finding the multiplicative inverse in a field Let $L/K$ be a field extension. Let $a\in L$ and $K[a]=\{p(a)\;|\; p\in K[x]\}$; then $K[a]$ is clearly an integral domain. I want to show that when $a$ is algebraic over $K$, then $K[a]$ is a field. 
So I want to know what is the multiplicative inverse of a non zero element  $p(a)\not =0$ for some $p\in K[X]$. Since $a$ is algebraic it has a minimum polynomial $m\in K[X]$ now $m$ and $p$ are coprime so there exists $\alpha$ and $\beta$ in $K[X]$ such that $\alpha m+\beta p=1$; passing to evaluation on $a$ we get $\beta(a).p(a)=1$ hence $p(a)$ is invertible with inverse $\beta(a)$. My only doubt is about $m$ and $p$ being coprime. I think a justification is that $p$ is irreducible in $K[X]$ and that $p$ can not be a multiple of $m$ is that correct? 
 A: To put the comments in a full answer.
two ways are presented here to answer why $K[a]$ is a field for a given  algebraic element $a$.
The direct way : Since $K[a]$ is clearly an integral domain, it is enough to show that any non zero element $p(a)\not = 0$ has a multiplicative inverse. Now since 
$a$ is an algebraic element then it has a minimal polynomial $m$. Now it is clear that $m$ and $p$ are coprime since $m$ is irreducible and $p$ can not be a multiple of $m$ because $p(a)\not = 0$. It follows that there exist $\alpha $ and $\beta$ in $K[X]$ such that $\alpha .m+\beta . p=1$. Now evaluate this identity on $a$ to get $\beta(a) . p(a)=1$ hence $p(a)$ is invertible with inverse $\beta(a)$.
The indirect way : consider the evaluation map $$ev_a:K[X]\rightarrow L;\; p\mapsto p(a)$$
the map $ev_a$ is clearly a ring homomorphism with image $K[a]$ and kernel the principal ideal $(m)$ generated by the minimal polynomial $m$ which is a maximal ideal because $m$ is irreducible, therefore we have a field isomorphism $k[x]/(m(x)) \cong k[a]$
