I've started reading Neukirch's Algebraic Number Theory book and at the beginning of Chapter II he starts his motivation for the $p$-adic numbers as follows:

"The idea originated from the observation made in the last chapter that the numbers $f\in\mathbb{Z}$ may be viewed in analogy with the polynomials $f(z)\in\mathbb{C}[z]$ as functions on the space $X$ of prime numbers in $\mathbb{Z}$, associating to them their "value" at the point $p\in X$, i.e., the element

$\begin{equation} f(p):=f \mod p \end{equation}$

in the residue class field $\kappa(p)=\mathbb{Z}/p\mathbb{Z}$. This point of view suggests the further question: whether not only the "value" of the integer $f\in\mathbb{Z}$ at $p$, but also the higher derivatives of $f$ can be reasonably defined."

My question is: If we define the higher derivatives of $f\in\mathbb{Z}$ at $p$ as the coefficientes of its $p$-adic expansion (as he does later on), can we give these derivatives an interpretation somehow analogous to the analytical one that goes with the $f(z)\in\mathbb{C}[z]$ polynomials?

I'm asking this because if there is no interpretation other than "coefficientes of the expansion", it seems like a very artificial analogy to me. I mean, if not, why would he even mention derivatives instead of just saying something like: "we expand functions this way, we expand integers this way and it's kind of similar."

Any explanations and/or motivations are welcome. Thanks in advance.

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    $\begingroup$ I think his point is that if one takes $X=\mathbb{A}^1_\mathbb{C}$ as a variety with sheaf $\mathcal{O}_X$, then for any point $p\in X$ one has that $\widehat{\mathcal{O}_{X,p}}=\mathbb{C}[[z-p]]$ and the map $\mathbb{C}[z]=\mathcal{O}_{X,x}\to \widehat{\mathcal{O}_{X,p}}$ is just taking $f\in\mathbb{C}[z]$ and expanding it as a Taylor series at $p$. Similarly, one may view the $p$-adic case as taking the scheme $X=\mathbb{Z}$, some global section $f\in\mathbb{Z}=\mathcal{O}_X(X)$ and mapping it to $\widehat{\mathcal{O}_{X,p}}=\mathbb{Z}_p$. So, yes, in some sense, it is like taking higher $\endgroup$ Jul 21, 2015 at 8:10
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    $\begingroup$ derivatives. One can see this if you think of $\widehat{\mathcal{O}_{X,p}}$ as $\varprojlim \mathcal{O}_{X,p}/\mathfrak{m}_p^n$. Mapping $f$ in $\mathcal{O}_X(X)$ to $\mathcal{O}_{X,p}/\mathfrak{m}_p$ is like taking the constant term, the ring $\mathcal{O}_{X,p}/\mathfrak{m}_p^2$ is like taking the linear term, etc. So, the inverse limit is like taking the Taylor expansion. $\endgroup$ Jul 21, 2015 at 8:12
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    $\begingroup$ What disturbs me most about trying to give meaning to a particular ”coefficient” in a $p$-adic expansion is that these coefficients depend so strongly on your choice of representatives in $\Bbb Z$ or $\Bbb Z_p$ for the elements of $\Bbb F_p$, the residue field. For $\Bbb Z_5$, for instance, maybe you want to use $\{-2,-1,0,1,2\}$ for your possible “coefficients” of your expansion. Makes perfectly good sense. Unless you want to go all the way into abstraction and represent a $p$-adic integer by a Witt vector, it seems to me that the analogy with power series is extremely weak. $\endgroup$
    – Lubin
    Jul 23, 2015 at 4:25
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    $\begingroup$ Well, one can argue forever on this, but I maintain that the only “natural” coefficients for $\Bbb Z_p$ are the Teichmüller representatives, which (except for $2$ and $3$) aren’t in $\Bbb Z$. $\endgroup$
    – Lubin
    Jul 23, 2015 at 12:38
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    $\begingroup$ Thanks guys! I didn't really understand much of what Alex said on his first comment beucase (forgot to mention) I'm still an 2-year undergrad and don't really understand varieties and sheaves yet, but I'll try to look it up on the internet. This question was more like a curiosity to me. I'll just keep reading and in the future hopefully I can read this post again and fully understand it. Thanks again to all of you. $\endgroup$
    – Shoutre
    Jul 23, 2015 at 14:16

1 Answer 1


"My question is: If we define the higher derivatives of f∈Z at p as the coefficientes of its p-adic expansion (as he does later on), can we give these derivatives an interpretation somehow analogous to the analytical one that goes with the f(z)∈C[z] polynomials?"

