I have a ring $R$ and a prime ideal $P$ of $S=R[t]$ with $t \in P$.
I'm trying to prove that if $\mathrm{ht}(P/tS)$ is finite then $\mathrm{ht}(P) > \mathrm{ht}(P/tS)$.
Here $\mathrm{ht}(P)$ denotes the height (i.e. the length of the longest chain of prime ideals 'topped' by $P$) of the prime ideal $P$.
I haven't had many opportunities to work with heights and now that I'm trying to get a taste of them. I honestly don't know how I would tackle this one.
Can you help me?
Thank you!
The reason is the following: $t$ is a non-zerodivisor in $R[t]$, so a chain of prime ideals containing $t$ (and contained in $P$) can't start with a minimal prime ideal. (See here.)
Added later. If $n=\operatorname{ht}(P/(t))$ then there is a chain of prime ideals $$P_0/(t)\subset P_1/(t)\subset\cdots\subset P_n/(t)=P/(t)$$ in with $t\in P_i$ for all $i$. This gives rise to a chain of primes $$P_0\subset P_1\subset\cdots\subset P_n=P.$$ Since $P_0$ is not minimal it must contain another prime ideal, so $\operatorname{ht}P>n$.
