The radical of the algebra $ A = T_n(F)$ is $N$, the set of all strictly upper triangular matrices. I read two questions about The maximal nilpotent ideal, but I have a problem.
Example:
Let $A$ be the set of all infinite, $\mathbb{N} \times \mathbb{N}$ matrices over $F$ that are upper triangular and have only finitely many nonzero entries. Note that $A$ is an algebra under the standard matrix operations. Let $I$ be the set of all matrices in $A$  that are strictly upper triangular. Then $I$ is a nil ideal of $A$ which is not nilpotent.

1: Why is $I$ a nil ideal of $A$ which is not nilpotent?

Definition: The maximal nilpotent ideal of a finite dimensional algebra $A$ is called the radical of $A$.
Example: Let $A = T_n(F)$ be the algebra of all upper triangular $n \times n$ matrices over a field $F$, i.e., matrices that have zeros below the main diagonal. Let $N$ be the set of all strictly upper triangular matrices, i.e., matrices in $A$ that have zeros also on the diagonal. Then $N $ is an ideal of $A$ such that $N^n = 0$ and $N^{n-1} \neq 0$. One can find a variety of other ideals of $A$ that are contained in $N$, and are therefore nilpotent. For instance, the set $I$ of all matrices whose  $(1, n)$ entry is arbitrary and all other entries are $0$ is an ideal of $A$ satisfying $I^2 = 0$.
The radical of the algebra $ A = T_n(F)$  is $N$, the set of all strictly upper triangular matrices. Indeed, $N$ is a nilpotent ideal of $A$, and any ideal of $A$  that properly contains $N$ cannot be nilpotent since it necessarily contains a nonzero diagonal matrix.

2: Why  is $ N$ an ideal of $A$ such that $N^n = 0$ and $N^{n-1} \neq 0$?
3: Can you explain "  One can find a variety of other ideals of $A$ that are contained in $N$, and are therefore nilpotent. For instance, the set $I$ of all matrices whose  $(1, n)$ entry is arbitrary and all other entries are $0$ is an ideal of $A$ satisfying $I^2 = 0$" ?
4:Does it mean "$N$ is a nilpotent ideal of $A$, and any ideal of $A$  that properly contains $N$ cannot be nilpotent since it necessarily contains a nonzero diagonal matrix" ?

 A: For 1: A nil ideal is one for which each element is nilpotent; a nilpotent ideal is one for which the ideal itself is nilpotent (in the sense that $I^k=\{0\}$ for some natural number $k$). Clearly any nilpotent ideal is nil, since for every $i\in I$, $i^k\in I^k$ hence $i^k=0$.
However, these things are not equivalent, since $I^k$ not only contains things like $i^k$ but also things like $i_1\ldots i_k$ where $i_1,\ldots, i_k\in I$. This provides one easy way to clarify the example: it is true that all the matrices in $I$ are clearly nilpotent (though I can expand on this if it is unclear), but there is no fixed $k$ such that all $k$-fold products are zero. 
In fact, this is easy to see by coming up with matrices $a_n\in I$ for all $n\geq 2$ such that $a_n^{n-1}\neq 0$ but $a_n^n=0$.
However, my favorite easy standard example would be: let $e_{i,j}$ be the matrix with exactly one
non-zero entry of $1$ in the $(i,j)$ position, and note that $e_{i,j}e_{k,\ell}=\begin{cases}e_{i,\ell}&\text{if }j=k\\0&\text{otherwise}\end{cases}$. Then $e_{1,2}e_{2,3}\ldots e_{k,k+1}=e_{1,k+1}\neq 0$ for any $k$ and $e_{1,2}e_{2,3}\ldots e_{k,k+1}\in I^k$, so $I^k\neq \{0\}$.

