# Transcendental elements over field extensions.

Let $E/F$ be a field extension, and suppose $a \in E$ is transcendental over $F$. I'm reading a proof that says $\dim_F F(a) \ge \dim_FF[x] = + \infty$ since the evaluation map $F[x] \to F [a]$, $p(x) \mapsto p(a)$ is one to one. However, I'm having trouble seeing why this is true.

My guess is that because $F[x]$ is isomorphic to a subspace of $F(a)$, the above claim is true. Some guidance would be appreciated.

## 2 Answers

If an element $p\in F[X]$ is in the kernel of the map, then $p(a)=0$, but $a$ is trascendental!

• Just think what it means for the element to be trascendental. – Mariano Suárez-Álvarez Feb 18 '13 at 5:11
• Well, you haven't said anything that I don't already know. I guess I didn't realize that the evaluation map is not just a ring homomorphism but a linear map. – user62761 Feb 18 '13 at 5:18
• I guessed you knew this. That often happens :-) – Mariano Suárez-Álvarez Feb 18 '13 at 5:29

The map $\psi: F[x] \rightarrow F[a]$ given by evaluation at $a$ is an isomorphism in fact. It's surjective by definition of $F[a]$ and the kernel is trivial, by virtue of transcendence of $a$. We notice that $\psi$ also fixes $F$ so $\dim_F F[x]=\dim_F F[a]$. Then we can pass this isomorphism up to the field of fractions of each ring $F(x)$ and $F(a)$. So finally we deduce that $\dim_F F(x)=\dim_F F(a)$.