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

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 $$\mathbb F (\alpha)$$?

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

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

1. 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.

• Thank you so much! Just some follow-up questions: 3. So starting with arbitrary $Q\in \mathbb F[x] / (m_\alpha)$ and writing $Q=R+P \cdot m_\alpha$ is essentially the same as starting with $P \in \mathbb F [x]$ and using remainder division of $P$ with $m_\alpha$ to get $R$? 4. Regarding the point that $B$ spans $\mathbb F (\alpha)$, just making sure: For all $a \in \mathbb F (\alpha)$ there exists $P \in \mathbb F [x] / (m_\alpha)$ such that $$a=P(\alpha)+0=\sum^{n-1}_{i=0}a_i \alpha^i$$ Hence $a \in \text{Span}(1,\alpha,\dots,\alpha^{n-1})$? (Btw you get the other phi using \varphi) Commented May 24 at 12:42
• (Also, the abuse of notation really is confusing - is $a=P(a)+0$ a legitimate expression? How do you "plug in" into an element of $\mathbb F [x] / (m_\alpha)$?) Commented May 24 at 12:44
• Yes, it is essentially the same. If $Q$ is an arbitrary element of $\mathbb{F}[x]/(m_{\alpha})$ then we have $Q = P + (m_\alpha)$ for some $P \in \mathbb{F}[x]$ and conversely any $P \in \mathbb{F}[x]$ determines an element $Q := P + (m_{\alpha}) \in \mathbb{F}[x]/(m_{\alpha})$. The remainder division is just used to write $P+(m_\alpha)=R+(m_\alpha)$ with deg$R <$ deg$m_{\alpha}$. For the point about $B$ spanning what you say is true, in my answer I was trying to say that $B$ is the isomorphic image of a spanning set, so it is also a spanning set (because we proved that for $\overline{B}$) Commented May 24 at 12:54
• We plug in the same way we do for elements of $\mathbb{F}[x]$. If $Q = P + (m_\alpha) \in \mathbb{F}[x]/(m_\alpha)$ then we "plug in" to P, i.e. $\varphi_{\alpha}(Q) = P(\alpha)$. The fact that $\alpha$ is a root of $m_\alpha$ guarantees that this is well-defined i.e if $P+(m_\alpha) = P' + (m_\alpha)$ then $P(\alpha) = P'(\alpha)$. $(P(\alpha) + 0$ is a legitimate expression it's just that you will usually not write "$+0$", just as you wouldn't normally write something like "$2 = 1+1+0$" even though it is legitimate). Commented May 24 at 13:01