Does $\mathcal{A}\otimes\mathbb{C}(t)\cong\mathcal{D}\otimes\mathbb{C}(t)$ imply an isomorphism of Lie algebras? Assume that $\mathcal{A}$ is a Lie $\mathbb{C}$-algebra. Also denote the Lie $\mathbb{C}$-algebra $\mathcal{D}=\mathbb{C}[x_1,\ldots, x_n]\partial_{x_1}\oplus\ldots\oplus\mathbb{C}[x_1,\ldots, x_n]\partial_{x_n}$. My questions are as follows
Let $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$, $\mathcal{D}\otimes_{\mathbb{C}}\mathbb{C}(t)$ are isomorphic $\mathbb{C}(t)$-Lie algebras.

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*Is it true that $\mathcal{A}, \mathcal{D}$ are isomorphic $\mathbb{C}$-Lie algebras?


*Is it true that $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}[t], \mathcal{D}\otimes_{\mathbb{C}}\mathbb{C}[t]$ are isomorphic $\mathbb{C}[t]$-Lie algebras?
 A: Put $\mathcal{A}'=\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$ and similarly for $\mathcal{D}'$.  Choose an isomorphism $f\colon\mathcal{D}'\to\mathcal{A'}$.
Choose a (countable) basis $\mathcal{D}_0=\{d_i:i\in\mathbb{N}\}$ for $\mathcal{D}$ over $\mathbb{C}$.  Then $f(\mathcal{D}_0)$ is a basis for $\mathcal{A}'$ over $\mathbb{C}(t)$ but $\dim_{\mathbb{C}}(\mathcal{A}')=\dim_{\mathbb{C}}(\mathcal{A})$ so we can also choose a countable basis $\mathcal{A}_0=\{a_i:i\in\mathbb{N}\}$ for $\mathcal{A}$ over $\mathcal{C}$.  We must have $[d_i,d_j]=\sum_kp_{ijk}d_k$ and $[a_i,a_j]=\sum_kq_{ijk}a_k$ for some structure constants $p_{ijk},q_{ijk}\in\mathbb{C}$.  Now let $K$ be a countable subfield of $\mathbb{C}$ containing all these structure constants, and also all the constants needed to ensure that $f(\mathcal{A}_0)\subseteq K(t).\mathcal{D}_0$ and $f^{-1}(\mathcal{D}_0)\subseteq K(t).\mathcal{A}_0$.  Put $\mathcal{A}_1=K.\mathcal{A}_0$ and $\mathcal{D}_1=K.\mathcal{D}_0$, so these are Lie algebras over $K$ that become isomorphic over $K(t)$.  If $\mathbb{C}$ were algebraic over $K$ then it would be countable, which is false.  Thus, we can choose an embedding $i\colon K(t)\to\mathbb{C}$ extending the identity on $K$.  By applying this to the coefficients of $f$, we obtain an isomorphism $\mathcal{D}\simeq\mathcal{A}$.
