Formal power series over a regular ring is regular I'm trying to prove that if $A$ is a regular ring then so is $A[[X]]$. 
The only proof I found of this statement is in Commutative Ring Theory by Matsumura, but it seems a bit over my knowledge so I'd like to know if there are simpler proofs that doesn't involve completion.
 A: 
Let $R$ be a commutative noetherian ring, $x\in R$ a non-zero divisor, and $x\in\mathfrak m-\mathfrak m^2$ for every maximal ideal $\mathfrak m$ of $R$. If $R/(x)$ is regular, then $R$ is regular.

This can be easily proved by reducing it to the local case.
Now show that $X\in M-M^2$ for every maximal ideal $M$ of $A[[X]]$, and then use that $A[[X]]/(X)\simeq A$ is regular.
A: After some days of thinking I may have a proof. I think there is something wrong in it but i can't find what. Here is the proof.
Let $B=A[[X]]$ and $A$ regular (and Noetherian). Then $B$ is Noetherian: to prove that it's regular I'll try to show that every localization on a maximal ideal has finite global dimension.
Let $f$ be the projection from $B$ to $A$. We know that $(X)$ is in the Jacobson so there is a corrispondence between every maximal $\mathfrak{M}$ and their image $f(\mathfrak{M})$.
If we chose a maximal $\mathfrak{M}$ we have a map given by $f$ from $B'=B_{\mathfrak{M}}$ to $A'=A_{f(\mathfrak{M})}$. That map means that every $A'$-module is also a $B'$-module. But $A'$ is a subring of $B'$ so every $B'$-module is a $A'$-module (by restricting the operations). Moreover the projective dimension of $A'$ as a $B'$-module is 1 because $0\rightarrow (X)B'\rightarrow B' \rightarrow A'\rightarrow 0$ is a projective resolution.
Now let $M$ be a module for both ring. I want to show that the $B'$-projective dimension of $M$ is finite. $A'$ is regular local so I know that the projective dimension in respect of $A'$ is finite. Let's call this number $n$. The proof works by induction on $n$.
If $n=0$ $M$ is free as an $A'$-module, so it's isomorphic to some $(A')^i$ an then
$$0\rightarrow ((X)B')^i\rightarrow (B')^i \rightarrow (A')^i\rightarrow 0$$
is still a projective resolution.
If $n>0$ let's consider $M$ as a $A'-module$. We then have a resolution
$$0 \rightarrow P_n \rightarrow \ldots \rightarrow P_1\rightarrow P_0 \xrightarrow{g} M\rightarrow 0$$
of $A'$-module. Then we have a short exact sequence
$$0 \rightarrow \ker g \rightarrow P_0 \rightarrow M\rightarrow 0$$
we know may see that sequence has a sequence of $B'$-module, where $\ker g$ and $P_0$ have the structure given by $A'$ and $M$ has the initial structure as a $B'$-module. With this definition those maps are $B'$-linear so it's still an exact sequence. By induction the projective dimension of $\ker g$ is finite and so is the projective dimension of $P_0$ and this implies that the projective dimension of $M$ is also finite.
The crucial part is the shifting between $A'$-module and $B'$-module so i think the error may be there but maybe i overlooked something else. Thanks for helping.
