# For a localization of a ring $S^{-1}R,$ the extension of a prime ideal $P^e$ has no units if $P \cap S = \emptyset.$

In my lecture notes, my professor wrote down that if a prime ideal $P$ of $R$ is such that $P \cap S = \emptyset$ then the extension of $P$ via a fixed homomorphism $\phi: R \rightarrow S^{-1}R$ contains no units as well. This confuses me. I know that $\phi(S) \subset (S^{-1}R)^\text{x}$ but couldn't $P^e = \phi(P)(S^{-1}R)$ contain a unit not in $\phi(S)$ then?

• I guess elements of $P^e$ look like $p/s$ where $p \in P$ and $s \in S$. Could this be a unit? If it had an inverse $q/t$ then $(p/s)(q/t) = 1 = (1/1$, so there must be a $u \in S$ such that $u(pq-st)=0$ in $R$. Now $ust \in S$ and $upq \in P$, but $ust = upq$. So this is a common element of $P$ and $S$, but they are supposed to be disjoint. – Zach Teitler Nov 15 '17 at 3:08
• $P^e$ is also a prime ideal (in the ring $S^{-1}R$). If such ideal has units, then it wouldn't be a prime ideal. Actually, as the other comment shows, the only case in which the extension of an (arbitrary) ideal $I$ of $R$ can have units is when $I\cap S\neq \emptyset$ and in such case $I^e=S^{-1}R$. – Xam Nov 15 '17 at 20:28