Basis of a field extension Let $K$ be a field, and let $A$ be a $K$-algebra such that $\alpha \in A$. Then the natural homomorphism $$ \phi: K[x] \to K[\alpha], \hspace{3mm} (x \mapsto \alpha )$$ has a kernel which is a principal ideal $ \langle f \rangle$ and so $$ K[x] / \langle f \rangle \cong K[\alpha]$$
Notice that $K[\alpha]$ is a field. The book then states that, if $n=$ deg $f$ we have that $\{1, \alpha, \alpha^2, \dots, \alpha^{n-1} \}$ are a $K$-basis of $K[\alpha]$. 
I am not sure how to convince myself that this set is indeed a basis of $K[\alpha]$, how would I go about showing this? 
 A: Towards showing it spans, let's prove the easier proposition that $\alpha^n$ is in the span of $\{1, \alpha,  . . ., \alpha^{n-1}\}$. (It's easy to get the rest of the way from this.) Write $f=c_nx^n+ . . . + c_0x^0$, with $c_n\not=0$. Then consider the element $$z=-{c_{n-1}x^{n-1}+c_{n-2}x^{x-2}+ . . . +c_0x^0\over c_n}$$ What can you say about the image of $z$ in $K[\alpha]$?
Towards showing it is linearly independent, suppose I have a nontrivial linear combination of $1, . . . , \alpha^{n-1}$ which equals zero in $K[\alpha]$; can I use that to get a polynomial $g\in K[x]$ with degree $<n$, whose $f$-image is $0$? Why is this a problem?
A: Every $g \in K[x]$ can be written as $g=fq+r$, with $r=0$ or $\deg r<n$.
Since $fq\in \ker\phi$, we have $\phi(g)=\phi(r)$.
In other words, the image of $\phi$ is generated by the image of $\{1, x, x^2, \dots, x^{n-1} \}$, with of course is $\{1, \alpha, \alpha^2, \dots, \alpha^{n-1} \}$.
That $\{1, \alpha, \alpha^2, \dots, \alpha^{n-1} \}$ is linearly independent comes from the minimality of $n$, the degree of the generator of $\ker\phi$.
