The nilradical and prime ideals
Posted on July 29, 2022. Tags: algebra, math, ring theoryBackground
A nilpotent element of a commutative ring \(R\) is an element that can be multiplied by itself some number of times to yield the zero element in the ring. More formally, it is an element \(r \in R\) such that there exists an \(n \in \mathbb{Z}_{>0}\) where \(r^n = 0_R\), where \(0_R\) is the zero element of \(R\).
If you’re too used to dealing with familiar things like numbers, you may think “Wouldn’t 0 be the only nilpotent? What other number raised to a power would be zero?” And you would be right, if we were working in a ring like \(\mathbb{Z}\), \(\mathbb{Q}\), or any field like \(\mathbb{R}\) or \(\mathbb{C}\). But we’re dealing with rings in general here, not just integral domains, so our rings may have zero divisors. Take, for example, the matrix \(A=\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}\), which is an element of the matrix ring \(M_2(\mathbb{Z})\). Clearly, \(A\) is not the zero matrix, yet \(A^2=0\). Thus \(A\) is an example of a nilpotent matrix.
The set of all nilpotent elements \(\mathfrak{N}_R = \{r \in R : r \textrm{ is nilpotent}\}\) is known as the nilradical of \(R\).
An ideal \(I\) is a non-empty subset of elements from \(R\) that is closed under addition (with its own elements) and closed under multiplication with any element from \(R\).
For example, take the integer ring \(\mathbb{Z}\). The set of all multiples of 6 (or any number, really) \(\{\ldots, -12, -6, 0, 6, 12, \ldots\}\) is an ideal of \(\mathbb{Z}\). Note how adding any two elements in this set yields another multiple of 6, and multiplying any element by any integer (not just other integers in this set) yields another multiple of 6.
For an ideal to be prime, we have an extra, stricter condition: if, for two elements \(a, b \in R\), their product \(ab \in I\), then either \(a \in I\) or \(b \in I\). (Also, the prime ideal cannot be the whole ring.)
Clearly, our example from before is not a prime ideal; 2 is not in that ideal, and neither is 3, but \(2*3=6\) is indeed in that set. As you may have guessed, in the integers, the only prime ideals that exist are those generated by multiples of prime numbers (e.g. all the multiples of 7 is a prime ideal), hence the name.
On the surface, these two concepts seem pretty unrelated. One deals with multiplying individual elements by themselves until they become zero; another deals with sets of elements and closure under multiplication. It turns out, though, that if you take the intersection of all of the prime ideals of a commutative ring, you get back exactly the nilradical of the ring. In other words, every single prime ideal contains every nilpotent element in the ring, and there is no other element that all the prime ideals share that is nonnilpotent.
As a corollary, we’ll also get the fact that the nilradical itself is an ideal, although it’s not necessarily prime itself.
Proof
Proposition. For a commutative ring \(R\), let \(\mathfrak{N}_R\) be its nilradical, and let \(\{P_1, P_2, \ldots, P_m\}\) be the set of all prime ideals \(P_i \subseteq R\). Then
\[\mathfrak{N}_R = \bigcap_{i=1}^m P_i.\]Proof. We will first show \(\mathfrak{N}_R \subseteq \bigcap_{i=1}^m P_i\). In other words, we wish to show that for any prime ideal \(P_i\), and any nilpotent element \(x \in \mathfrak{N}_R\), \(x \in P_i\).
Note that \(0 \in P_i\), as for any element \(p \in P_i\), \(0\cdot p \in P_i\) as \(0 \in R\) and ideals are closed under multiplication from the ring.
We also have that, since \(x\) is nilpotent, there exists some \(n>0\) such that \(x^n=0\). Thus \(x^n=x\cdot x^{n-1} \in P_i\). Since \(P_i\) is prime, this implies either \(x\) or \(x^{n-1}\) is in \(P_i\).
But this forces \(x\in P_i\), as if \(x\notin P_i\), then \(x^{n-1}\in P_i\), but \(x^{n-1} = x\cdot x^{n-2}\), and we could continue expanding the terms until we arrive at a contradiction where \(x\) must be in \(P_i\) (eventually, we’ll arrive at the fact that \(x\cdot x^{n-n} = x\cdot 1 \in P_i\), but 1 is definitely not in \(P_i\), as prime ideals cannot be the whole ring, so \(x\) would have to be in \(P_i\)).
Since for all \(x \in \mathfrak{N}_R\), \(x \in P_i\), \(\mathfrak{N}_R \subseteq P_i\). Because this holds true for all \(P_i\), \(\mathfrak{N}_R \subseteq \bigcap_{i=1}^m P_i\).
We will now show the reverse inclusion also holds, i.e. \(\bigcap_{i=1}^m P_i\subseteq \mathfrak{N}_R\).
