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Let $k$ be a finite field and write $k[[t]]$ for the set of formal power series $\sum_{i=0}^{\infty}a_{i}t^{i}$ with coefficients $a_{i}$ in $k$.

Prove that $k[[t]]$ is the completion of the polynomial algebra $k[t]$ with respect to the family of powers of the ideal generated by $t$; thus $k[[t]]$ is a pro-$p$ ring, where $p = \text{char} \; k$.

I've made a basic start on this question but no idea if I'm going in the right direction. My work so far:

Let $I$ be the family of powers of the ideal generated by $t$. Then $I$ is a non-empty filter base of normal subgroups.

Let $\hat{G} = s\varprojlim_{I}k[x]/K$ for $K \in I$ and let $j$ be the map $g \mapsto (Kg)$ from $k[x]$ to $\hat{G}$. Then $\hat{G}$ has the properties of the completion of $k[x]$ with respect to $I$.

Then it suffices to show that $s\varprojlim_{I}k[x]/K = k[[x]]$.

I'm not entirely sure where to go from here (or if I'm even close), any input would be appreciated.

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    I'm sorry, what does this mean and how did you get this? I'm somewhat new to this field and could use some extra detail. – 2017-01-17
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    Oh... I guess $k[[x]]\cong\prod_{\mathbb{N}}k$ only as $k$-vector spaces. What I wanted to do is to construct a morphism $k[[x]]\to \varprojlim_n k[x]/(x^n)$. Using the universal property, we can give compatible maps $k[[x]]\to k[x]/(x^n)$ instead. I think $\sum_{i=0}^\infty a_ix^i\mapsto a_0+a_1x+\ldots + a_{n-1}x^{n-1} + (x^n)$ should give the isomorphism. – 2017-01-17
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    In such an inverse system, what would the maps $k[x]/(x^{j}) \rightarrow k[x]/(x^{i})$ between each space be defined as? (assuming $i \leq j$) – 2017-02-15
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    I suppose it would just be the set of maps $k[x]/(x^{j}) \rightarrow k[x]/(x^{i}) ; a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} + (x^{n}) \mapsto a_{0} + a_{1}x + ... + a_{n-2}x^{n-2} + (x^{n-1})$ – 2017-02-15
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    Yeah, the map $k[x]/(x^j)\to k[x]/(x^i)$ for $i\leq j$ is just the canonical projection. Note that $(x^j)\subset (x^i)$. – 2017-02-16

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