Lemma 54.2.1. Let $\varphi : R[x]/(x^ p - a) \to R[y]/(y^ p - b)$ be an $R$-algebra homomorphism. Then $\text{Tr}_ x = \text{Tr}_ y \circ \varphi $.

## 54.2 A trace map in positive characteristic

Some of the results in this section can be deduced from the much more general discussion on traces on differential forms in de Rham Cohomology, Section 50.19. See Remark 54.2.3 for a discussion.

We fix a prime number $p$. Let $R$ be an $\mathbf{F}_ p$-algebra. Given an $a \in R$ set $S = R[x]/(x^ p - a)$. Define an $R$-linear map

by the rule

This makes sense as $\Omega _{S/R}$ is a free $R$-module with basis $x^ i\text{d}x$, $0 \leq i \leq p - 1$. The following lemma implies that the trace map is well defined, i.e., independent of the choice of the coordinate $x$.

**Proof.**
Say $\varphi (x) = \lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1}$ with $\lambda _ i \in R$. The condition that mapping $x$ to $\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1}$ induces an $R$-algebra homomorphism $R[x]/(x^ p - a) \to R[y]/(y^ p - b)$ is equivalent to the condition that

in the ring $R$. Consider the polynomial ring

with the element $a = \lambda _0^ p + \lambda _1^ p b + \ldots + \lambda _{p - 1}^ pb^{p - 1}$ Consider the universal algebra map $\varphi _{univ} : R_{univ}[x]/(x^ p - a) \to R_{univ}[y]/(y^ p - b)$ given by mapping $x$ to $\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1}$. We obtain a canonical map

sending $b, \lambda _ i$ to $b, \lambda _ i$. By construction we get a commutative diagram

and the horizontal arrows are compatible with the trace maps. Hence it suffices to prove the lemma for the map $\varphi _{univ}$. Thus we may assume $R = \mathbf{F}_ p[b, \lambda _0, \ldots , \lambda _{p - 1}]$ is a polynomial ring. We will check the lemma holds in this case by evaluating $\text{Tr}_ y(\varphi (x)^ i\text{d}\varphi (x))$ for $i = 0 , \ldots , p - 1$.

The case $0 \leq i \leq p - 2$. Expand

in the ring $R[y]/(y^ p - b)$. We have to show that the coefficient of $y^{p - 1}$ is zero. For this it suffices to show that the expression above as a polynomial in $y$ has vanishing coefficients in front of the powers $y^{pk - 1}$. Then we write our polynomial as

and indeed the coefficients of $y^{kp - 1}$ are all zero.

The case $i = p - 1$. Expand

in the ring $R[y]/(y^ p - b)$. To finish the proof we have to show that the coefficient of $y^{p - 1}$ times $\text{d}b$ is $\text{d}a$. Here we use that $R$ is $S/pS$ where $S = \mathbf{Z}[b, \lambda _0, \ldots , \lambda _{p - 1}]$. Then the above, as a polynomial in $y$, is equal to

Since $\frac{\text{d}}{\text{d}y}(y^{pk}) = pk y^{pk - 1}$ it suffices to understand the coefficients of $y^{pk}$ in the polynomial $(\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1})^ p$ modulo $p$. The sum of these terms gives

Whence we see that we obtain after applying the operator $\frac{\text{d}}{p\text{d}y}$ and after reducing modulo $y^ p - b$ the value

for the coefficient of $y^{p - 1}$ we wanted to compute. Now because $a = \lambda _0^ p + \lambda _1^ p b + \ldots + \lambda _{p - 1}^ pb^{p - 1}$ in $R$ we obtain that

in $R$. This proves that the coefficient of $y^{p - 1}$ is as desired. $\square$

Lemma 54.2.2. Let $\mathbf{F}_ p \subset \Lambda \subset R \subset S$ be ring extensions and assume that $S$ is isomorphic to $R[x]/(x^ p - a)$ for some $a \in R$. Then there are canonical $R$-linear maps

for $t \geq 0$ such that

for $\eta _ i \in \Omega _{R/\Lambda }$ and such that $\text{Tr}$ annihilates the image of $S \otimes _ R \Omega _{R/\Lambda }^{t + 1} \to \Omega _{S/\Lambda }^{t + 1}$.

