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Let $\mathbb{Z}_p$ the ring of $p$-adic numbers and $\mathbb{F}_p|[t]|= \{ \sum_j ^{\infty} a_j t^j \}$ and the ring of formal power series.

I don't see how to show that $\mathbb{Z}_p$ and $\mathbb{F}_p|[t]|$ aren't isomorphic as rings.

user267839
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    A ring isomorphism preserves characteristic. – Keenan Kidwell Feb 18 '18 at 18:54
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    Contrary to what the notation suggests, $\mathbb{Z}_p$ contains a subring isomorphic to $\mathbb{Z}$ – Maxime Ramzi Feb 18 '18 at 18:55
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    @Max: Some paths through algebra give one the expectation that $\mathbb{Z}_{\mathrm{subscript}}$ is some sort of extension of $\mathbb{Z}$, such as a localization or a completion; containing $\mathbb{Z}$ is very much what the notation does suggest! The use of $\mathbb{Z}_p$ for the finite field really irritates me since $\mathbb{F}_p$ is even better (it generalizes to any finite field, not just of prime order), and if its use were more common we wouldn't have this conflict of conventions. –  Feb 18 '18 at 19:32
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    While elements of $\Bbb Z_p$ might be represented as infinite sums $\sum_{n \geq 0} a_n p^n$ with $0 \leq a_n < p$, it is very different from $\Bbb F_p[[t]]$. For instance $\underbrace{1 \cdot p^0 + \cdots + 1 \cdot p^0}{p \text{ times}} = 1 \cdot p^1$ in $\Bbb Z_p$, while $\underbrace{1 \cdot t^0 + \cdots + 1 \cdot t^0}{p \text{ times}} = 0 \in \Bbb F_p[[t]]$. – Watson Feb 18 '18 at 19:43
  • @Hurkyl : of course, but often when one begins in algebra, things with a subscript $p$ are of characteristic $p$.. but actually my comment was also half-ironic, precisely because of what you're saying – Maxime Ramzi Feb 18 '18 at 19:52
  • Possibly related: https://math.stackexchange.com/questions/2436717 – Watson Feb 18 '18 at 19:59
  • Somehow, in $\Bbb Z_p$, there is a "carry" (looking in terms of the sums $\sum_n a_n p^n$ as above), which does not happen for $\Bbb F_p[[t]]$. – Watson Feb 18 '18 at 20:19

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$\renewcommand{\power}{[\![t]\!]}$ Assume that $\phi : \Bbb F_p\power \to \Bbb Z_p$ is a ring morphism (it might not even be an isomorphism). Then $$p = p \cdot 1_{\Bbb Z_p} = p \cdot \phi(1_{\Bbb F_p}) = \phi(p \cdot 1_{\Bbb F_p}) = \phi(0) = 0,$$ which is not possible. Indeed, the ring $\Bbb Z_p$ has characteristic $0$ (even if it is an inverse limit of rings of positive characteristic [which actually grows to infinity, that's why the inverse limit has characteristic $0$, somehow]). Said differently, $\Bbb Z_p$ contains a copy of $\Bbb Z$, via the injective ring morphism $$ \begin{array}{lrcl} & \Bbb Z & \longrightarrow & \Bbb Z_p \hookrightarrow \prod\limits_{m \geq 0} \Bbb Z /p^m \Bbb Z \\ & n & \longmapsto & ([n]_{p^m})_{m \geq 0}. \end{array} $$

Some remarks:

– Actually, $\Bbb F_p\power$ is not even isomorphic to $\Bbb Z_p$ as additive group, since $\Bbb Z_p$ is torsion-free.

– However, we have a ring isomorphism $\Bbb Z_p \cong \Bbb Z\power / (t - p)$.

Watson
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  • Actually, $( \Bbb F_p[[t]], +)$ is a group of finite exponent (namely $p$), so the only group morphism $( \Bbb F_p[[t]], +) \to (\Bbb Z_p, +)$ is the zero morphism, since $\Bbb Z_p$ is torsion-free. So there are no unital ring morphism $\Bbb F_p[[t]] \to \Bbb Z_p$. – Watson Feb 19 '18 at 14:59
  • (Recall that the additive group of an integral domain of characteristic $0$ is torsion-free). – Watson Feb 19 '18 at 16:52