TITLE: How to prove $W(k)$ is a complete discrete valuation ring?
QUESTION [7 upvotes]: I am trying to prove the fact that the ring of Witt Vectors $W\left(k\right)$
  is a complete discrete valuation ring, where $k$
  is a perfect field of characteristic $p$
 , but I'm stuck. Theorem 2 on http://ncatlab.org/nlab/show/ring+of+Witt+vectors mentions that the maximal ideal of $W\left(k\right)$
  is $pW\left(k\right)$
  because of the property that $p\left(a_{0},a_{1},\ldots\right)=\left(0,a_{0}^{p},a_{1}^{p},\ldots\right)$
 , but I don't see how they get this property. Any help would be appreciated.

REPLY [2 votes]: Consider this:
Define $a=(a_0,a_1,\ldots)$ and $W_n(a)=\sum_{i=0}^np^ia_i^{p^{n-i}}$ are the witt polynomials
Then we get the following
$$\begin{alignat*}{1}
W_n(pa)&=W_n(a+\ldots+a)\\
&=pW_n(a)\\
&=p\sum_{i=0}^np^ia_i^{p^{n-i}}\\
&=\sum_{i=0}^np^{i+1}a_i^{p^{n-i}}\\
&=\sum_{i=1}^{n+1}p^{i}a_i^{p^{n-i+1}}\\
&=\sum_{i=1}^{n+1}p^{i}(a_i^p)^{p^{n-i}}\\
&=W_n((0,a_0^p,a_1^p,\ldots,a_{n-1}^p+pa_n))\\
&=W_n((0,a_0^p,a_1^p,\ldots,a_{n-1}^p))
\end{alignat*}
$$
with the last equality holding because $char(k)=p$
This means $pa=(0,a_0^p,a_1^p,\ldots)$.
The other arguments to get a dvr should follow easily from that.