Jacobson radical of rings

Question

I have some question about the Jacobson radical of rings.

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?

Let $R=C([0,1])$, the ring of real continuous functions on the interval $[0,1]$. Then what is $J(R)$?
Let $R$ be a commutative ring. $J(R[[x]])$?
$J(\mathbb Z_n)$?
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?

Where are you stuck? Surely you can add a few thoughts on each question? — amWhy, Feb 20 '14 at 13:35
For sure there are a lot of PID with nontrivial J(R), take for example all DVR's — Ferra, Feb 20 '14 at 14:36

score 2 · Accepted Answer · edited Dec 24 '14 at 20:21

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 $J(C([0,1]))\subseteq \bigcap_{a\in [0,1]}I_a=\{0\}$.
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 $a_0\in J(R)$

1 Answers1