## 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

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

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

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

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 $J(R) = N$.