Given a multiplicative subset $U \subset R$ and let $L_U$ be the localization homomorphism. I am trying to figure out why $ J \subset L_U^{-1}(U^{-1}J)$.
As I understand it, when J is prime it would be equal, since $L_U^{-1}(U^{-1}J) = \{r \in R : ur \in J \text{ for some u } \in U\}$ and since $ur$ is in $J$ then $r$ must be in J since we know $u$ isn't and J is prime, but when $J$ isn't prime, I'm struggling to visualise why it produces a subset.
On
The result
Given a ring $R$, a multiplicative subset $U\subset R$ and a prime ideal $\mathfrak p\subset R$ we have the equivalence $$L_U^{-1}(\mathfrak p\cdot U^{-1}R) =\mathfrak p \iff \mathfrak p\cdot U^{-1}R \; \operatorname {is prime in}\: U^{-1}R\iff U\cap \mathfrak p=\emptyset \quad \quad (\bigstar) $$ If contrariwise $U\cap \mathfrak p\neq \emptyset$, then $\mathfrak p\cdot U^{-1}R=R$, so that $L_U^{-1}(\mathfrak p\cdot U^{-1}R)=R$
An example
If $R=\mathbb Z$ and $U=p^\mathbb N=\{1,p, p^2,\cdots\}$ for some prime integer $p$, then the only prime ideals $\mathfrak p\subset \mathbb Z$ for which $(\bigstar)$ holds are $(0)$ and $p\mathbb Z$.
The good news
If on the other hand we start with a prime ideal $\mathfrak q\subset U^{-1}R$, then we always have $$\mathfrak q=L_U^{-1}(\mathfrak q) \cdot U^{-1}R \quad \quad (\bigstar \bigstar)$$ Anthropomorphic mnemonic
"A prime in $R$ wanting to survive in $U^{-1}R \:$ shouldn't touch $U$"
By definition, any $x\in J$ is sent under $L_{U}$ to $\frac{x}{1}\in U^{-1}J$. So we have indeed $J\subseteq L_{U}^{-1}(U^{-1}J)$ for any subset $J\subseteq R$.
But it is not true that this is an equality for $J$ a prime ideal. If your set $U$ contains nilpotent elements, for example, the map $L_{U}$ is just the zero map (because $U^{-1}R$ is the zero ring), so the preimage of the image of any nonempty set is the whole $R$.