Let $k$ be a field, $L$ be a complete nonarchimedean field and $\mathcal{O}_L$ its valuation ring. Suppose we have a ring homomorphism $k \rightarrow \mathcal{O}_L$. Fix an $\omega \in L$ with $0<\mid w\mid <1$. We then have the well defined homomorphism of $k$-algebras: \begin{align} k[[X]] \longrightarrow \mathcal{O}_L, f(X) \longmapsto f(\omega). \end{align} Is this homorphism always injective?

  • $\begingroup$ A ring homoorphism from $k \rightarrow \mathcal{O}_L$ isn't what you meant, is it? $\endgroup$ – sharding4 May 14 '17 at 18:22
  • $\begingroup$ Yes it is, we also need it to have a map $k[[X]]⟶\mathcal{O}_L$ (for evaluating the coefficients of the power series. $\endgroup$ – Layer Cake May 14 '17 at 18:26

Let $f(X) = a_0 + a_1X + a_2X^2 + \cdots$, with $a_i \in k$. Fix $\omega \in \mathcal O_L$, with $0< |\omega| < 1$.

Since $k$ is contained in $\mathcal O_L$, all its nonzero elements must lie in $\mathcal O_L^{\ast}$. In particular, $|a_i| = 0$ or $1$. Then $f(\omega)$ converges absolutely, since

$$|a_0| + |a_1 \omega| +|a_2\omega^2| + \cdots \leq 1 + |\omega| + |\omega|^2 + \cdots < \infty$$

as a geometric series.

Suppose that $f(\omega) = 0$. We want to show that all the coefficients of $f$ are zero.

Suppose that $a_0 \neq 0$, so $|a_0| = 1$. For each $N$, and each $1 \leq i \leq N$, we have $|a_i \omega^i| \leq |\omega|^i < 1$.

This implies that $$1 = |a_0 + a_1 \omega + \cdots + a_N\omega^N|$$

And so $1 = \lim\limits_N |a_0 + a_1 \omega + \cdots + a_N\omega^N| = |f(\omega)| = 0$, contradiction.

So $a_0$ has to be zero. Let $n\geq 1$ be the smallest integer such that $a_n \neq 0$. Then

$$f(\omega) = a_n\omega^n + a_{n+1}\omega^{n+1} + \cdots = \omega^n g(\omega)$$

where $g(X) = a_n + a_{n+1} X + \cdots$. Then $g(\omega)$ converges absolutely, and has nonzero constant term. By what we just proved, $g(\omega) \neq 0$, hence $f(\omega) = \omega^n g(\omega)$ is not zero either.


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