The Jacobson Radical

The Jacobson radical \(J(R)\) of a ring \(R\) is the intersection of the maximal ideals of \(R\). Since all maximal ideals are prime, the nilradical is contained in the Jacobson radical.

Theorem: Let \(x\in R\). Then \(x\) lies in the Jacobson radical \(J(R)\) of \(R\) if and only if \(1 - x y\) is a unit for all \(y\in R\).

Proof: Suppose \(1 - x y\) is a nonunit for some \(y \in R\). Then \(1-x y\in M\) for some maximal ideal \(M\). If \(x\in J(R)\) then \(x\in M\) since \(J(R)\) is the intersection of all the maximal ideals, thus \(1 = (1-x y) + x y \in M\), which is a contradiction.

Conversely, suppose \(x \notin J(R)\). Then \(x \notin M\) for some maximal ideal \(M\). Then

\[ R = \langle M \cup \{x\}\rangle = \{m+ x y|m\in M, y \in R \} \]

In particular \(1 = m + x y\) for some \(m \in M, y\in R\). Hence \(m = 1 - x y\) cannot be a unit since \(M\) is a proper ideal.


Let \(R = F[G]\), a group ring where \(F\) is a field of prime characteristic and \(G\) is an abelian \(p\)-group. Then we shall show that

\[ J(R) = N = {\{\sum\alpha_g g | \sum\alpha_g = 0\}} \]

Define \(\phi:R\rightarrow F\) by \(\sum \alpha_g g \mapsto \sum \alpha_g\). Then \(\phi\) is a surjective ring homomorphism. It is known as the augmentation map. We have

\[ ker \phi = {\{\sum \alpha_g g | \sum \alpha_g = 0\}} \]

This is known as the augmentation ideal. By the Fundamental Homomorphism Theorem, \(R/ker \phi \cong F\) hence \(ker \phi\) is a maximal ideal of \(R\). Thus \(J(R) \subset ker\phi\).

Consider \(x=\alpha_1 g_1 +...+ \alpha_n g_n \in ker\phi\). Then \(\alpha_1 +...+ \alpha_n = 0\) and for some sufficiently high power \(m = p^k\) we have \(g_i^m = 1\) for \(i=1,...,n\). Then

\[ \array { x^m &=& (\alpha_1 g_1 + ... + \alpha_n g_n)^m \\ &=& (\alpha_1 g_1)^m +...+ (\alpha_n g_n)^m \\ &=& \alpha_1^m +...+ \alpha_n^m \\ &=& (\alpha_1+...+\alpha_n)^m \\ &=& 0 } \]

Thus \(x \in N\), and we have \(J(R) \subset ker\phi \subset N \subset J(R)\), implying the result. 3. Take \(R = \mathbb{R}[[x]]\). Then \(N = \{0\}\). \(R\) is local with a unique maximal ideal \(M = J(R)\) consisting of power series with zero constant term. 4. Suppose \(R\) is a finite ring. We shall show the nilradical and Jacobson radical coincide.

For any \(x \in R\), we must have \(x^m = x^n\) for some \(m \lt n\) since there are only a finite number of elements. Then we may add arbitrary multiples of \(n - m\) to the exponent on the right-hand side to obtain \(x^m = x^{n'}\) where \(n' \ge 2m\), say \(n' = 2m + k\). Lastly multiplying both sides by \(x^k\) shows that some power of \(x\) is idempotent.

Suppose \(0\ne e = e^2 \in R\). Then \((1-e)^2 = 1 - e - e + e^2 = 1 - e\) hence \(1 -e\) is idempotent. Then if \(1-e\) is a unit, then \((1-e)y = 1\) for some \(y \in R\). Squaring this gives \(1 = (1-e)^2 y^2 = (1-e)y \cdot y = y\), which implies \(1 - e = 1\), a contradiction since \(e \ne 0\). Hence \(1-e\) is a nonunit if \(0\ne e =e^2\).

Now suppose \(x \in J(R)\). Then \(1 - x y\) is a unit for all \(y \in R\). For some \(n\) we have \(x^n\) idempotent. Pick \(y = x^{n-1}\). Then \(1 - x^n\) is also idempotent, and as it is also a unit, we must have \(x^n = 0\), thus \(x \in N\). Hence \(J(R) = N\).

Ben Lynn 💡