# Isomorphic field extensions have the same degree

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.

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 $$k_1$$.
• @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. – Matt Samuel Feb 24 at 12:44
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 $$k_2$$.
• Yes. Any set of $k_1$-linearly independent elements is sent to a $k_2$-linearly independent elements. – Berci Feb 24 at 17:33