Problems understanding proof of a theorem within field theory. Theorem: if $K$ is any field and $m$ is any irreducible monic polynomial over $K$, then there exists an extension $K(\alpha):K$ such that $\alpha$ has minimum polynomial $m$ over $K$.
Proof: There is a natural monomoprhism $i:K \rightarrow K[t]$. Let $I$ be the ideal of $K[t]$ consisting of all multiples of $m$, let $S = K[t]/I$, and let $v$ be the natural homomorphism $K[t] \rightarrow S$. By some previous lemmas, $S$ is a field and the composite map $v\circ i$ is a monomorphism. Identify $K$ with its image $v(i(K))$, and let $\alpha = I + t$. Clearly $S = K(\alpha)$. Since $m \in I$ we have  $m(\alpha) = I$, and $I$ is the zero element of $S$. Since $m$ is irreducible and monic it must be the minimum polynomial of $\alpha$. For if $p$ is the minimum polynomial, then $p\vert m$. Therefore $p = m$.
My problems: 


*

*$m = 1$ is also irreducible monic polynomial, isn't it? If it is, then $v \circ i$ isn't injective, hence not a monomoprhism. 

*Clearly $S = K(\alpha)$. Since $m \in I$ we have  $m(\alpha) = I$, and $I$ is the zero element of $S$. - this whole part is not that clear for me.

 A: Regarding your first problem: You should exclude units from the definition of irreuducible, so $1$ is not irreducible. 
Regarding the second, it says that, since $m\in I$, the element $m$ is mapped to zero under the canonical map $\phi:K[t]\to S$. That follows because $I$ is the kernel of this map. Now, evaluating $m=\sum_{k=0}^d m_k t^k\in K[t]$ at the element $\alpha=\phi(t)\in S$ yields:
$$
m(\alpha) = \sum_{k=0}^d m_k \alpha^k = \sum_{k=0}^d m_k \phi(t)^k
 = \phi\left(\sum_{k=0}^d m_k t^k\right) = \phi(m) = 0.
$$
Let us show that $S=K(\alpha)$. First, we remark that  $S$ is a field extension of $K$ because $I$ is a maximal ideal and $S=K[t]/I$. Since $K(\alpha)\subseteq S$, we are left to show that $S\subseteq K(\alpha)$. In other words, we are left to prove that $S$ is generated as a field by $K$ and $\alpha=\phi(t)$. Let $\beta\in S$, then there is some $f\in K[t]$ with $\beta=\phi(f)$. Write $f=\sum_{k=0}^d f_k t^k$ with $f_k\in K$ for all $k$ and we get 
$$
\beta = \phi(f) = \phi\left(\sum_{k=0}^d f_k t^k\right) = \sum_{k=0}^d f_k \phi(t)^k = \sum_{k=0}^d f_k \alpha^k \in K(\alpha)
$$
As $\beta$ was arbitrary, this proves $S\subseteq K(\alpha)$.
