A commutative diagram for $F(\alpha) \cong F[x] / \langle p(x) \rangle$

Question

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:

1. What can we use to map from $F[x]/\langle p(x) \rangle$ to $\phi_\alpha(F[x])$?
2. 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.

Answer 1

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.

Answer 2

The explicit formula of the isomorphism $\tau: F[x]/\langle p(x)\rangle\overset{\cong}{\to} F(\alpha)$ is \begin{equation} g(x)+ \langle p(x)\rangle\mapsto g(\alpha). \end{equation}

You can directly check that it is an isomorphism. Be careful that we need to show the well-definedness. In fact, if $g_1(x)+ \langle p(x)\rangle=g_2(x)+ \langle p(x)\rangle$, then $p\mid g_1-g_2$. Therefore $g_1(\alpha)-g_2(\alpha)=0$, i.e. $g_1(\alpha)=g_2(\alpha)$, and this proves the well-definedness. That $\tau$ is bijective and is a ring homomorphism is easy.

Answer 3

I'm gonna convert my comment as an answer beacuse it addresses directly your second question.

How is it using the first isomorphism theorem to show that $F(\alpha)\cong\phi_\alpha(F[x])$?

Note that by definitition $\phi_\alpha(F[x]) = F[\alpha]$ and $F(\alpha)$ is the smallest field which contains both $F$ and $\alpha$ (so clearly it is always true that $F[\alpha]\subseteq F(\alpha)$). Since $p(x)$ is irreducible and $F[x]$ is a PID, $F[\alpha] \cong F[x]/ \langle p(x) \rangle$ is a field (containing both $F$ and $\alpha$), it follows that $F[\alpha]=F(\alpha)$.