# height of contracted prime ideals in power series rings

Let $K$ be a field of characteristic zero and let $R_n = K[[X_1,\ldots,X_n]]$ be the power series ring in $n$-variables. Let $P$ be a prime ideal in $R_n$. Let $Q = P \cap R_{n-1}$. Then is $height(Q) \geq height(P) - 1$ ?

In the case of polynomial rings this can be easily proved as follows. Let $S_n = K[X_1,\ldots,X_n]$ and let $W$ be a prime in $S_n$. Let $L = W\cap S_{n-1}$. Notice we have an inclusion of domains

$$\frac{S_{n-1}}{L} \rightarrow \frac{S_n}{W}.$$

Considering the transcendence degree over $K$ of the corresponding quotient fields we get

$$\dim \frac{S_{n-1}}{L} \leq \dim\frac{S_n}{W}.$$

From this we can conclude $height(L) \geq height(W) - 1$. Note we did not require $K$ to be of characteristic zero.

In the case of power series rings we do have an inclusion $$\frac{R_{n-1}}{Q} \rightarrow \frac{R_n}{P}.$$ If we show $$\dim \frac{R_{n-1}}{Q} \leq \dim\frac{R_n}{P}$$ then we are done. But how to show this?

• I think this is false. Actually there is a prime P such that Q = 0. See the paper of Heinzer ''Generic formal fibre'', which HI am now studying. Pierre Matsumi – Pierre MATSUMI Mar 24 '16 at 5:50