Isomorphic field extensions have the same degree

Question

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.

Answer 1

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 $k_1$.

Answer 2

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 $k_2$.