Lemma 49.4.8. Let $k$ be a field and let $A$ be a finite $k$-algebra. Assume $A$ is local with residue field $k'$. The following are equivalent

1. $\text{Trace}_{A/k}$ is nonzero,

2. $\tau _{A/k} \in \omega _{A/k}$ is nonzero, and

3. $k'/k$ is separable and $\text{length}_ A(A)$ is prime to the characteristic of $k$.

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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