I know that in $K(R)$, the set of maximal ideals is the set of associated primes of $K(R)$ and that an ideal is maximal if and only if it is the localization of a maximal associated prime of $R$.
So, we know there are only finitely many maximal ideals in $K(R)$.
I'm not sure if that is helpful, but, We want to show that if $R$ is a reduced Noetherian ring, then every prime ideal in $K(R)$ is in fact maximal.
I'm not sure how to use assumption that $R$ is reduced. All I know is that this means that there are no nilpotent elements in $R$.
Best Answer
First, note that $Q(R)$ is Noetherian and reduced since $R$ is, and second, every element of $Q(R)$ is either a unit or a zero divisor. Once this is established, we can effectively forget about $R$. With that in mind, all ideals introduced are ideals of $Q(R)$.
Let $\mathfrak{p}_1,\ldots,\mathfrak{p}_k$ be the minimal prime ideals of $Q(R)$ (there are finitely many since $Q(R)$ is Noetherian). Recall that the nilradical (the set of nilpotent elements, or equivalently, the radical of the zero ideal) of $Q(R)$ is equal to the intersection of all primes ideals of $Q(R)$. Since $Q(R)$ is reduced, this means that the intersection of the prime ideals is the zero ideal. Furthermore, because every prime ideal contains a minimal prime ideal, the intersection of all prime ideals equals $\cap_{i=1}^k\mathfrak{p}_i$ and therefore $\cap_{i=1}^k\mathfrak{p}_i=(0)$.
We now show that $\mathfrak{p}_1,\ldots,\mathfrak{p}_k$ are all maximal ideals. Let $j_1\in\{1,\ldots,k\}$ and let $I$ be an ideal such that $\mathfrak{p}_{j_1}\subseteq I\subsetneq Q(R)$. Let $x\in I$. Then $x$ is not a unit since $I\neq Q(R)$ which implies $x$ is a zero divisor. So, there exists nonzero $y\in Q(R)$ such that $xy=0$. Since the $y$ is nonzero and $\cap_{i=1}^k\mathfrak{p}_i=(0)$, we see $y\notin\cap_{i=1}^k\mathfrak{p}_i$ and therefore there exists $j_2\in\{1,\ldots,k\}$ such that $y\notin \mathfrak{p}_{j_2}$. However, $xy=0\in\mathfrak{p}_{j_2}$, so $x\in\mathfrak{p}_{j_2}\subseteq\cup_{i=1}^k\mathfrak{p}_i$. Thus, we conclude $I\subseteq\cup_{i=1}^k\mathfrak{p}_i$.
Next, by prime avoidance, we deduce that $I\subseteq \mathfrak{p}_{j_3}$ for some $j_3\in\{1,\ldots,k\}$ and hence $\mathfrak{p}_{j_1}\subseteq\mathfrak{p}_{j_3}$. Because $\mathfrak{p}_{j_3}$ is a minimal prime ideal, this implies $\mathfrak{p}_{j_1}=\mathfrak{p}_{j_3}$ and therefore $I=\mathfrak{p}_{j_1}$. Thus, we conclude $\mathfrak{p}_1,\ldots,\mathfrak{p}_k$ are maximal ideals. Finally, if $\mathfrak{p}$ is any prime ideal, then it contains a minimal prime $\mathfrak{p}_j$ for some $j\in\{1,\ldots,k\}$ and hence $\mathfrak{p}=\mathfrak{p}_j$ by maximality of $\mathfrak{p}_j$, so every prime ideal is maximal.
I have edited my answer to fill in a lot more details, but let me know if you need further clarification on any points.