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Let $f = a_0 + a_1t + a_2t^2 + \cdots$ be a power series with coefficients in $\mathbb{Z}$. Let $p$ be a prime number, and suppose $p$ does not divide all of the coefficients of $f$. That is to say, $p$ does not divide $f$ in the ring $\mathbb{Z}[[t]]$.

Let $P(X)\in\mathbb{Z}[X]$ be a polynomial, such that $p$ does not divide $P(X)$. Could $p$ divide $P(f)$ (in $\mathbb{Z}[[t]]$?

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  • $\begingroup$ Easier to answer when both $f$ and $P$ have constant term zero. $\endgroup$
    – Lubin
    Sep 15, 2017 at 2:26

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How about $p=2$, $f=1+2t$, $P=1+x^2$? Or even just $f=1$.

Neither $f$ nor $P$ is divisible by $2$. But $$ P(f)=1+(1+2t)^2=1+1+4t+4t^2=2+4t+4t^2 $$ is divisible by $2$. So is $P(1)=2$.

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  • $\begingroup$ Yes when $f$ is constant in $\mathbb{F}_p[[t]]$ it is trivial $t \mapsto f(t)$ is not injective. But what if $f$ is non-constant ? $\endgroup$
    – reuns
    Sep 15, 2017 at 2:02
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  • If $g(t) \in \mathbb{F}_p[[t]]^*$ then $\phi : t \mapsto t\, g(t)$ defines a ring homomorphism $\mathbb{F}_p((t))\to \mathbb{F}_p((t))$. And since $\mathbb{F}_p((t))$ is a field then $\phi$ is injective $\mathbb{F}_p[[t]]\to \mathbb{F}_p[[t]]$ and $\mathbb{F}_p[t]\to \mathbb{F}_p[[t]]$.

  • If $f(t)\in \mathbb{F}_p[[t]]$ is non-constant then $f(t)= f(0)+t\, g(t)$ with $g(t), \phi$ as above. Let $\psi : t \mapsto t+f(0)$ an injective homomorphism $\mathbb{F}_p[t]\to \mathbb{F}_p[t]$ so that $\phi \circ \psi : t \mapsto f(t)$ is an injective homomorphism $\mathbb{F}_p[t]\to \mathbb{F}_p[[t]]$.

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I think your question is not sufficiently sharply focused. My example is $f(t)=1+pt$, $P(t)=-1+t$, so that $(P\circ f)(t)=pt$.

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    $\begingroup$ Ah, okay, but I suppose the statement would be true if we assume that $f$ has nonconstant coefficients which are not divisible by $p$. In that case, $f\mod p$ would is not a constant in $\mathbb{F}_p[[t]]$, and so the only way it satisfies a nonzero polynomial mod $p$ is if it is algebraic over $\mathbb{F}_p$, which it is not, since the algebraic closure of $\mathbb{F}_p$ in $\mathbb{F}_p[[t]]$ is just $\mathbb{F}_p$, and $f$ is not a constant in $\mathbb{F}_p[[t]]$. $\endgroup$
    – user355183
    Sep 16, 2017 at 20:03
  • $\begingroup$ My own approach is to restrict to power series with no constant term at all; or to power series whose constant term is in the maximal ideal ($p\Bbb Z_p$ in this case). $\endgroup$
    – Lubin
    Sep 17, 2017 at 3:20

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