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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 $X$.

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You can't really get anywhere by only using the fact that nilpotent ideals are nil (like you are in the last line.) In fact, the sum of all nil ideals in a ring is another radical called the "upper nilradical," and it must contain $X$. (The intersection of prime ideals is called the "lower nilradical". You can read about these and more here) You need to use the full power of "nilpotent."

By definition, a nilpotent ideal $N$ is one for which $N^n=\{0\}$ for some positive integer $N$.

If $P$ is any given prime ideal, then $N^n\subseteq P$ implies $N\subseteq P$ by definition of primality and induction. Thus $N$ is contained in $X$.

Since all nilpotent ideals are contained in $X$, their sum $N(R)$ is contained in $X$.

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    $\begingroup$ Ahh damn that's what I was missing. Thanks for the help. $\endgroup$ – Henry Mar 24 '17 at 17:25

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