tensor product of modules over tensor product

Question

Let $K$ and $L$ be fields of characteristic zero such that $K \hookrightarrow L$. Let us denote the tensor product $L \otimes_K L$ by $L_1 \otimes_K L_2$. Assume that $M_1, M_2$ are vector spaces over $L_1, L_2$, respectively. Assume further that $N$ is a (faithful) module over $L_1 \otimes_K L_2$. Then $N$ is a module over both $L_1$ and $L_2$ through the natural maps $L_i \hookrightarrow L_1 \otimes_K L_2$. Is it the case that $$M_1 \otimes_{L_1} (M_2 \otimes_{L_2} N) = (M_1 \otimes_K M_2) \otimes_{L_1 \otimes_K L_2} N?$$ I found a similar result for the case when $N$ is itself a tensor product of an $L_1$- and an $L_2$-module, but I don't necessarily want to assume this.

Answer 1

Yes, this is true, even if $N$ is not faithful and you can also take $L_1\neq L_2$ Here's an abstract-nonsense argument. Tensor products commute with direct sums and every vector space is a direct sum of copies of the base field. Hence it suffices to check this for $M_1=L_1, M_2=L_2$, but then both sides are just isomorphic to $N$.