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.

Example:

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 blynn@cs.stanford.edu 💡