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 $R$.

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2 Answers2


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.

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  • Why can you use prime avoidance on $I$? How is $I$ contained in the union of the minimal primes? – user46372819 Jun 19 '20 at 03:54
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    @AlJebr $I$ is contained in the union because $I$ does not contain any units (since $I$ is a proper ideal) and, as established, every nonunit is contained in the union of the minimal primes. Prime avoidance then tells us that since $I$ is contained in the union of the minimal primes, it must be contained in some individual prime. – Anonymous Jun 19 '20 at 07:01
  • It feels like there is a lot of steps missing here. Are you implying that $Q(R)$ is reduced? How do you know that if $xy = 0$, then there is some minimal prime of $\frak p$ of $Q(R)$ such that $xy \in \frak p$? – user46372819 Jun 19 '20 at 15:35
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    $R$ is reduced, but it seems like you are working with $Q(R)$ as the reduced ring. – user46372819 Jun 19 '20 at 15:35
  • @AlJebr Yes, $Q(R)$ is clearly reduced since $R$ is. I didn't want to fill in all the details for you so you could work out the details yourself, but if the answer is too hard to follow, I can certainly edit it to fill in some more details. – Anonymous Jun 19 '20 at 16:31
  • I don't see how if $R$ is reduced, then $Q(R)$ is reduced. If $R$ is reduced, then intersection of all prime ideals is $(0)$. We want to show that after extending every prime ideal in $R$, the intersection in $Q(R)$ remains $(0)$. Why is this true? – user46372819 Jun 19 '20 at 16:50
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    @AlJebr Let $\frac{x}{y}\in Q(R)$ for $x,y\in R$ and suppose $\left(\frac{x}{y}\right)^n=0$ for some $n$. Then $x^n=0\cdot y^n=0$ and hence $x=0$ since $R$ is reduced. – Anonymous Jun 19 '20 at 17:29

Maybe you are tasked with proving it from scratch, but I figured I'd give another theoretical answer that's useful.

A reduced Noetherian ring is a (special case of a) Goldie ring. Goldie's theorem says that the classical ring of quotients is an Artinian ring, and you probably already know that prime ideals in an Artinian ring are maximal.

There's another useful theorem:

Johnson's theorem TFAE:

  1. $R$ is a right nonsingular ring
  2. $Q^r_{max}(R)$ is von Neumann regular

A reduced ring is nonsingular, and a von Neumann regular ring is $0$ dimensional. The only question is whether or not the classical ring of quotients coincides with the maximal ring of quotients in this case (I don't recall and I can't see that it is the case.) One might expect it to hold especially for a commutative Noetherian ring.

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