1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How do you define the intersection $V_1 \cap V_2$? $\endgroup$ – VorKir Mar 11 '17 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 Mar 11 '17 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 Mar 11 '17 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 Mar 11 '17 at 1:48

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.