If $p(X)$ is irreducible polynomial with $\alpha$ being a root in field K then $p(X)=min(K,\alpha)$, is it right? I think that my question is very stupid but I just wanna ask that:
If $p(X)$ is monoic and irreducible polynomial with coefficient in K and $\alpha$ being a root in field $K(\alpha)$ then $p(X)=min(K,\alpha)$, is it right? And I suppose that it is not right because $p(X)$ have to has minimal degree. I mean that it still has at least one more irreducible polynomial with $\alpha$ being a root.
 A: If $\,\alpha\,$ is an algebraic over $\,K\,$ with minimal polynomial $\,g\,$ then $\,f(\alpha) = 0\iff g\mid f,\,$ since the polynomials $\,\in K[x]\,$ having $\,\alpha\,$  as a root form an ideal $\,I.\,$ $\,K[x]\,$ is Euclidean $\Rightarrow$ PID, so $I$ is generated by any element $\,g\,$ of minimal degree - a "minimal" polynomial (for, if not, then $\,g\nmid f\,$ so $\,0\neq  f\ {\rm mod}\ g = f - q\,g\in I\,$ and it has smaller degree than $\,g,\,$ contra to minimality of $\,g).$ We can choose $\,g\,$ monic to make it unique.
Note that any irreducible $\,h\in I = (g)$ necessarily has minimal degree in $I$ (else $\,g\mid h\,$ properly, contra $\,h\,$ irreducible). For this reason a minimal polynomial for $\,\alpha\,$ is sometimes called an irreducible polynomial for $\,\alpha.$
The same argument works in any Euclidean domain: $ $ a nonzero ideal $I$ is generated by any element $i\in I$ of minimal Euclidean value (else $\ i\nmid j\in I\ \Rightarrow\ j\ {\rm mod}\ i\ =\ j-r\ i\in I\,$ and has smaller value than $\,i,\,$ contra minimality of $\,i$)
A: Let $K$ a field and $L/Q$ a field extension s.t. $\alpha \in L$. If $p(X)$ is irreducible and monoic, and that $p(\alpha )=0$, then $$p(X)=\min(\alpha ,K).$$
In what you wrote (i.e. $p(X)=\min(K,\alpha )$ and $p(\alpha )=0$ with $\alpha \in K$), you get $p(X)=X-\alpha $.
