Jacobson radicals of $R$ and $R/I$ where $I$ is a nilpotent ideal.

Question

Out of interest

If i have the map $\phi: R \longrightarrow R/I $ where $R$ is a ring and $I$ is a nilpotent ideal ?

then would i be right in saying that if i were to apply this map to the jacobson radical of $R$ it would take me to the jacobson radical of $R/I$

i.e. is the following true: $\phi(J(R)) = J(R/I)$

I am guessing this is right but can't be certain

Also if $\phi$ is surjective with kernel $I$ then this would imply R is artinian right? with R/I as semi simple?

any help would be great, thank you !

Answer 1

A nil ideal (in particular a nilpotent ideal) is contained in every maximal right ideal. Indeed, if $I$ is a nil ideal and $\mathfrak{m}$ is a maximal right ideal with $I\not\subseteq\mathfrak{m}$, we have $r+x=1$ with $r\in I$ and $x\in\mathfrak{m}$. But, as $r$ is nilpotent, say $r^n=0$, we have $$ (1-r)(1+r+r^2+\dots+r^{n-1})=1 $$ so $x\in\mathfrak{m}$ would be invertible: absurd.

Therefore your nilpotent ideal $I$ is certainly contained in $J(R)$ and so $$ J(R/I)=J(R)/I. $$

The quotient $R/I$ need not be artinian nor semisimple. Just consider $R=\mathbb{Z}$ and $I=\{0\}$.

Answer 2

Firstly, any maximal right ideal contains any nilpotent ideal. We want to prove this because it match the maximal right ideals of both rings exactly.

If $M$ were a maximal right ideal not containing $I$, then $M+I=R$, and in particular $m+i=1$ for some $m\in M$, $i\in I$. Pick $n$ such that $i^n=0$. Then $1=1^n=(m+i)^n\in M$, since all terms except $i^n$ have a multiple of $m$ are in $M$, and the last term $i^n=0$. But $M$ can't contain $1$ because it is proper.

Therefore by the correspondence of maximal right ideals, the two rings have "the same" maximal right ideals, and $\cap\{M/I\mid M\text{ maximal right ideal in } R\}=\cap\{M\mid M\text{ maximal right ideal in } R\}/I$

or in other words, $\mathrm{rad}(R/I)=\mathrm{rad}(R)/I$.

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This is not always the case, though, if $I$ isn't nilpotent (or nil). In general, if $f:R\to S$ is a surjective ring homomorphism, then only $f(\mathrm{rad}(R))\subseteq \mathrm{rad}(S)$.

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Also if $ϕ$ is surjective with kernel $I$ then this would imply $R$ is artinian right? with $R/I$ as semi simple?

No, that is non-sequitur. Even if $R/I$ is a field, $R$ need not be Artinian.

For example, let $\Bbb F_2$ be the field of two elements, and $V$ be an infinite dimensional $\Bbb F_2$ vector space. Define a ring structure on $R=\Bbb F_2\times V$ by using coordinatewise addition and by making multiplication $(a,v)(a',v')=(aa',av'+a'v)$. It's easy to see the following facts:

1. for any subspace $W<V$, $\{0\}\times W$ is an ideal of this ring, and since there is an infintely descending chain of subspaces, there's an infinitely descending chain of ideals. So $R$ is not Artinian. (Actually, it's not even Noetherian.)

2. $R$ has a unique maximal ideal $\{0\}\times V$ which has square zero, so that is the Jacobson radical.

3. $R/\mathrm{rad}(R)\cong \Bbb F_2$ is semisimple Artinian.