Ring endomorphism of $p$-adic integers

Question

I am doing an individual study of an abstract algebra for number theory course online. I just started, so I hope my question just note come off as too trivial. The lecture notes state that the ring of $p$-adic integers does not have a ring endomorphism.

Questions:

1. Does not the identity mapping work as a counterexample?

Then, assuming they meant: "no endomorphism except the trivial case", so the entire thing is not just a mistake:

2. I still cannot convince myself that there is no other ring endomorphism of $p$-adic integers. Could you please give me a hint how to prove it or point me to literature where such a proof is shown?

The first thing to note is that any ring endomorphism $f : \mathbf Z_p \to \mathbf Z_p$ must send $1\mapsto 1$ and hence send the integers inside of the domain identitcally to the integers inside the codomain.
The intuition is then that the integers are a dense subset of $\mathbf Z_p$ topologically (coming from the $p$-adic metric), and so your morphism that fixes a dense subset must fix everything.

To make this precise we have to arguing some version of the statement that any ring isomorphism must be continuous for this topology. — Alex J Best, Mar 28 '21 at 19:18
@AlexJBest Why not make that an answer? (And it's any ring endomorphism which is continuous, not necessarily isos, although of course in the end it turns out the only endo is the identity iso.) — Torsten Schoeneberg, Mar 28 '21 at 21:15
@TorstenSchoeneberg Does a complete DVR with $v(p)=0$ for all $p$ necessary have non-trivial endomorphisms? — reuns, Mar 28 '21 at 22:00
@TorstenSchoeneberg I ran out of time to complete it :) before I had to do something else, if someone else wants to turn it into an answer please do. — Alex J Best, Mar 28 '21 at 22:04
$f(p^k \Bbb{Z}_p)\subset p^k \Bbb{Z}_p$ so continuity is immediate @AlexJBest — reuns, Mar 28 '21 at 22:20
@reuns I agree but would write more than just that sentence as an answer to this question, based on the OPs background. — Alex J Best, Mar 28 '21 at 22:30
@AlexJBest sorry, but I do not see how the lack of ring endomorphisms follows from your comment. Would you care to elaborate more? — tamboo, Mar 29 '21 at 18:13

score 4 · Answer 1 · edited Oct 25 '24 at 14:02

Re Question 1) Yes, the identity $id: \mathbb Z_p \rightarrow \mathbb Z_p$ is of course a ring endomorphism.

Re Question 2) To show that there is no other, assume $f: \mathbb Z_p \rightarrow \mathbb Z_p$ is any ring endomorphism. We have $f(1)=1$ hence $f(1+1)=1+1$ etc. as well as $f(-1)=-1$ etc., so that

$$(*) \qquad \qquad f(x) =x \text{ at least for all } x \in \mathbb Z.$$

We now will show it for all $x \in \mathbb Z_p$. Generally one could say that $\mathbb Z$ is dense in $\mathbb Z_p$, so it suffices to show continuity of $f$. But we'll be a bit more down to earth.

$(*)$ in particular means $f(p^n)=p^n$ and hence, by $f$ being an endomorphism (edit_corrected, thanks @KCd), $f(p^n \mathbb Z_p) \subseteq p^n\mathbb Z_p$ for all $n \in \mathbb N$. By the definition of the $p$-adic value, this implies that

$$\lvert f(x) \rvert_p \le \lvert x \rvert_p$$ for all $x \in \mathbb Z_p$, hence by $f$ being a ring endomorphism also

$$\lvert f(x)-f(y) \rvert_p \le \lvert x-y\rvert_p$$

for any $x,y \in \mathbb Z_p$, i.e. $f$ is necessarily (uniformly) continuous.

Now let $x \in \mathbb Z_p$. There exists some sequence of integers (!) $x_n \in \mathbb Z$ such that $\lim_{n \to \infty} x_n =x$. (This is the "density" argument in a nutshell.) Then

$$f(x) \stackrel{cont.}= \lim_{n \to \infty}f(x_n) \stackrel{(*)}=\lim_{n\to \infty} x_n \stackrel{def.}=x.$$

Since $x \in \mathbb Z_p$ was arbitrary, we just showed $f=id$.

Being an endomorphism does not let you deduce that $f(p^n\mathbf Z_p) = p^n\mathbf Z_p$ as soon as you did. You can only say at that point $f(p^n\mathbf Z_p) \subset p^n\mathbf Z_p$. Then if $x \not= 0$, so $|x|_p = 1/p^n$ for some $n$, we have $f(x) \in f(p^n\mathbf Z_p) \subset p^n\mathbf Z_p$, so $|f(x)|_p \leq 1/p^n = |x|_p$. Thus $|f(x)|_p \leq |x|_p$ for all $x$ (it's obvious at $x = 0$), so for $x, y \in \mathbf Z_p$ we have $|f(x) - f(y)|_p = |f(x-y)|_p \leq |x-y|_p$. Thus $f$ is (uniformly) continuous, so from $f$ fixing the dense subset $\mathbf Z$, $f$ fixes all of $\mathbf Z_p$. — KCd, Apr 16 '21 at 02:45
$\def\H{\operatorname{Hom}} \def\Z{\mathbf{Z}}$Here's the proof in a nutshell: Any ring endomorphism $\phi$ of $\Z_p$ is continuous for $p^n\Z_p\subset\phi^{-1}(p^n\Z_p)$. Thus precomposition by $\iota:\Z\to\Z_p$ induces a bijection $$ \iota^*:\H(\Z_p,\Z_p)\to\H_{\mathsf{TopRing}}(\Z,\Z_p) $$where $\mathsf{TopRing}$ is the category of topological rings and $\Z$ has the $p$-adic topology (Lemma 6 from here). But there is a unique continuous ring homomorphism $\Z\to\Z_p$ when $\Z$ has the $p$-adic topology. — Elías Guisado Villalgordo, Oct 25 '24 at 14:11

score 2 · Answer 2 · answered Jan 18 '22 at 01:42

For your reference, there is a more general perspective from the theory of "Witt vectors". Throughout, we let $p$ denote a fixed prime and $\mathbb{F}_p$ the finite field with $p$ elements.

The p-adic integers $\mathbb{Z}_p$ are an example of a strict $p$-ring, which is a ring $A$ such that

$p$ is not a zero divisor in $A$,
$A$ is $p$-adically complete and Hausdorff, and
$A/(p)$ is a perfect ring (for example, a finite field).

Then we have the following theorem: if $A$ and $B$ are strict $p$-rings, then there is a natural "mod $p$" bijection between ring homomorphisms $A \to B$ and ring homomorphisms $A/(p) \to B/(p)$ (see theorem 1.2 in https://arxiv.org/abs/1409.7445, which is an expository paper on Witt vectors, and/or Chapter II of Serre's Local Fields). Taking $A = \mathbb{Z}_p = B$, we have a bijection between endomorphisms of $\mathbb{Z}_p$ and endomorphisms of $\mathbb{Z}_p/(p) = \mathbb{F}_p$. The only ring endomorphism of $\mathbb{F}_p$ is the identity, so the same holds for $\mathbb{Z}_p$.

Ring endomorphism of $p$-adic integers

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