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

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?

• This is really nonsequitur: "infinitely many nilpotents implies the nil ideal prime" Commented Oct 29, 2013 at 17:08
• @rschwieb- For every non-nilpotent, there is a prime ideal which does not contain it - The intersection of all such prime ideals will not contain any non-nilpotent- If there are an infinite number of such non-nilpotents, then we will have to take an infinite intersectin of prime ideals to arrive upon a set which does not contain any non-nilpotent. How is this nonsequitur?
– user67803
Commented Oct 30, 2013 at 4:08
• Dear @AyushKhaitan : Sure, I agree with all that, but you left out the "implies the nil ideal is prime" part, which still seems nonsequitur after your last comment. Commented Oct 30, 2013 at 12:38
• This is however true if you have intersection of a descending chain of prime ideals.
– mez
Commented Dec 17, 2014 at 21:43

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.

• Is the infinite intersection of ideals an ideal?
– user67803
Commented Oct 30, 2013 at 4:10
• @AyushKhaitan Sure: you can prove this easily by a direct check. Commented Oct 30, 2013 at 12:33
• $(0,0)$ is prime in $F \times F$, isn't it? An easy example can be $(2) \cap (3) = (6)$ in $\mathbb{Z}$. Commented Feb 28, 2014 at 17:48
• Dear @PedroA.Castillejo $((0,0))$ isn't prime in $F\times F$ for any commutative ring $F$ with unity: $(1,0)\cdot (0,1)\in ((0,0))$. Commented Feb 28, 2014 at 18:06
• @William well, actually "never" would be the right choice. Commented Oct 22, 2016 at 18:25

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 $\mathfrak{r}$.

• @user: The OP called the ring $A$... but I changed it anyway. Commented Oct 29, 2013 at 16:55
• Is the infinite intersection of ideals an ideal? I have no reason to suspect not, but I can't be sure.
– user67803
Commented Oct 30, 2013 at 4:09
• To take this to the extreme, if your ring is a dimension $1$ domain, then the intersection of any two unequal primes is never a prime. Commented Oct 30, 2013 at 9:05

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.

• Is the infintie interection of ideals an ideal?
– user67803
Commented Oct 30, 2013 at 4:10