ac.commutative algebra - Is it true that, as $Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other?

I give this problem each year in a problem-solving seminar. Here is the solution that I wrote up. I am using $f$ instead of $varphi$ and $e_n$ instead of $x^n$.

Let $x=(x_1,x_2,dots)$. Since $2^n$
and $3^n$ are relatively prime, there are integers $a_n$ and $b_n$
for which $x_n=a_n2^n+b_n3^n$. Hence $f(x)=f(y)+f(z)$, where $y = (2a_1, 4a_2, 8a_3,dots)$ and $z=(3b_1,9b_2,27b_3,dots)$. Now for
any $kgeq 1$ we have
$$f(y) = f(2a_1,4a_2,dots,2^{k-1}a_{k-1},0,0, dots)$$
$$qquad + f(0,0,dots,0,2^ka_k,2^{k+1}a_{k+1},dots)$$
$$qquad= 0+2^kf(0,0,dots,0,a_k,2a_{k+1},4a_{k+2},dots).$$
Hence $f(y)$ is divisible by $2^k$ for all $kgeq 1$, so
$f(y)=0$. Similarly $f(z)$ is divisible by $3^k$ for all $kgeq 1$, so $f(z)=0$. Hence $f(x)=0$.

Now let $a_i=f(e_i)$. Define integers $0< n_1 < n_2 <cdots$ such that for all $kgeq 1$,
$$sum_{i=1}^k|a_i|2^{n_i} < frac 12 2^{n_{k+1}}.$$
(Clearly this is possible --- once $n_1,dots,n_k$ have been chosen,
simply choose $n_{k+1}$ sufficiently large.) Consider $x=(2^{n_1}, 2^{n_2}, dots)$. Then
$$f(x) = f(a_1e_1 + cdots + a_k e_k +2^{n_{k+1}} (e_{k+1}+2^{n_{k+2}-n_{k+1}}e_{k+2}+cdots))$$ $$qquad= sum_{i=1}^ka_i 2^{n_i}+2^{n_{k+1}}b_k,$$
where $b_k=f(e_{k+1}+2^{n_{k+2}-n_{k+1}}e_{k+2}+cdots)$. Thus by the
triangle inequality,
$$left| 2^{n_{k+1}}b_kright| < left| sum_{i=1}^k a_i 2^{n_i}right| + |f(x)|$$ $$qquad < frac 12 2^{n_{k+1}} + |f(x)|.$$
Thus for sufficiently large $k$ we have $b_k=0$ [why?]. Since
$$b_j - 2^{n_{j+2}-n_{j+1}}b_{j+1}=f(e_{j+1}) mbox{[why?]},$$
we have $f(e_k)=0$ for $k$ sufficiently large.}

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