It is false that any nilpotent element of a ring is contained in every ideal. For example, in the ring $R[x,y]/(x^n)$ where $R$ is any ring and $n>1$, the element $x^n=0$ is every ideal but $x^r$ is not in any ideal of the form $(g)$ for $g\in R[y]$ when $r<n$, for example. To be even more concrete, consider the example of $\mathbb{Z}[x]/(x^2)$, in which $x^2=0$ but $x$ is not in the ideal generated by $2$, for example.
Also, "smallest non-zero ideal of the ring" doesn't make sense for most rings. For example, in $\mathbb{Z}$, there is no minimal non-zero ideal; given any non-zero ideal $(n)$, the ideal $(2n)$ is smaller, and still non-zero. So there are no minimal non-zero ideals in $\mathbb{Z}$. In general, the poset of non-zero ideals may have many minimal elements: if $n\in\mathbb{Z}$ is a product of $k$ distinct prime numbers $q_1,\ldots,q_k$, then $\mathbb{Z}/n\mathbb{Z}$ has $k$ distinct minimal non-zero ideals, the $a_i\mathbb{Z}/n\mathbb{Z}$ where $a_i=q_1q_2\cdots q_{i-1}q_{i+1}\cdots q_k$.
What is true is that the nilradical of $R$ is the intersection of all the prime ideals of the ring $R$ (and this includes the zero ideal if $R$ is an integral domain).
This is because, if $P\subset R$ is a prime ideal, then $x^n=0\in P$ does imply that $x\in P$ (by the definition of prime ideal), so $$x\in\text{nil}(R)\implies x\in\bigcap_{P\subset R \text{ prime} }P,$$
and if $x\notin\text{nil}(R)$, then the collection $\Sigma$ of ideals of $R$ not containing $1,x,x^2,\ldots$ is a partially ordered set under inclusion, and it has some maximal element $M$ (using Zorn's lemma). This maximal element $M$ must be a prime ideal, because if $a\notin M$ and $b\notin M$, then $M+(a)$ and $M+(b)$ are both ideals of $R$ strictly containing $M$, hence containing powers of $x$ (because $M$ is maximal among ideals not containing powers of $x$). Thus the ideal $M+(ab)\supseteq (M+(a))(M+(b))$ contains a power of $x$, hence $M+(ab)$ strictly contains $M$, hence $ab\notin M$. Thus we have shown that $a,b\notin M\implies ab\notin M$, so $M$ is a prime ideal not containing $x$, and therefore we have shown
$$x\notin\text{nil}(R)\implies x\notin\bigcap_{P\subset R \text{ prime} }P.$$
Therefore
$$\text{nil}(R)=\bigcap_{P\subset R \text{ prime} }P.$$
Amitesh's exercises are also excellent, as usual :)
The upper nilradical $Nil^\ast(R)$ is defined to be the sum of nil ideals.
The lower nilradical $Nil_\ast(R)$ is defined to be the intersection of prime ideals of the ring.
In the context of Artinian rings, nil radicals are nilpotent, so the sum of all nilpotent ideals is again a nilpotent ideal, and this is sometimes referred to as the Wedderburn radical. But I'm not sure if anyone uses the sum of all nilpotent ideals outside this context.
We have $N(R)\subseteq Nil_\ast(R)\subseteq Nil^\ast(R)$, and I don't think the first two are always equal.
No, it is not the set of nilpotent elements. Take $M_2(F)$ for any field $F$, and $N(R)$ is the zero ideal, yet there are nilpotent elements.
Yes.
Every nilpotent ideal is a nilpotent left ideal. So the set of nilpotent ideals is a subset of the set of nilpotent left ideals. Therefore the sum of the former is contained in the sum of the latter. There is nothing hard here.
Best Answer
The sum of two (or finitely many) nilpotent ideals is nilpotent. However, infinite sums of nilpotent ideals are not necessarily nilpotent.
Consider, however, the ring $$ \Bbb R[x_1,x_2,x_3,\ldots]/(x_1,x_2^2,x_3^3,\ldots) $$ Here the nilradical $(x_1,x_2,x_3,\ldots)$ is indeed nil but not nilpotent.