I have some question about the Jacobson radical of rings.

  1. What is $J(R)$ when $R$ is a Principal Ideal Domain but not a field?

e.g. I know that $\mathbb Z$ is a PID and why is $J(\mathbb Z)=0$ but can we say that is true for every Principal Ideal Domain but not a field? why?

  1. Let $R=C([0,1])$, the ring of real continuous functions on the interval $[0,1]$. Then what is $J(R)$?

  2. Let $R$ be a commutative ring. $J(R[[x]])$?

  3. $J(\mathbb Z_n)$?
    I know if $n$ is a prime then $\mathbb Z_p$ is a field and $J(\mathbb Z_p)=0$ also $J(\mathbb Z_{p^2})=(p)$ because it is a local ring. Is that true for example $J(\mathbb Z_{12})=( (2, 3)\mathbb Z/12 \mathbb Z ) = 6\mathbb Z/12\mathbb Z$? why?

  • $\begingroup$ Where are you stuck? Surely you can add a few thoughts on each question? $\endgroup$ – Namaste Feb 20 '14 at 13:35
  • $\begingroup$ For sure there are a lot of PID with nontrivial J(R), take for example all DVR's $\endgroup$ – Ferra Feb 20 '14 at 14:36
  1. Note that all ideals of the form $I_a=\{f\in C([0,1])\colon f(a)=0\}$, where $a\in[0,1]$, are maximal. In fact you can check easily that $C([0,1])/I_a\simeq \mathbb R$. Therefore $J(C([0,1]))\subseteq \bigcap_{a\in [0,1]}I_a=\{0\}$.

  2. If your $R$ is unitary, you need to prove that $f=\sum_{n\geq 0}a_nx^n$ is invertible iff $a_0$ is invertible in $R$. Now use the fact that the $J(A)$ for any ring $A$ is given by $\{x\in A\colon 1-xy\in A^*\,\forall y\in A\}$. From these two facts you deduce easily that $f\in J(R[[x]])$ iff $a_0\in J(R)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.