Wedderburn's Theorem on nilpotent Ideals

Question

Let $A$ be a finite dimensional algebra over a field $\mathbb{k}$, and $I \trianglelefteq A$ a two sided ideal. There is a theorem of Wedderburn which states

(*) If $I$ is generated by nilpotent elements (as $\mathbb{k}$-vector space) then $I$ is already nilpotent.

There is an outline of a proof in [Farb, Dennis - Noncommutative Algebra] (Theorem 3.21) which uses induction on the dimension of $I$. However, I think it is possible to simplify this proof by avoiding induction. I would be happy if anyone can confirm my argument or point out an error. Thank you in advance!

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Proof of (*): We assume the statement is wrong, so let $I$ be a non-nilpotent ideal of a $\mathbb{k}$-algebra $A$ which is generated by nilpotent elements. The following steps show that we can build more counterexamples from $(A,I)$ with additional properties, so that we will finally get a contradiction.

1. $\overline{\mathbb{k}} \otimes I$ is a non-nilpotent ideal of $\overline{\mathbb{k}} \otimes A$ which is generated by nilpotent elements, so we may assume that $\mathbb{k}$ is algebraically closed.
2. The image of $I$ in $A/J(A)$ is non-nilpotent but generated by nilpotent elements, so we may assume that $A$ is semisimple.
3. Now $A \cong \prod_i M_{n_i}(\mathbb{k})$ is a product of matrix rings over $\mathbb{k}$, and $I$ is a product of a nonempty subcollection of these. As $I$ is generated by nilpotents, there is an $i$ such that $M_{n_i}(\mathbb{k})$ is generated by nilpotents. But this is absurd since nilpotent matrices have zero trace, and so have all linear combinations of them.

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Added later: If my argument is correct, then I don't see why the statement (*) shouldn't be valid for left (or right) ideals as well. Suppose there is a non-nilpotent left ideal $L \leq A$ which is generated by nilpotents. Then we can adopt the arguments 1. and 2. verbatim and assume that $\mathbb{k}$ is algebraically closed and $A$ is semisimple. But then $L$ is a direct sum of left ideals of matrix rings which are all generated by nilpotents. This is a contradiction, as there are no nonzero left ideals of $M_n(\mathbb{k})$ consisting of matrices of zero trace.

At least now there should be an error somewhere. :)

Answer 1

I just realized that Wedderburn's theorem is actually true for left ideals, as it can be deduced from (*):

Let $L \leq A$ be a left ideal which is generated by nilpotents. Then $L' := 0 \oplus L$ is a two sided ideal in $B := \mathbb{k} \oplus L$, the unitalization of $L$ (viewed as a non-unital algebra). Since $L'$ is generated by nilpotents, it must be nilpotent by (*), and so must be $L$.

Indeed this theorem of Wedderburn is sometimes stated as follows:

If a non-unital algebra $A$ has a basis of nilpotent elements, then $A$ is nilpotent.