Lemma 15.46.3. Let $k \subset K$ be a field extension. Let $\{ K_\alpha \} _{\alpha \in A}$ be a collection of subfields of $K$ with the following properties

1. $k \subset K_\alpha$ for all $\alpha \in A$,

2. $k = \bigcap _{\alpha \in A} K_\alpha$,

3. for $\alpha , \alpha ' \in A$ there exists an $\alpha '' \in A$ such that $K_{\alpha ''} \subset K_\alpha \cap K_{\alpha '}$.

Then for $n \geq 1$ and $V \subset K^{\oplus n}$ a $K$-vector space we have $V \cap k^{\oplus n} \not= 0$ if and only if $V \cap K_\alpha ^{\oplus n} \not= 0$ for all $\alpha \in A$.

Proof. By induction on $n$. The case $n = 1$ follows from the assumptions. Assume the result proven for subspaces of $K^{\oplus n - 1}$. Assume that $V \subset K^{\oplus n}$ has nonzero intersection with $K_\alpha ^{\oplus n}$ for all $\alpha \in A$. If $V \cap 0 \oplus k^{\oplus n - 1}$ is nonzero then we win. Hence we may assume this is not the case. By induction hypothesis we can find an $\alpha$ such that $V \cap 0 \oplus K_\alpha ^{\oplus n - 1}$ is zero. Let $v = (x_1, \ldots , x_ n) \in V \cap K_\alpha$ be a nonzero element. By our choice of $\alpha$ we see that $x_1$ is not zero. Replace $v$ by $x_1^{-1}v$ so that $v = (1, x_2, \ldots , x_ n)$. Note that if $v' = (x_1', \ldots , x'_ n) \in V \cap K_\alpha$, then $v' - x_1'v = 0$ by our choice of $\alpha$. Hence we see that $V \cap K_\alpha ^{\oplus n} = K_\alpha v$. If we choose some $\alpha '$ such that $K_{\alpha '} \subset K_\alpha$, then we see that necessarily $v \in V \cap K_{\alpha '}^{\oplus n}$ (by the same arguments applied to $\alpha '$). Hence

$x_2, \ldots , x_ n \in \bigcap \nolimits _{\alpha ' \in A, K_{\alpha '} \subset K_\alpha } K_{\alpha '}$

which equals $k$ by (2) and (3). $\square$

There are also:

• 2 comment(s) on Section 15.46: Field extensions, revisited

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).