Clarification on the proof that $[\mathbb F (\alpha) : \mathbb F] = \deg_{\mathbb F} \alpha$

Question

Let $\mathbb E / \mathbb F$ be an extension of fields, $\alpha \in \mathbb E$ algebraic over $\mathbb F$. Then $[\mathbb F (\alpha) : \mathbb F] = \deg_{\mathbb F} \alpha$.

I could use some help understanding the proof of the claim above. Here's the proof that I've seen in class:

Denote $m_{\alpha}=\sum^{n}_{i=0} a_i x^i$ where $m_{\alpha}$ is the minimal polynomial of $\alpha$, such that $a_n=1$ and $\deg m_{\alpha}=n$. We'll prove the set $B=\{1,\alpha,\dots,\alpha^{n-1}\}$ forms a basis for $\mathbb F (\alpha)$ over $\mathbb F$.

1. For linear independence, assume for sake of contradiction that $B$ is linear dependent. This implies that there exist $b_0,\dots,b_{n-1} \in \mathbb F$ not all zero such that $\sum_{i=0}^{n-1}b_i\alpha^i=0$. Denote $P=\sum_{i=0}^{n-1}b_i x^i$. So $P(\alpha)=0$ and $\deg P < n$, and so it's not possible that $P=m_\alpha \cdot Q$ for some $Q \neq 0$, in contradiction to that $P\in (m_\alpha)$.
2. Define $\varphi_\alpha:\mathbb F[x]\to\mathbb E$ as $\varphi(P)=P(\alpha)$. This is a ring homomorphism. Observe that $B$ is the image under $\varphi_\alpha$ of $\overline B=\{1,\overline x,\dots ,\overline x^{n-1}\}\subset\mathbb F[x]/(m_\alpha)$, where $\overline x^i=x^i+(m_\alpha)$. We'll show $\overline B$ spans $\mathbb F[x]/(m_\alpha)$. Let $P\in\mathbb F[x]$. By remainder division of $P$ by $m_\alpha$, we get that $P=S\cdot m_\alpha+R$ such that $R=0$ or $\deg R < n$. If $R=\sum_{i=0}^{n-1}r_ix^i$, then $$P+(m_\alpha)=R+(m_\alpha)\in\text{Span}\overline B$$ and so $\overline B$ spans $\mathbb F[x]/(m_\alpha)$, therefore $B$ spans $\mathbb F(\alpha)$ over $\mathbb F$.

My questions are:

1. In the linear independence part of the proof, why is $P\in (m_\alpha)$ in the first place? Is it because $m_\alpha$ is a monic generator of the ideal of $\{P|P(\alpha)=0\}\lhd\mathbb F[x]$ and $P\in \{P|P(\alpha)=0\}$?
2. In the spanning part of the proof, why is $B$ the image under $\varphi_\alpha$ of $\overline B=\{1,\overline x,\dots ,\overline x^{n-1}\}\subset\mathbb F[x]/(m_\alpha)$?
3. Again in the spanning part of the proof, we want to show $\overline B$ spans $\mathbb F[x]/(m_\alpha)$, so shouldn't we start with an element of $F[x]/(m_\alpha)$ and show it's a linear comination of elements of $\overline B$?
4. Again in the spanning part of the proof, I can't wrap my head around the last three implications - why is $R+(m_\alpha) \in \text{Span}\overline B$? How does this imply that $\overline B$ spans $\mathbb F[x] /(m_\alpha)$? And lastly, how does this imply $B$ spans $\mathbb F (\alpha)$?

Answer 1

1. Yes, your answer is correct. The minimal polynomial $m_{\alpha}$ divides any polynomial which has $\alpha$ as a root. Since $P(\alpha)=0$ by assumption, $m_{\alpha}$ divides $P$ which means that $P \in (m_{\alpha})$.

2. Maybe you are confused because there is a slight abuse of notation. We are now viewing $\varphi_{\alpha}$ as a function from $\mathbb{F}[x]/(m_{\alpha})$, where now $\varphi_{\alpha}(P+(m_{\alpha})) = P(\alpha)$. This is well-defined because $\varphi_{\alpha}(m_{\alpha}) = m_{\alpha}(\alpha)=0$. If you are convinced that $\varphi_{\alpha}$ is still a ring homomorphism over $\mathbb{F}[x]/(m_{\alpha})$ then you only need to see that $\varphi_{\alpha}(1) = 1$ and $\varphi_{\alpha}(\overline{x}) = \alpha$.

3. This is what has been done. It might be clearer if you write

$$R+(m_\alpha)=\sum_{i=1}^{n-1} r_ix^i + (m_\alpha) = \sum_{i=1}^{n-1}r_i\left[x^i+(m_\alpha)\right]=\sum_{i=1}^{n-1}r_i\overline{x}^i.$$

4. The point above should show why $R + (m_\alpha) \in \text{Span}\overline{B}$. This shows that $\overline{B}$ spans $\mathbb{F}[x]/(m_{\alpha})$ because $P+(m_{\alpha})$ was an arbitrary element and it was shown that

$$P+(m_{\alpha}) = R+(m_{\alpha}) \in \text{Span}\overline{B}.$$ The last point follows from that fact that $\mathbb{F}[x]/(m_{\alpha}) \cong \mathbb{F}(\alpha)$ and $\varphi_{\alpha}(\overline{B})=B$. If you have not seen this isomorphism before, convince yourself that $(m_\alpha)$ is the kernel of $\varphi_{\alpha}$ and use the first isomorphism theorem for rings.