Theorem: Let $f\in k[x]$ be irreducible. There exists a simple algebraic extension $K/k$ with $[K:k]=deg(f)$, such that $f$ has a zero in $K$.
Well, I read the following lecture notes about how to proof this theorem but I did not understand.
Proof:This is a formal, almost tautological construction. We will use f to construct the field. As $f$ is irreducible and $k[x]$ is a PID, we know that $K:=k[x]/<f>$ is field. Let $\pi:k[x]\rightarrow K$ be the canonical projection. $\pi|_{k}$ is surjective and the image is a subfield of $K$(which we identify with $k$). Thus $K/k$ is a field extension. We claim that $f$ has a zero in $K$, namely the class [x]: Write $f:=\sum_{i=0}^n a_ix^i$ with $a_i\in k$ .Then $$0=[f]=\sum_{i=0}^n[a_i][x^i]=\sum_{i=0}^na_i[x]^i=f([x])$$ Thus $[x]$ is a zero of $f$ in $k$, and $K=k([x])$.
Finally we determine $[K:k]$. By dividing $f$ by its leading coefficient we can assume that $f$ is monic. So $f$ is an irreducible monic polynomial with $f([x])=0$ that is $f$ is the minimal ploynomialof [x] over $k$. Thus $[K:k]=deg(f).$\
Here comes my question: I didnt understand how could one derive the equation in the middle of the proof. I am not sure if I understand the proof correctly: $[f]$ is the additive identity in $K$ and hence it is zero. And $[x]$ is the coset of $<f>$, namely $x+<f>$. But how could one factor out the $i$the power from the square bracket. I guess I didnt understand correctly what the notation means in the equation and also didnt see clearly how can one get the equation.
Thank you.