Exercise 1.4

Question

In the ring $A[x]$, the Jacobson radical is equal to the nilradical.

Amit Awasthi · Accepted Answer

\begin{proof}
  Since every maximal ideal is prime, Prop.~1.8 implies $\mathfrak{N} \subset \mathfrak{R}$. Conversely, suppose $f = a_0 + a_1x + \cdots + a_nx^n \in \mathfrak{R}$. By Prop.~1.9, $1-fx$ is a unit. By Exercise $2i)$, the $a_i$ are nilpotent, and so $f \in \mathfrak{N}$ by Exercise $2ii)$.
\end{proof}

Exercise 1.4

Answers

Comments

Exercise 1.4

Answers

Comments

Add answer