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Let $K/k$ be a field extension and A a finite type (finitely generated) k-module. Further, let $R := A \otimes_k k (\cong A)$ and $L:=A\otimes_kK$. Is L always a finite type R-module?

It is known that (1) $dim_{Krull}(L)=dim_{Krull}(R)(=dim_{Krull}(A))$. But my instincts tell me to be cautious here:

Take $a_1,..., a_m$ generators for A (as $k$-module). Then any $a\otimes\alpha \in R$ can be expressed as $k$-linear combination of the $a_1\otimes1,...,a_m\otimes1$, and since the $a\otimes\alpha$ generate $R$, one obtains finitely many generators for $R$ (as $k$-module). The trouble starts when $N:=[K:k]$ isn't finite:

$K$ can be viewed as $k^N$. But if $N$ isn't finite, I do not know how to find or how to see the existence of finitely many generators for $L$ as $R$-module.

Does anyone have a counter-example or an idea on how to demonstrate the assertion (should it actually hold)?

EDIT As suspected, the answer below shows that the statement doesn't hold. However, one has the following:

Using "Noether Normalisation", we obtain $k[X_1,...,X_n] \subset A$ is finite. This consequently tells us $k[X_1,...,X_n] \otimes_k K \rightarrow A \otimes_k K$ is injective and finite, thus yielding (1).

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No, not necessarily. Take for example $k = \mathbb{Q}$, $K = \mathbb{R}$, and let $R = A = \mathbb{Q}$, which is indeed a finitely generated $\mathbb{Q}$-module. Then $L = R \otimes_k K = \mathbb{R}$ is not a finitely generated $R$-module, because it contains an infinite algebraically independent subset, e.g. $\{ \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7} \dots \}$.

However, if $K$ is a finitely generated $k$-module, then this is true. Suppose that $\{\lambda_1, \dots, \lambda_d\}$ is a finite generating subset of $K$ as a $k$-module, i.e. every element of $K$ can be written as a polynomial in the $\lambda_i$ with coefficients in $k$. Then $\{ 1_R \otimes \lambda_1, \dots, 1_R \otimes \lambda_d \}$ is a generating subset of $L$ as an $R$-module.

Indeed every element $w$ of $L = R \otimes_k K$ can be written as a sum: $$w = \sum r_i \otimes x_i$$ with $r_i \in R$ and $x_i \in K$. Each $x_i$ can itself be written as $P_i(\lambda_1, \dots, \lambda_d)$, where $P_i \in k[X_1, \dots, X_d]$. It follows that $$w = \sum r_i \cdot (1_R \otimes P_i(\lambda_1, \dots, \lambda_d)) = (\sum r_i P_i)(1_R \otimes \lambda_1, \dots, 1_R \otimes \lambda_d)$$ is a polynomial in terms of the $1_R \otimes \lambda_i$ with coefficients in $R$.