A field extension with same transcendental degree is algebraic

I am trying to do this exercise of Fulton's Algebraic Curves:

6.45 Let $C, C'$ be curves, $F$ a rational map from $C'$ to $C$. Prove: (a) Either $F$ is dominating, or $F$ is constant (i.e., for some $P\in C$, $F(Q)=P$, all $Q\in C'$). (b) If $F$ is dominating, then $k(C')$ is a finite algebraic extension of $\hat{F}(k(C))$.

To conclude the problem i have to show the following:

If $E/F$ is an extension of fields over $k$, such that $E$ and $F$ have the same transcendental degree(in this case 1) then $E/F$ is algebraic.

I would like a hint to prove that. Thank you.

• If $x\in E$ is not algebraic over $F$, then it is transcendental over $F$ and hence over $k$. Then the transcendental degree of $E$ should be larger than the transcendental degree of $F$. – Levent Mar 6 '18 at 14:49
• Thank you very much Levent, was simpler than I thought! – John Mar 6 '18 at 14:54