Theorem 20.8 in "Commutative ring theory" states that if $A$ is a regular UFD then so is $A[[X]]$.

Here is the proof. He has to prove that the intersection of principal ideals $\mathfrak{b}=uB\cap vB$ where $B=A[[X]]$ is principal and considers WLOG the case $\mathfrak{b}\nsubseteq XB$.

I don't understand why $\mathfrak{b}:XB=\mathfrak{b}$.

The containment $\supseteq$ is obvious. If I consider $h\in \mathfrak{b}:XB$ then $hX\in \mathfrak{b}$ so if $P$ is a minimal prime divisor of $\mathfrak{b}$ I get $h\in P$ because $X\notin P$. So $h$ is in the radical of $\mathfrak{b}$ but from this point I can't prove that in fact $h\in \mathfrak{b}$.

Update: i understood this point but now i don't know why $I$ doesn't have embedded primes.


$\mathfrak b$ is projective, hence locally principal. Since $B$ is regular all the associated primes of $\mathfrak b$ are of height one. (If $P$ is associated to $B/\mathfrak b$ then $PB_P$ is associated to $B_P/\mathfrak bB_P$. But $B_P$ is a Cohen-Macaulay integral domain (since it is regular and local), and $\mathfrak bB_P$ is principal, so $B_P/\mathfrak bB_P$ is also Cohen-Macaulay. Now use Matsumura, Theorem 17.3(i).)

Now let $\mathfrak b=\bigcap_{i=1}^n\mathfrak q_i$ be a reduced primary decomposition, and set $\mathfrak p_i=\sqrt{\mathfrak q_i}$. Since $\mathfrak b\nsubseteq XB$ we have $\mathfrak p_i\ne XB$ for all $i=1,\dots,n$. Furthermore $(\mathfrak b:XB)=\bigcap_{i=1}^n(\mathfrak q_i:XB)$. But it's easily seen that $(\mathfrak q_i:XB)=\mathfrak q_i$: if $bX\in\mathfrak q_i$ then $b\in\mathfrak q_i$ (since $X\notin\mathfrak p_i$, otherwise $XB\subseteq\mathfrak p_i$ and both are prime ideals of height one, hence they are equal, a contradiction). Thus we get $(\mathfrak b:XB)=\mathfrak b$.

  • $\begingroup$ ok, thanks. But now i can't understand why all associated primes are of height 1. If $P$ is minimal i can localize in $P$ and since $I_P$ is principal $P_P$ has height 1 and then $P$ has height 1 using the principal ideal theorem. The question is: why aren't there embedded primes? $\endgroup$ – karmalu Aug 15 '15 at 15:42
  • $\begingroup$ Since $B$ is regular all associated primes of $I$ have the same height. $\endgroup$ – user26857 Aug 15 '15 at 17:22
  • $\begingroup$ Well, i simply didn't know it. And how can it be proven? $\endgroup$ – karmalu Aug 15 '15 at 17:23
  • $\begingroup$ If $P$ is associated to $R/I$ then $PR_P$ is associated to $R_P/IR_P$. But $R_P$ is a Cohen-Macaulay integral domain (since it is regular and local), and $IR_P$ is principal, so $R_P/IR_P$ is also Cohen-Macaulay. Now use Matsumura, Theorem 17.3(i). $\endgroup$ – user26857 Aug 15 '15 at 17:32

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