I've been trying to understand the following proposition:
Let $E$ be a field extension of $F$ and $\alpha \in E$ be algebraic over $F$. Then, $F(\alpha) \cong F[x]/\langle p(x) \rangle$ where $p(x)$ is the minimal polynomial of $\alpha$ over $F$
The book's proof states:
Let $\phi_\alpha: F[x] \rightarrow E $ be the evaluation homomorphism. The kernel of this map is $\langle p(x) \rangle$, where $p(x)$ is the minimal polynomial of $\alpha$. By the first isomorphism theorem for rings, the image of $\phi_\alpha$ in E is isomorphic to $F(\alpha)$ since it contains both $F$ and $\alpha$.
Here is what I understand:
- The kernel of $\phi_\alpha$ is $\langle p(x) \rangle$ since it consists of all multiples of $p(x)$
- We can use the first isomorphism theorem to state that there is an isomorphism $\phi_\alpha(F[x]) \cong F[x]/\langle p(x) \rangle$ since $\langle p(x) \rangle$ is the kernel of $\phi_\alpha$.
I've made a small diagram to help myself:
where $\Phi_{F[x]}$ is the canonical homomorphism from $F[x]$ to $F[x]/\langle p(x) \rangle$.
Now, here are the two things I don't get:
- What can we use to map from $F[x]/\langle p(x) \rangle$ to $\phi_\alpha(F[x])$?
- How is it using the first isomorphism theorem to show that $F(\alpha) \cong \phi_\alpha(F[x])$?
Ideally, I would like to get to a commutative diagram showing $F(\alpha) \cong F[x] / \langle p(x) \rangle$.
Thanks a lot for the help!
edits: removed $E = \phi_\alpha(F[x])$, updated question 2.
Best Answer
The map you are missing is nothing but the induced map from the Fundamental Theorem of Homomorphisms. Given a coset $g(x)+\langle p(x)\rangle \in F[x]/\langle p(x)\rangle$, the map sends it to $g(\alpha)$. It's the only map that could possibly fit, which you can see by starting with $g(x)$ in the upper left hand corner.
The fact that $p(\alpha)=0$ ensures the map is well defined. The fact that $p(x)$ is the minimal polynomial of $\alpha$ ensures it is an isomorphism onto its image.
Note that no one is claiming that $E$ equals the image. So your second point is a misunderstanding.