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Let $K_1, K_2$ be subfields of $K$, let $k = K_1 \bigcap K_2$, let $V_1$ be a $K_1$-vector space, $V_2$ be a $K_2$-vector space, both of them subsets of a $K$-vector space $V$.

How can I compute a $k$-basis for the intersection $V_1 \bigcap V_2$?

I'm mainly interested in the case $K = \mathbb{C}(x,y)$, $V = K^N$, $K_1 = \mathbb{C}(x)$, $K_2 = \mathbb{C}(y)$, an explicit $K_1$-basis $b_1,\ldots,b_m$ is known for $V_1$, as well as an explicit $K_2$-basis $B_1,\ldots,B_n$ for $V_2$.

The problem looks so elementary that one might hope that there should be a nice and short procedure for this.

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  • $\begingroup$ How do you define the intersection $V_1 \cap V_2$? $\endgroup$
    – VorKir
    Commented Mar 11, 2017 at 1:08
  • $\begingroup$ $V_1$ and $V_2$ are subsets of a $K$-vector space $V$. The intersection is the set-theoretic intersection, and is also a $K_1 \bigcap K_2$ vector space. $\endgroup$
    – Mark
    Commented Mar 11, 2017 at 1:11
  • $\begingroup$ Probably I am not familiar with the notations. What is $\mathbb{C}(x,y)$? Is $\mathbb{C}(x,y)$ equal to $\mathbb{C}^2$? $\endgroup$
    – VorKir
    Commented Mar 11, 2017 at 1:18
  • $\begingroup$ $\mathbb{C}(x,y)$ is the field of all fractions $P/Q$ where $P,Q$ are polynomials in $x,y$ with coefficients in $\mathbb{C}$. $\endgroup$
    – Mark
    Commented Mar 11, 2017 at 1:48

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