Commutative Algebra - The Jacobson Radical

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 = ⟨ M \cup {x} ⟩ = {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 $R$ be a PID. Then $J (R) = N = {0}$ .
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
$J (R) = N = {\sum α_{g} g ∣ \sum α_{g} = 0}$
Define $ϕ : R \to F$ by $\sum α_{g} g \mapsto \sum α_{g}$ . Then $ϕ$ is a surjective ring homomorphism. It is known as the augmentation map. We have
$\ker ϕ = {\sum α_{g} g ∣ \sum α_{g} = 0}$
This is known as the augmentation ideal. By the Fundamental Homomorphism Theorem, $R / \ker ϕ ≅ F$ hence $\ker ϕ$ is a maximal ideal of $R$ . Thus $J (R) \subset \ker ϕ$ .

Consider $x = α_{1} g_{1} + . . . + α_{n} g_{n} \in \ker ϕ$ . Then $α_{1} + . . . + α_{n} = 0$ and for some sufficiently high power $m = p^{k}$ we have $g_{i}^{m} = 1$ for $i = 1, . . ., n$ . Then
$\begin{matrix} x^{m} & = & (α_{1} g_{1} + . . . + α_{n} g_{n})^{m} \\ = & (α_{1} g_{1})^{m} + . . . + (α_{n} g_{n})^{m} \\ = & α_{1}^{m} + . . . + α_{n}^{m} \\ = & (α_{1} + . . . + α_{n})^{m} \\ = & 0 \end{matrix}$
Thus $x \in N$ , and we have $J (R) \subset \ker ϕ \subset N \subset J (R)$ , implying the result. 3. Take $R = ℝ [[x]]$ . 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 $m < n$ 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 $n' \geq 2 m$ , say $n' = 2 m + k$ . Lastly multiplying both sides by $x^{k}$ shows that some power of $x$ is idempotent.

Suppose $0 \neq e = e^{2} \in R$ . 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 $e \neq 0$ . Hence $1 - e$ is a nonunit if $0 \neq e = e^{2}$ .

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 $J (R) = N$ .