degree of quotient $K[X]/P_{\alpha}$ Let $k$ be a field and $K$ be a finite extension of $k$.
Let $\alpha \in K$ and let $P_{\alpha}$ be the minimal polynomial of $\alpha$ of degree $d$.
Let $f : k[X] → K$ be defined by $f(P)=P(\alpha)$
My course says:
1-  $f(k[X])$ is a subfield of $K$ that is isomorphic to $k[X]/<P_{\alpha}>$.
2 -$f(k[X])$ it is an extension of degree $d$
I could understand and prove the 1- but not the 2-.
$f(k[X])$ is an extension of which field and why its degree is $d$?
What I did:
I tried to understand the degree.
if $Q\in k[X]/<P_{\alpha}>$ we have $Q=AP_{\alpha}, A\in k[X]$, so the dimension of $k[X]/<P_{\alpha}>$ depends on the dimension of $k[X]$,not the degree of P_{\alpha} which doesn't make sense.
Could you please help me understand the 2- or point me to any useful links to understand this property?
Many thanks!
 A: Edit: everywhere below, a bar over a polynomial in $k[X]$ means its equivalence class in the quotient.
$k[X]/P_{\alpha}$ is an extension of $k$ since the morphism $f$ is injective when restricted to $k$ (you can either check this directly or note that it's true simply because $k$ is a field).
Now, since $P_{\alpha}$ is a polynomial of degree $d$, this implies that you can write $\alpha^d$ as a linear combination of $1,\alpha,\dots,\alpha^{d-1}$. Using this, you can prove that $(1,\overline{X}, \dots, \overline{X^{d-1}})$ is a basis of the $k$-vector space $k[X]/P_{\alpha}$ and thus the degree of the extension is $d$.
Hint: we know that $P_{\alpha}(\overline{X}) = \overline{0}$ in $k[X]/P_{\alpha}$ and that any element of $k[X]/P_{\alpha}$ is the class $\overline{Q}$ of some polynomial $Q\in k[X]$. Also, $\overline{X^d}$ is a linear combination of $1,\overline{X},\dots,\overline{X^{d-1}}$ thus by induction, one can prove that $\overline{X^k}$ is also a linear combination of $1,\overline{X},\dots,\overline{X^{d-1}}$ for all $k\geq d$.
