Is the infinite intersection of prime ideals also a prime ideal?

Question

I'm currently reading the proof of the fact that the intersection of all the prime ideals of the commutative ring $A$ is nilradical of $A$.

The proof takes every non-nilpotent element $f$, and proves that there exists a prime ideal which does not contain $f$. Clearly the intersection of all such prime ideals will not contain any non-nilpotent element.

However, $A$ may contain an infinite number of non-nilpotent elements. Does this not imply that an infinite intersection of prime ideals should also be a prime ideal?

Thanks in advance!

Answer 1

Certainly not. If we assumed that the intersections of infinitely many prime ideals always had to be prime, then the nilradical of any ring with infinitely many prime ideals would have to be prime, but the nilradical of a commutative ring is prime iff the ring has a unique minimal prime ideal. There are lots of counterexamples to that, though, so the original assumption was wrong.

Michalis pointed out that the intersection of two primes does not have to be prime. A simple example would be $F\times F$ for a field $F$. $F\times \{0\}$ and $\{0\}\times F$ are both maximal, hence prime, ideals, and their intersection is $\{0\}\times\{0\}$ which is certainly not prime.

This idea generalizes to an infinite ideal example. Let $R=\prod_{i=1}^\infty F$ For a field $F$. The ideal $M_i$ which is $F$ on every coordinate except on the $i$th coordinate, and is zero on the $i$th coordinate is a maximal, hence prime ideal of $R$. The intersection of these ideals is the zero ideal, but again, the zero ideal isn't prime.

Finally, let's get one with a nonzero nilradical. Take the same ring $R$ as above, and construct the triangular matrix ring of matrices that look like this: $\begin{bmatrix}a&b\\0&a\end{bmatrix}$ for $a,b\in R$. The intersection of maximal ideals of this ring is equal to the set of matrices of the form $\begin{bmatrix}0&b\\0&0\end{bmatrix}$ for $a,b\in R$, which is again not prime. There are infinitely many such matrices if your field is infinite.

However, there is an important case when the intersection of a family of prime ideals is prime: if they form a descending chain.

Answer 2

The intersection of two primes is in fact rarely a prime ideal: Suppose $\mathfrak{q}\cap \mathfrak{p}=\mathfrak{r}$ is prime for two distinct prime ideals such that neither of them contains the other (if e.g. $\mathfrak{q}\subseteq \mathfrak{p}$ the intersection of the two, $\mathfrak{p}$, is trivially a prime ideal). Take $x\in \mathfrak{p}\setminus \mathfrak{q}$ and $y\in \mathfrak{q}\setminus \mathfrak{p}$. Then $xy\in \mathfrak{r}$ but neither $x$, not $y$ is in $\mathfrak{r}$.

Answer 3

Let $R = \mathbb{Z}\times\mathbb{Z}$. Then $(1,0)(0,1) = (0,0)$ and is in the nilradical. However, neither $(1,0)$ nor $(0,1)$ are nilpotent. In other words, the nilradical is not in general a prime ideal.