**Proof.**
Conditions (1) and (2) are equivalent by Lemma 49.4.3. Let $\mathfrak m \subset A$. Since $\dim _ k(A) < \infty $ it is clear that $A$ has finite length over $A$. Choose a filtration

\[ A = I_0 \supset \mathfrak m = I_1 \supset I_2 \supset \ldots I_ n = 0 \]

by ideals such that $I_ i/I_{i + 1} \cong k'$ as $A$-modules. See Algebra, Lemma 10.52.11 which also shows that $n = \text{length}_ A(A)$. If $a \in \mathfrak m$ then $aI_ i \subset I_{i + 1}$ and it is immediate that $\text{Trace}_{A/k}(a) = 0$. If $a \not\in \mathfrak m$ with image $\lambda \in k'$, then we conclude

\[ \text{Trace}_{A/k}(a) = \sum \nolimits _{i = 0, \ldots , n - 1} \text{Trace}_ k(a : I_ i/I_{i - 1} \to I_ i/I_{i - 1}) = n \text{Trace}_{k'/k}(\lambda ) \]

The proof of the lemma is finished by applying Fields, Lemma 9.20.7.
$\square$

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