How to show that $P(R)$ is contained in $\operatorname{Nil}(R)$ (where $R$ is a noncommutative ring with identity)?
Definitions I am using:
- A nil right ideal is one whose elements are all nilpotent.
- A prime ideal is a two sided ideal which satisfies the property: $aRb$ contained in $I$ implies $a$ is in $I$ or $b$ is in $I$.
- $P(R)$ is the intersection of all prime ideals.
- $\operatorname{Nil}(R)$ is the sum of all right nil ideals of $R$.
This is easy since $P(R)$ is a nil ideal. Happily, this can be seen using ideas from the commutative theory.
Proof
Suppose to the contrary that $x\in P(R)$ isn't nilpotent. Then $\{1,x,x^2,x^3\dots\}$ is a multiplicatively closed system disjoint from $\{0\}$. Using the usual Zorn's lemma argument, find an ideal $I$ which is maximal with respect to being disjoint from the powers of $x$.
We first contend that $I$ is a prime ideal. Let $aRb\subseteq I$ with $a,b\notin I$. By maximality, $x^n\in RaR+I$ for some $n$, and $x^k\in RbR+I$ for some $k$. But then $x^{n+k}\in(RaR+I)(RbR+I)\subseteq I$, a contradiction. Thus at least one of $a,b$ is in $I$, proving the primeness of $I$.
But the existence of this prime ideal $I$ disjoint from the powers of $x$ witnesses that $x\notin P(R)$, which is a contradiction. Therefore $x$ is nilpotent.
Thus $P(R)$ is a nil ideal, and is thus trivially contained in $Nil(R)$.