$R[[x]]$ for a Noetherian ring $R$?

Let $R$ be a Noetherian ring. How can one prove that the ring of the formal power series $R[[x]]$ is again a Noetherian ring?

It is well-known that the ring of polynomials $R[x]$ is Noetherian. I try imitating the standard proof of the fact by replacing "leading coefficients" by "lowest coefficients", but it does not work.

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Also, see Matsumura, Commutative Ring Theory, p 16. You can see his argument at the preview: books.google.com/books/about/… – John Myers Jan 26 '13 at 4:48

Step 1: Let $f\in R[[x]]$ such that the leading coefficient of $f$ is a unit. Then $f$ is a unit. It's an easy exercise to construct $f^{-1}$ term-by-term.

Step 2: Show that if $I$ is an ideal, the set of lowest nonzero terms of elements of $I$ is an ideal of $R$.

Step 3: Consider an ascending chain of ideal. $I_1\subset I_2\subset\ldots$, and let $J_1\subset J_2\subset\ldots$ be the corresponding chain of ideals in $R$, where $J_i$ is the ideal of lowest nonzero terms of elements of $I_i$.

Step 4: $J_1\subset J_2\subset\ldots$ can only be strict inclusion at finitely many places. The only other way to have strict inclusion in $I_1\subset I_2\subset\ldots$ is via the lowest degree in $x$ of nonzero terms, which can only decrease finitely many times.

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I think here you meant $A$ to be your base ring $R$ – massy255 Mar 5 '14 at 18:24
@massy255 Thanks. Just fixed it! – Brett Frankel Apr 14 '14 at 1:44
@BrettFrankel How does it follow that $I_i \subset I_{i+1}?$ – The Substitute May 21 '14 at 2:12
@TheSubstitute The ascending chain $I_i$ is given. – Brett Frankel May 21 '14 at 3:02

Hilbert's basis theorem can be adapted for formal power series. I found a .pdf on the internet that describes the process well, if you look at section 8.2.3: here.

Basically, you simply replace the degree of the polynomial with the lowest degree in the power series.

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