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 $.
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
\[ a = \lambda _0^ p + \lambda _1^ p b + \ldots + \lambda _{p - 1}^ pb^{p - 1} \]
in the ring $R$. Consider the polynomial ring
\[ R_{univ} = \mathbf{F}_ p[b, \lambda _0, \ldots , \lambda _{p - 1}] \]
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
\[ R_{univ} \longrightarrow R \]
sending $b, \lambda _ i$ to $b, \lambda _ i$. By construction we get a commutative diagram
\[ \xymatrix{ R_{univ}[x]/(x^ p - a) \ar[r] \ar[d]_{\varphi _{univ}} & R[x]/(x^ p - a) \ar[d]^\varphi \\ R_{univ}[y]/(y^ p - b) \ar[r] & R[y]/(y^ p - b) } \]
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
\[ (\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1})^ i (\lambda _1 + 2 \lambda _2 y + \ldots + (p - 1)\lambda _{p - 1}y^{p - 2}) \]
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
\[ \frac{\text{d}}{(i + 1)\text{d}y} (\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1})^{i + 1} \]
and indeed the coefficients of $y^{kp - 1}$ are all zero.
The case $i = p - 1$. Expand
\[ (\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1})^{p - 1} (\lambda _1 + 2 \lambda _2 y + \ldots + (p - 1)\lambda _{p - 1}y^{p - 2}) \]
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
\[ \frac{\text{d}}{p\text{d}y} (\lambda _0 + \lambda _1 y + \ldots + \lambda _{p - 1}y^{p - 1})^ p \]
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
\[ \lambda _0^ p + \lambda _1^ py^ p + \ldots + \lambda _{p - 1}^ py^{p(p - 1)} \bmod p \]
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
\[ \lambda _1^ p + 2\lambda _2^ pb + \ldots + (p - 1)\lambda _{p - 1}b^{p - 2} \]
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
\[ \text{d}a = (\lambda _1^ p + 2 \lambda _2^ p b + \ldots + (p - 1) \lambda _{p - 1}^ p b^{p - 2}) \text{d}b \]
in $R$. This proves that the coefficient of $y^{p - 1}$ is as desired. $\square$
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