Problem 5 - Algebra

Question

(a) Let \(J\) be a two-sided nilpotent ideal of \(R\). Show that \(J\) is contained in the radical. (b) Conversely, assume that \(R\) is Artinian. Show that its radical is nilpotent, i.e.. that there exists an integer \(r \geqslant 1\) such that \(N^{r}=0 .\) [Hint: Consider the descending sequence of powers \(N^{r}\), and apply Nakayama to a minimal finitely generated left ideal \(L \subset N^{*}\) such that \(N^{*} L+0\).

Step-by-Step Solution

Answer

In this exercise, we investigated the relationship between two-sided nilpotent ideals and the radical of a ring. For part (a), we showed that any two-sided nilpotent ideal J is contained within the radical by proving that for any maximal left ideal M of R, \(J \subseteq M\). For part (b), we assumed that R is Artinian and showed that the radical is nilpotent by applying Nakayama's lemma to a minimal finitely generated left ideal L contained in N. We concluded that the radical themselves are nilpotent.

Step 1: Part (a): Showing J is contained in the Radical

Recall that an ideal J is called a two-sided nilpotent ideal if there exists a non-negative integer n such that \(J^n = 0\), and the radical (or Jacobson radical) of a ring R, denoted \(N\), is the intersection of all the maximal left ideals of R. To show that J is contained in the radical, we need to prove that for any maximal left ideal \(M\) of R, \(J \subseteq M\). Let \(j\) be an element in J. Since J is a two-sided nilpotent ideal, there exists an integer \(n \geq 1\) such that \(J^n = 0\). In particular, we have that \(j^n \in J^n = 0\) for all \(j \in J\). Thus, \(j^n = 0\) for all \(j \in J\). Now consider any maximal left ideal \(M\) of R. Since \(R/M\) is a left simple \(R\)-module and \(j^n = 0\), we have that \((1+M)(j+M) = j+M\). Therefore, for any \(m \in M\), we have \((1+m)j = j\). This implies that \(j - j = m \cdot j\), and since j is an element of J and J is an ideal, we have that \(m \cdot j\) is an element of J as well. Now, since any element in J can be represented as an element of M times an element in J, it follows that \(J \subseteq M\). This is true for every maximal left ideal M of R. Hence, we conclude that \(J \subseteq N\), where N is the radical.

Step 2: Part (b): Showing the Radical is Nilpotent when R is Artinian

Lets assume that R is Artinian. This means that every descending chain of left ideals in R eventually stabilizes. We want to show that there exists an integer \(r \geqslant 1\) such that \(N^{r} = 0\), where N is the radical. Consider the descending sequence of powers of the radical: \(N \supseteq N^2 \supseteq N^3 \supseteq \cdots\). Since R is Artinian, this chain must stabilize, i.e., there exists an integer \(r \geq 1\) such that \(N^r = N^{r+1} = \cdots\). Now, take a minimal finitely generated left ideal \(L \subset N\) such that \(N^r L = 0\). We will apply Nakayama's lemma to L. Nakayama's lemma states that if M is a finitely generated left R-module, I is an ideal contained in the radical, and \(IM = M\), then M = 0. In our case, let \(I = N^{r}\) and \(M=L\), we have that \(IM = N^{r} L = 0\). By Nakayama's lemma, it follows that L = 0. Hence we have \(N^r = N^{r+1} = \cdots = 0\). Therefore, the radical N is nilpotent.

Key Concepts

Jacobson Radical

Understanding the Jacobson radical is fundamental when studying ring theory. Imagine a ring as a city, and within this city, we find the Jacobson radical, a special neighborhood made up of elements that behave quietly when interacting with the rest of the ring's elements.

In mathematical terms, the Jacobson radical, often denoted by 'N', is the intersection of all maximal left ideals of a ring 'R'. These maximal left ideals can be thought of as the largest 'blocks' in our metaphorical city that don't contain the whole city itself. Just as every neighborhood is part of the city, the nilpotent ideal, is always nestled within the Jacobson radical. This ideal is like a sleepy neighborhood where, if you walk through it enough times—mathematically speaking, multiply enough times—you'll find yourself at the point of 'zero', signifying the elements eventually have no effect.

Examining the Nilpotent Ideal within the Radical

When we prove that a two-sided nilpotent ideal 'J' is contained within the Jacobson radical, we consider an element within the neighborhood of J. As it's nilpotent, walking through it 'n' times, where 'n' is the level of its nilpotency, leads us to the point of zero. Since this happens with all elements of the neighborhood 'J', and since every maximal left ideal comprises part of the city's boundaries, 'J' must be situated inside the radical's confines.

Artinian Ring

If you think of a ring as a vast library with infinite shelves of books, an Artinian ring makes sure that if you start removing books from the top shelf, eventually you will reach a point where removing more books doesn't change the shelf. Symbolically, it's a ring where any descending sequence of left ideals will eventually stabilize—you can't keep going down the shelves forever without eventually hitting the same spot.

Stability in Descending Ideal Sequences

An Artinian ring is essential for our stability-themed story, since it ensures that our exploration of the radical's powers won't keep us descending eternally. We are guaranteed to find a 'ground level' (an integer 'r') where the powers of the radical don't diminish any further. This aspect of Artinian rings plays a crucial role in verifying the nilpotency of the Jacobson radical, as it stops the 'fall' within the ideal's descending sequence, providing a foundation we can use to further apply other concepts like Nakayama's lemma.

Nakayama's Lemma

Much like a secret magic spell in a fantasy novel, Nakayama's lemma is a powerful tool in commutative algebra and module theory that often reveals surprising results. It tells us, quite astonishingly, that if a finitely generated module over a ring is entirely captured by an ideal within the Jacobson radical, then the module itself must be zero.

Applying Nakayama's Magic

In our textbook problem, Nakayama's lemma comes to the rescue by asserting that if we have a minimal finitely generated left ideal 'L' sitting snugly within our Jacobson radical, then 'L' cannot be anything but zero after we've wandered down through 'r' levels of our radical's power. Intuitively, it's like saying if every book on a shelf is a copy of the same book, removing that book means there are no books left—an empty shelf. So, when we apply Nakayama's lemma to our scenario, we find that the ideal 'L' simply vanishes, which in turn confirms the radical's nilpotency in an Artinian ring. This insinuates a profound relationship between the elements of the ring and the architecture of its ideals and radicals.