I was able to prove "$\geq$" by showing that every prime ideal $p \subset R$ can be extended to $p' \subset R[[X]]$, with $p'$ being a prime ideal in $R[[X]]$. For a chain $p_1 \subsetneq … \subsetneq p_n$ of prime ideals in $R$, we also have a chain $p'_1 \subsetneq … \subsetneq p'_n \subsetneq (p_n,X)_{R[[X]]}$ of prime ideals in $R[[X]]$, showing $\dim (R[[X]]) \geq \dim (R) + 1$.
Now i want to show the converse, but I am not sure how. For $p \subset R[[X]]$ prime ideal, we can always find a maximal ideal $M \subset R[[X]]$ with $p \subset M$, but the maximal ideals of $R[[X]]$ can be written as $M = (m,X)_{R[[X]]}$ with $m \subset R$ being a maximal ideal. Since $R$ is noetherian, $m$ gets generated by some $r_1,…,r_k \in R$, so $M$ is generated by $r_1,…,r_k,X \in R[[X]]$. By Krull's Principal Ideal Theorem we hence see $ht(M) \leq k+1$, so $ht(p) \leq ht(M) \leq k+1$, but what now? I would like to show that $k \leq \dim(R)$, but I don't see how. I saw similar questions to mine, but I got the tip that I should use Krull's Principal Ideal Theorem, so I want to apply it. Does someone have some advice?
Best Answer
Lemma: If $R$ is a Noetherian ring, and $P<R$ is a prime of height $n$, then there exist $x_1,\dots,x_n\in P$ such that $P$ is a minimal prime over $\langle x_1,\dots,x_n\rangle$.
Proof: By induction on $n$. The case $n=0$ is clear, since then $P$ is minimal over the zero ideal, which is generated by $\varnothing$. For the inductive step, suppose $n>0$. Now, since $R$ is Noetherian, it has finitely many minimal prime ideals; let $Q_1,\dots,Q_k$ be a complete list. Since $\operatorname{ht}Q_i=0<n=\operatorname{ht}P$ for each $i$, we have $P\nsubseteq Q_i$ for each $i$, and hence by prime avoidance we may find $x_n\in P\setminus\bigcup_{i=1}^nQ_i$. Then the prime ideal $P\big/\langle x_n\rangle$ of the Noetherian ring $R\big/\langle x_n\rangle$ has height strictly smaller than $n$, so by the inductive hypothesis there exist $x_1,\dots,x_{n-1}\in R$ such that $P\big/\langle x_n\rangle$ is a minimal prime over $\langle \bar{x}_1,\dots,\bar{x}_{n-1}\rangle$. Now $P$ is a minimal prime over $\langle x_1,\dots,x_n\rangle$, so we are done. $\blacksquare$
Now, for convenience denote $A=R[[X]]$. To show $\dim A\leqslant 1+\dim R$, it suffices to show $\operatorname{ht}M\leqslant 1+\dim R$ for any maximal ideal $M<A$, so let $M$ be a maximal ideal of $A$. As you have noted, we then have $M=\mathfrak{m}A+\langle X\rangle$, where $\mathfrak{m}=M\cap R$ is a maximal ideal of $R$. In particular, $k:=\operatorname{ht}\mathfrak{m}\leqslant\dim R$. By the lemma, there exist $r_1,\dots,r_k\in\mathfrak{m}$ such that $\mathfrak{m}$ is a minimal prime over $\langle r_1,\dots,r_k\rangle$. But now $M$ is a minimal prime over $\langle r_1,\dots,r_k,X\rangle$, so by Krull's height theorem $\operatorname{ht}M\leqslant k+1\leqslant 1+\dim R$, so we are done.