Equivalently, we can just show that for any nonnilpotent element in \(x \in R\), there exists at least one prime ideal that does not contain \(x\).
Let us take the set of all ideals that do not contain any element of the form \(x^n\) for any \(n\in \mathbb{Z}_{>0}\). We know that this set is non-empty, as it contains the zero ideal \(\{0\}\). The set is partially ordered by \(\subseteq\), and we apply Zorn’s lemma to take the maximal element \(M\) from the set.
Assume to the contrary that this maximal element is not prime, i.e. there exist some \(a, b \in R \setminus M\) such that \(ab \in M\). Then \(M\) is strictly contained within each of the ideal sums \((a) + M\) and \((b) + M\).
But \(M\) is known to be the maximal ideal not containing any element that is a power of \(x\); thus \((a) + M\), which is a strict superset of \(M\), contains at least some element of the form \(x^j\) for some \(j \in \mathbb{Z}_{>0}\), and likewise there is some \(x^k \in (b) + M\). But then \(x^jx^k = x^{j+k}\in (ab) + M = M\), as \(ab \in M\). This is a contradiction, since \(M\) was taken from the set of ideals not containing any power of \(x\).
Thus \(M\) must be prime, and we have found a prime ideal not containing any nonnilpotent element \(x \in R\). Thus \(\bigcap_{i=1}^m P_i\subseteq \mathfrak{N}_R\).
Both \(\mathfrak{N}_R\) and \(\bigcap_{i=1}^m P_i\) are subsets of each other, thus they are equivalent. \(\blacksquare\)
A Corollary: The nilradical is an ideal
With the above characterization of the nilradical, we can easily prove that the nilradical itself is always an ideal.
Proposition. The nilradical \(\mathfrak{N}_R\) of a commutative ring \(R\) is an ideal of \(R\).
Proof. We first prove an almost trivial lemma, which shows that intersections of ideals are themselves ideals.
Let \(I = \bigcap_{k \in K} J_k\), where each \(J_k\) is an ideal of \(R\), and \(K\) is just an index set.
Then for all \(x,y \in I\), \(x,y \in J_k\) for all \(k \in K\), as \(I\) is the set intersection of all \(J_k\). Then \(x+y \in J_k\), as \(J_k\) is an ideal, so \(x+y \in I\). Thus \(I\) is closed under addition.
For all \(x \in I\), \(r \in R\), \(x \in J_k\) for all \(k\) for the same reason as above. Then \(rx \in J_k\), as \(J_k\) is an ideal. Thus \(rx \in I\), and \(I\) is closed under multiplication from the ring. Thus \(I\) is an ideal, and this concludes the proof of the lemma.
Since \(\mathfrak{N}_R = \bigcap_{i=1}^m P_i\), then by the above lemma \(\mathfrak{N}_R\) is an ideal. \(\blacksquare\)
Of course, if we were just trying to show the nilradical is an ideal, we can do so without going through all the work of showing that it is the intersection of all prime ideals. Here is a alternate, direct proof of this phenomenon, which doesn’t need such heavy machinery:
Proposition. The nilradical \(\mathfrak{N}_R\) of a commutative ring \(R\) is an ideal of \(R\).
Proof. We need to show that both properties of ideals are satisfied, i.e. it is closed under addition and multiplication from \(R\).
Let \(x,y \in \mathfrak{N}_R\). \(x\) and \(y\) are both nilpotent by definition of being in \(\mathfrak{N}_R\), so \(x^n=0\) for some \(n\), and \(y^m=0\) for some \(m\). We examine
\[(x+y)^{n+m} = \sum_{i=0}^{n+m} {n + m \choose i}x^iy^{m+n-i},\]by the binomial theorem. Note when \(i \leq n\), \(y^{m+n-i}=y^my^{n-i}=0\cdot y^{n-i}=0\). Likewise when \(i \geq n\), the \(x^i\) term is zero, so the whole product is zero.
Thus, every term in the summation is forced to be zero, and the whole sum is zero. Since \((x+y)^{n+m}=0\), \(x+y\) is nilpotent, so \(x+y\in \mathfrak{N}_R\), and \(\mathfrak{N}_R\) is closed under addition.
Closure under multiplication is a bit simpler. For all \(x\in \mathfrak{N}_R\), \(r\in R\), \(x^n = 0\) for some \(n\), as \(x\) nilpotent. Then \((rx)^n = r^nx^n = r^n \cdot 0 = 0\), so \(rx \in \mathfrak{N}_R\).
Both properties hold, so \(\mathfrak{N}_R\) is an ideal. \(\blacksquare\)
Conclusion
I thought this bit of structure connecting nilpotent elements and prime ideals was pretty interesting, and the corollary we got out of it was quite non-obvious as well. We’ve shown that the nilradical is the intersection of all of the prime ideals in a commutative ring, and that the nilradical itself is also an ideal.
