Nilpotent ideals of algebra over a field. I have 2 questions about the nilpotent ideals of an algebra $A$ over a field $K$.

*

*Each non-nilpotent left ideal of $A$ contains a nonzero idempotent element of $A$.

*The sum of all left nilpotent ideals of $A$ is still nilpotent.

For the first question, I tried to prove by contradiction but I have no idea how to show that the ideal is nilpotent.
For the second question, I can prove it when the number of left nilpotent ideals is finite, but I can't deal with the case that there are infinitely many left nilpotent ideals or show that's impossible.
 A: Counterexamples in the infinite dimensional case
The first proposition is not even true for commutative algebras in general: in $F[x,y]/(xy, x^2)$, the ideal generated by $y$ is not nilpotent (you can check it looks just like polynomials in $y$ over $F$) and contains no nonzero idempotents.
And, as you noted, the sum of finitely many nilpotent left ideals is nilpotent, but it is not so clear for infinite sums. For the second proposition, consider $R=F[x_1, x_2,\ldots]/(\{x_i^i\mid i\geq 2\})$. The sum $\sum _{i=2}^\infty (x_i)$ isn't nilpotent.
This leads me to suspect that the intended context was finite dimensional algebras. If $A$ is left Artinian both can be done, and finite dimensionality would do that simply.
The two things mentioned are properties of left Artinian rings: the former because of idempotent lifting (I think) and the latter because a sum of left ideals is equal to a finite sum of left ideals.
For complete proofs
It seems like an elementary proof that a non-nilpotent left ideal contains a nonzero idempotent is fiddly and I am reluctant to do a transcription. The one I have in mind is here

Curtis, Charles W., and Irving Reiner. Representation theory of finite groups and associative algebras. Vol. 356. American Mathematical Soc., 1966.

on page 160, Theorem 24.2.  Presently I can view the pages in google books
Then a few pages later, theorem 24.4 is that the sum of nilpotent left ideals is an ideal.
In both cases, it is postulated that the ring "has minimum condition" which means that it is Artinian.
A: With the claim that $\dim A$ should be finite and the hints from @rschwieb, I tried to work out a way to complete the proof.
For the first proposition
Given a left ideal $I$ of $A$, obviously $I$ is closed under addition. For any $k \in K$ and $x \in I$ we also have
$$
kx = 1 \cdot (kx) = (k1) \cdot x \in I
$$
Hence $I$ becomes a $K$-linear subspace of $A$.
Note that $\dim I \leq \dim A < +\infty$, we can take $x_1, \cdots, x_k$ as basis of $I$ and then
$$
I = Ax_1 \oplus Ax_2 \oplus \cdots \oplus Ax_k
$$
For non-nilpotent left ideal $I$, there must be some $i$ such that $Ax_i$ is also non-nilpotent. So we just need to consider the left ideal given by $J = Ax_1$.
Obviously $\dim J = 1$. The fact that $J$ is non-nilpotent leads to $Jx_1 = J$, which is to say $Ax_1^2 = Ax_1$, so we can find some $a \in A$ such that
$$
ax_1^2 = x_1
$$
Then one can check that $ax_1$ is idempotent.
For the second proposition
It seems can also be proved by observing left ideals as subspaces to show that a sum of left ideals is equal to a finite sum of left ideals.