Answer: At the following link


you will find an elementary construction of a "Taylor morphism"

T1. $T^l: E \rightarrow J^l(E)$

for any commutative ring $A$ and any left $A$-module $E$. If $A:=k[x_1,..,x_n]$ with $k$ a field of characteristic zero, you get a map

T2. $T^l: A \rightarrow J^l(A):=A\otimes_k A/I^{l+1}$

defined by $T^l(f(x_1,..,x_n)):=f(y_1,..,y_n)$. Let $dx_i:=y_i-x_i$. You may prove that there is an isomorphism

T3. $J^l(A)\cong A\{dx_1^{l_1}\cdots dx_n^{l_n}: \sum_i l_i\leq l\}$.

Hence the left $A$-module $J^l(A)$ is a free $A$-module on the set $\{dx_1^{l_1}\cdots dx_n^{l_n}: \sum_i l_i\leq l\}$. In this situation it follows the element $T^l(f(x_1,..,x_n))$ is the Taylor expansion of the polynomial $f$ in the variables $x_i$.

Comment: "I think his point is that if one takes X=A1C as a variety with sheaf OX, then for any point p∈X one has that OX,pˆ=C[[z−p]] and the map C[z]=OX,x→OX,pˆ is just taking f∈C[z] and expanding it as a Taylor series at p. Similarly, one may view the p-adic case as taking the scheme X=Z, some global section f∈Z=OX(X) and mapping it to OX,pˆ=Zp."

Answer: The fiber of the $A$-module $J^l(A)$ at a $k$-rational point $\mathfrak{m}$ is the local ring $A/\mathfrak{m}^{l+1}$, hence this construction may be related to the construction mentioned in the comments above.

In Neukirch, Prop 2.6 page 114 the following isomorphism is constructed:

There is for every non-zero prime $p$ a canonical isomorphism

$T_p: \mathbb{Z}_{p}\cong \mathbb{Z}[[x]]/(x-p)$ and a canonical injective map $i_p: \mathbb{Z} \rightarrow \mathbb{Z}_p$.

There is a canonical map $T: \mathbb{Z}\rightarrow \mathbb{Z}[[x]]$ and we recover the map $i_p$ by passing to the quotient $P(p): \mathbb{Z}[[x]] \rightarrow \mathbb{Z}[[x]]/(x-p)$.

$i_p:=P(p)\circ T: \mathbb{Z}\rightarrow \mathbb{Z}[[x]]/(x-p)\cong \mathbb{Z}_p$.

Hence all the injections $i_p$ can be constructed from one "global" map $T$. We may pass to "spectra" $C:=Spec(\mathbb{Z}[[x]])$ and $S:=Spec(\mathbb{Z})$ and the canonical map

$\pi: C \rightarrow S$.

The ideal $I(p):=(x-p) \subseteq \mathbb{Z}[[x]]$ corresponds to a closed subscheme $C_p:=Z(x-p)\subseteq C$ and we get a canonical morphism $\pi_p: C_p\rightarrow S$. There is a corresponding map of structure sheaves

$\pi_p^{\#}: \mathcal{O}_S \rightarrow (\pi_p)_*\mathcal{O}_{C_p}$

whose global sections is the canonical map $i_p:\mathbb{Z}\rightarrow \mathbb{Z}_p$. Hence all maps $i_p$ may be given a "geometric" construction using the map $\pi$: You restrict the map $\pi$ to the closed subscheme $C_p$ and pass to global sections.


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