(x_1, \ldots , x_ r)[Y_1, \ldots , Y_ r]$ and that $L = k(x_1, \ldots , x_ r)[Y_1, \ldots , Y_ r]/(P_1, \ldots , P_ r)$. If $h \in k[x_1, \ldots , x_ d]$ is a polynomial such that $P_ i \in k[x_1, \ldots , x_ d, 1/h, Y_1, \ldots , Y_ r]$, then we see that $P_1, \ldots , P_ r$ is a regular sequence in $k[x_1, \ldots , x_ d, 1/h, Y_1, \ldots , Y_ r]$ and $A = k[x_1, \ldots , x_ d, 1/h, Y_1, \ldots , Y_ r]/(P_1, \ldots , P_ r)$ is a global complete intersection. After adjusting our choice of $h$ we may assume $E \subset A$ and we win.
$\square$

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