Dimension of $(A[[X]])_\mathfrak{m} \geq k+1$, regular ring, chain of prime ideals Assume $A$ is a regular ring and $m$ a maximal ideal of $A$. We define $R= A[[X]]$.Then $\mathfrak{M}=mR + XR$ is a maximal ideal of $R$. Lets assume $h(mA_m) = k$.
I want to show, that
\begin{align*}
\dim(R_\mathfrak{M}) = h(m R_\mathfrak{M} + X R_\mathfrak{M}) \geq k+1
\end{align*}
holds.
My first try:
Lets assume $p_0,\ldots,p_k$ are prime ideals in $A_m$, such that
\begin{align*}
p_0 \subsetneq p_1 \subsetneq...\subsetneq p_k = m A_m.
\end{align*}
It follows immediately:
\begin{align*}
p_0 R_\mathfrak{M} \subsetneq p_1 R_\mathfrak{M}\subsetneq...\subsetneq p_kR_\mathfrak{M} \subsetneq \mathfrak{M}R_\mathfrak{M}.
\end{align*}
The proof would be finished in showing that $p_j R_\mathfrak{M}$ is a prime ideal in $R_\mathfrak{M}$ or equally to show that $B := R_\mathfrak{M}/ p_j R_\mathfrak{M}$ is an integral domain. My idea is to show some isomorphism between $B$ and something with $A_m$, since $p_j$ is prime in $A_m$. However, I am not able to get a solution.
It would be perfect to get a hint, not a full solution. This way I'd learn more.
 A: Can you see a mistake in this idea?
We know there is a one-to-one correspondence between prime ideals in $S^{-1}A$ and prime ideals in $A$ that have an empty intersection with $S$ (use the mapp $P \mapsto S^{-1}P$).
Since $p_j$ is a prime ideal of $A_m$ we know there exists a prime ideal $p_j' \subset A$, such that $S^{-1}p_j' = p_j$ in $A_m$ (with $S= A \setminus m$).
We don't use $p_0 R_\mathfrak{M},\ldots, p_k R_\mathfrak{M}, \mathfrak{M} R_\mathfrak{M}$ as a chain of prime ideals in $R_\mathfrak{M}$, as it was written in my question.
Instead we use the corresponding prime ideals in $A$ (which also work as prime ideals in $R$). Thus, our chain would be $p_0' R_\mathfrak{M}, \ldots, p_k' R_\mathfrak{M},\mathfrak{M} R_\mathfrak{M}$.
Left to show is, that the ideals $p_j' R_\mathfrak{M}$ indeed are prime ideals:
Say $S = R\setminus \mathfrak{M}$ and $S' $ being the image of $S$ in $R/p_j'R$.
Then
\begin{align*}
R_\mathfrak{M}/ p_j' R_\mathfrak{M} = S^{-1}R/ p_j' S^{-1}R \cong  S'^{-1}(R/p_j'R).
\end{align*}
Now $R/p_j'R \cong A/p_j' [[X]]$. It follows $p_j'R$ is a prime ideal in $R$, i.e $(R/p_j'R)$ is an integral domain. Any localization of an integral domain (as example: $S'^{-1}(R/p_j'R)$) is also an integral domain and we are done.