**Proof.**
For $t = 0$ we use the composition

where the second map is Lemma 54.2.1. There is an exact sequence

(Algebra, Lemma 10.134.4). The module $\Omega _{S/R}$ is free over $S$ with basis $\text{d}x$ and the module $H_1(L_{S/R})$ is free over $S$ with basis $x^ p - a$ which $\delta $ maps to $-\text{d}a \otimes 1$ in $\Omega _{R/\Lambda } \otimes _ R S$. In particular, if we set

then we see that $\mathop{\mathrm{Coker}}(\delta ) = M \otimes _ R S$. We obtain a canonical map

Now, since the image of the map $\text{Tr} : \Omega _{S/R} \to \Omega _{R/\Lambda }$ of Lemma 54.2.1 is contained in $R\text{d}a$ we see that wedging with an element in the image annihilates $\text{d}a$. Hence there is a canonical map

mapping $\overline{\eta }_1 \wedge \ldots \wedge \overline{\eta }_ t \wedge \omega $ to $\eta _1 \wedge \ldots \wedge \eta _ t \wedge \text{Tr}(\omega )$. $\square$

Remark 54.2.3. Let $\mathbf{F}_ p \subset \Lambda \subset R \subset S$ and $\text{Tr}$ be as in Lemma 54.2.2. By de Rham Cohomology, Proposition 50.19.3 there is a canonical map of complexes

The computation in de Rham Cohomology, Example 50.19.4 shows that $\Theta _{S/R}(x^ i \text{d}x) = \text{Tr}_ x(x^ i\text{d}x)$ for all $i$. Since $\text{Trace}_{S/R} = \Theta ^0_{S/R}$ is identically zero and since

for $a \in \Omega ^ i_{R/\Lambda }$ and $b \in \Omega ^ j_{S/\Lambda }$ it follows that $\text{Tr} = \Theta _{S/R}$. The advantage of using $\text{Tr}$ is that it is a good deal more elementary to construct.

Lemma 54.2.4. Let $S$ be a scheme over $\mathbf{F}_ p$. Let $f : Y \to X$ be a finite morphism of Noetherian normal integral schemes over $S$. Assume

the extension of function fields is purely inseparable of degree $p$, and

$\Omega _{X/S}$ is a coherent $\mathcal{O}_ X$-module (for example if $X$ is of finite type over $S$).

For $i \geq 1$ there is a canonical map

whose stalk in the generic point of $X$ recovers the trace map of Lemma 54.2.2.

**Proof.**
The exact sequence $f^*\Omega _{X/S} \to \Omega _{Y/S} \to \Omega _{Y/X} \to 0$ shows that $\Omega _{Y/S}$ and hence $f_*\Omega _{Y/S}$ are coherent modules as well. Thus it suffices to prove the trace map in the generic point extends to stalks at $x \in X$ with $\dim (\mathcal{O}_{X, x}) = 1$, see Divisors, Lemma 31.12.14. Thus we reduce to the case discussed in the next paragraph.

Assume $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ with $A$ a discrete valuation ring and $B$ finite over $A$. Since the induced extension $L/K$ of fraction fields is purely inseparable, we see that $B$ is local too. Hence $B$ is a discrete valuation ring too. Then either

$B/A$ has ramification index $p$ and hence $B = A[x]/(x^ p - a)$ where $a \in A$ is a uniformizer, or

$\mathfrak m_ B = \mathfrak m_ A B$ and the residue field $B/\mathfrak m_ A B$ is purely inseparable of degree $p$ over $\kappa _ A = A/\mathfrak m_ A$. Choose any $x \in B$ whose residue class is not in $\kappa _ A$ and then we'll have $B = A[x]/(x^ p - a)$ where $a \in A$ is a unit.

Let $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$ be an affine open such that $X$ maps into $\mathop{\mathrm{Spec}}(\Lambda )$. Then we can apply Lemma 54.2.2 to see that the trace map extends to $\Omega ^ i_{B/\Lambda } \to \Omega ^ i_{A/\Lambda }$ for all $i \geq 1$. $\square$

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