1
$\begingroup$

Let $k_1 \subseteq k'_1$ and $k_2 \subseteq k_2'$ be field extensions. Suppose there is a field isomorphism $\phi: k'_1 \to k'_2$ where $\phi(k_1)=k_2$. Show that $[k_1':k_1]=[k'_2:k_2]$.

Now my first instinct was to try and show $k_1'$ is isomorphic to $k'_2$ as vector spaces but this is nonsense since the base fields aren't equal.

Therefore I want to somehow show that $k_1=k_2$ but I am not really sure this is even true.

I can't seem to understand why this ought to be true.

$\endgroup$
0
$\begingroup$

It may help that since $\phi(k_1)=k_2$, the restriction $\phi:k_1\to k_2$ is a field isomorphism, since $\phi$ is injective. Thus you can treat $k_2'$ as a vector space over $k_1$ as follows: if $x\in k_1$ and $y\in k_2'$, then $x\cdot y = \phi(x)y$. Using this it is possible to prove that they are isomorphic as vector spaces over the new base field $k_1$.

$\endgroup$
  • $\begingroup$ And I suppose I would have to also show that this new vector space has the same dimension as the old one? $\endgroup$ – thundergraduate Feb 24 at 12:43
  • 1
    $\begingroup$ @thundergraduate Yes. Essentially it's true that "$k_1=k_2$", but we have the formality of pushing isomorphisms around. Easy, but we have that small barrier. $\endgroup$ – Matt Samuel Feb 24 at 12:44
0
$\begingroup$

Let $1_{k_1},a_1,\dots, a_n$ a linear basis of $k_1'$ over $k_1$.
Show that $\phi(1_{k_1}),\phi(a_1),\dots, \phi(a_n)$ is a linear basis of $k_2'$ over $k_2$.

$\endgroup$
  • $\begingroup$ Does this work even when the spaces aren't finite dimensional? $\endgroup$ – thundergraduate Feb 24 at 16:10
  • $\begingroup$ Yes. Any set of $k_1$-linearly independent elements is sent to a $k_2$-linearly independent elements. $\endgroup$ – Berci Feb 24 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.