**Proof.**
Denote $f : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphism. Assertion (1) really means that $\omega _ X^\bullet $ and $\omega _ X$ are as in Lemma 53.4.1 for the morphism $f' : X \to \mathop{\mathrm{Spec}}(k')$. In the proof of Lemma 53.4.1 we took $\omega _ X^\bullet = a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ where $a$ be is the right adjoint of Duality for Schemes, Lemma 48.3.1 for $f$. Thus we have to show $a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) \cong a'(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ where $a'$ be is the right adjoint of Duality for Schemes, Lemma 48.3.1 for $f'$. Since $k' \subset H^0(X, \mathcal{O}_ X)$ we see that $k'/k$ is a finite extension (Cohomology of Schemes, Lemma 30.19.2). By uniqueness of adjoints we have $a = a' \circ b$ where $b$ is the right adjoint of Duality for Schemes, Lemma 48.3.1 for $g : \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. Another way to say this: we have $f^! = (f')^! \circ g^!$. Thus it suffices to show that $\mathop{\mathrm{Hom}}\nolimits _ k(k', k) \cong k'$ as $k'$-modules, see Duality for Schemes, Example 48.3.2. This holds because these are $k'$-vector spaces of the same dimension (namely dimension $1$).

Proof of (2). This holds because we have base change for $a$ by Duality for Schemes, Lemma 48.6.2. See discussion in Duality for Schemes, Remark 48.12.5.
$\square$

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