For 2: Using the above notations, let $D=e_{1,2}+e_{2,3}+\ldots +e_{n-1,n}$. Then $D\in N$, and one computes that $D^{n-1}=e_{1,n}$ (since the only non-zero product after expanding $D^{n-1}$ is $e_{1,2}e_{2,3}\ldots e_{n-1,n}$). Since $D^{n-1}\in N^{n-1}$ by definition, $N^{n-1}\neq \{0\}$.
It is intuitively clear that any $n$-th power of an element of $N$ is zero: by computing a few examples, it is easy to see that the $k$th power of an element of $N$ only has non-zero entries on the $k$th off-diagonal and above. It's also pretty easy to convince yourself this holds for products of $k$ elements. What follows is one way to prove this, but certainly isn't the only way.
Any element $a$ of $N$ is of the form $a=\sum_{i<j} a_{i,j}e_{i,j}$ where $a_{i,j}\in F$, and given $b=\sum_{i<j}b_{i,j}e_{i,j}$, we can
explicitly determine the product: 
$$ab=\sum_{i<j} \sum_{i<k<j} (a_{i,k}b_{k,j}) e_{i,j}.$$ We could define a statistic $\deg(a)=\min(j-i:a_{i,j}\neq 0)$
and $\deg(0)=\infty$, so e.g. $\deg(e_{1,2}+e_{1,3})=\min(2-1,3-1)=1$. An easy observation is if $a\neq 0$, $1\leq \deg(a)\leq n-1$. 
On the other hand, $\deg(ab)\geq \deg(a)+\deg(b)$: if $ab=0$, this is obvious since $\deg(ab)=\infty$. 
Otherwise,  $\deg(ab)=j-i$ for some $j,i$ where $(ab)_{i,j}\neq 0$. Since $(ab)_{i,j}=\sum_{i<k<j} (a_{i,k}b_{k,j})$
then some $a_{i,k}b_{k,j}\neq 0$ for some $i<k<j$, so $a_{i,k},b_{k,j}\neq 0$. But then $\deg(a)\leq k-i$, $\deg(b)\leq j-k$, so $\deg(a)+\deg(b)\leq j-i=\deg(ab)$.
But then for any product $a_1\ldots a_n$ where $a_i\in N$, $$\deg(a_1\ldots a_n)\geq \deg(a_1)+\ldots+\deg(a_n)\geq 1+\ldots+1=n,$$ so $\deg(a_1\ldots a_n)=\infty$ thus $a_1\ldots a_n=0$. Since any $n$-fold product is zero
and $N^n$ is by definition the span of such products, $N^n=\{0\}$.

For 3: Again, in the above notation, they are referring to $I=\{xe_{1,n}\mid x\in F\}$ (for $n=2$, this means $I=\left\{\begin{bmatrix}0&x\\0&0\end{bmatrix}\mid x\in F\right\}$). This is certainly an ideal of $N$: for any $a\in N$, $a=\sum_{i<j}a_{i,j}e_{i,j}$ and $ae_{1,n}=0$
since if $1\leq i<j\leq n$, $j>1$ so $e_{i,j}e_{1,n}=0$. In particular, since $e_{1,n}^2=0$, we see $I^2=\{0\}$. There are other similar ideals we could define, like let $I_k=\{a\in N\mid \deg(a)\geq k\}$, so $I=I_{n-1}$ and $N=I_1$.
Then $I_k$ is nilpotent, with $I_k^{n+1-k}=\{0\}$.

For 4: What they mean is that if if we have an proper ideal strictly containing $N$, say $A\supset I\supset N$, then there is some $a\in I\setminus N$, which means $a$ can't be strictly upper-triangular.
In particular, we can subtract off the strictly upper triangular part (which is an element of $N$, hence $I$) and assume $a$ is diagonal. Since it is not in $N$, it must be non-zero, thus $a$ is a non-zero diagonal matrix.
Then we can write $a=\sum a_{i} e_{i,i}$ where $a_i\in F$. Since $a\neq 0$,
there is some $j$ such that $a_{j}\neq 0$. Then for any $k$, $a^k=\sum a_i^k e_{i,i}$, and since $a_j\neq 0$, $a_j^k\neq 0$, thus $a^k\neq 0$. Since $a^k\in I^k$ for any $k$, we see $I^k\neq \{0\}$ and thus $I$ cannot be nilpotent.
