Is there an injective $\mathbb{Z}_p$-ring homomorphism from $\mathbb{Z}_p[[x,y]]$ to $\mathbb{Z}_p[[x]]$?
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1$\begingroup$ Do you want the homomorphism to be continuous? (I'm not sure if that's automatic.) $\endgroup$– LSpiceCommented Apr 7 at 22:52
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1$\begingroup$ But then wouldn't $(y-x)\mapsto 0$ imply your map isn't injective? $\endgroup$– kindasortaCommented Apr 7 at 23:01
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1$\begingroup$ Is there such a homomorphism from $\mathbb{Z}[[x,y]]$ to $\mathbb{Z}[[x]]$? $\endgroup$– user525759Commented Apr 7 at 23:07
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3$\begingroup$ @KevinCasto Of course, just embed formal series into the algebraic closure of the fraction field. It is abstractly isomorphic to complex numbers by categoricity of theory of alg. closed fields. $\endgroup$– Denis TCommented Apr 7 at 23:44
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1$\begingroup$ @DenisT I guess I was referring to $\mathbb C$-algebra homomorphisms (taking 1 to 1), which are thus the identity on $\mathbb C \cdot 1$ and so can't embed the domain's copy of $\mathbb C$ in a strict subfield of the codomain's copy of $\mathbb C$. $\endgroup$– Kevin CastoCommented Apr 8 at 0:26
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1 Answer
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Surprisingly, at least to me, yes.
Let $f$ and $g$ be two power series in $\mathbb Z_p[[x]]$ that, modulo $p$, are algebraically independent. Send $x$ to $pf$ and $y$ to $pg$. To check the map $\mathbb Z_p[[x,y]] \to \mathbb Z_p[[x]]$ is injective, it suffices to check that the induced map
$$\mathbb Z[x,y]/(x,y,p)^n = \mathbb Z_p[[x,y]]/( x,y,p)^n \to \mathbb Z_p[[x]]/(p^n)$$
is injective, or, by induction on $n$, that the induced map
$$(x,y,p)^{n-1} \mathbb Z[x,y]/(x,y,p)^n \to (p^{n-1}) \mathbb Z_p[[x]]/(p^n)$$
is injective. But the right-hand side is just $\mathbb F_p[[x]]$ and the left-hand side is sent by this isomorphism to polynomials in $f,g$ of degree $\leq n-1$, so the map is injective by the assumed algebraic independence.
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