Endomorphisms of p-adics

Question

I know that there is a unique unital homomorphism $\mathbb{R}\rightarrow \mathbb{R}$ [1]. I also know that there is a unique surjective unital homomorphism $\mathbb{Q}_p\rightarrow \mathbb{Q}_p$ for any prime number $p$ [2]. For which primes, if any, do there exist non-surjective unital homomorphisms $\mathbb{Q}_p\rightarrow \mathbb{Q}_p$?

For finite extensions of prime fields, I can prove that every unital homomorphism $F\rightarrow F$ is surjective but $\mathbb{Q}_p$ is not such a field so I do not know what to do.

Both $\mathbb{Q_p}$ and $\mathbb{R}$ have only trivial endomorphisms of the kind taking one to one. — Parthiv Basu, Aug 03 '19 at 12:02
This is a fact; one shows that they have to be continous. You'll definitely find a proof if you google. — Parthiv Basu, Aug 03 '19 at 12:07
Because the endomorphism is the identity on $\mathbb{Q}$ by additivity. And $\mathbb{Q}$ is dense in both $\mathbb{R}$ and $\mathbb{Q}_p$ — Parthiv Basu, Aug 03 '19 at 12:24

score 1 · Accepted Answer · answered Aug 03 '19 at 12:40

Let $f$ be a field homomorphism $\Bbb{R} \to \Bbb{R}$. The usual order can be defined algebraically by $a \ge 0$ iff $x^2-a$ has a root in $\Bbb{R}$, and $f$ preserves that algebraic property, thus $f$ preserves $\ge$, it is continuous and since $\Bbb{Q}$ is dense in $\Bbb{R}$, $f$ is the identity.
Let $p \ne 2$ and $f$ a field homomorphism $\Bbb{Q}_p \to \Bbb{Q}_p$. The $p$-adic valuation can be defined algebraically by $v(a) = n$ iff for some $b \in 1\ldots p-1$, $b p^{-n} a \in 1 + p \Bbb{Z}_p$ iff for all $m \in \Bbb{Z}, 1+ m (b p^{-n} a-1)$ has a square root in $\Bbb{Q}_p$. This algebraic property is preserved by $f$, thus $f$ preverses the valuation, it is continuous and since $\Bbb{Q}$ is dense in $\Bbb{Q}_p$, $f$ is the identity.
For $p = 2$ it is the same with 3rd roots.

Endomorphisms of p-adics

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