If $R$ is a ring (not assumed to be commutative or containing 1)
We define the nilpotent radical of $R$, $N(R)$, to be the sum of all nilpotent ideals of R.
Suppose $R$ has at least one prime ideal. Let $X$ be the intersection of all prime ideals of $R$.
Then $N(R)\subset X$ (with equality, i.e. $X\subset N(R)$, if $R$ is commutative).
In my lecture notes, the proof of the first part, $N(R)\subset X$, just says "easy". I've puzzled over this for a good while and I just can't see why. Sorry if i'm missing something very obvious here.
Things i've tried. $N(R)$ is a nil ideal (but need not be nilpotent). $X$ is semiprime and nil.
If I take $x\in N(R)$, then $0=x^n\in P$ for every prime ideal P, so $x^n\in X$, but I don't see a way to prove $x$ is in $X$.