1

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.

Thế Long Lê
  • 215

2 Answers2

2

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 $\,i$)

Bill Dubuque
  • 272,048
1

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 $.

Surb
  • 55,662
  • I am sorry. That is not my mean. $p(X)=X-\alpha$ if $[K(\alpha):K]=1$ and the other case is my problem. – Thế Long Lê Jun 27 '16 at 13:28
  • I mean that if $[K(\alpha):K]=a$ with $a>1$ then we might have$ 2$ monoic and irreducible polynomial with $\alpha$ being a root, is it right? – Thế Long Lê Jun 27 '16 at 13:32
  • If $\alpha $ is a root of $p(X)$ in $K$, then $\min(\alpha ,K)=X-a$ i.e. $P(X)=X-a$. If $[K(a):K]>1$, then $a$ is not a root of $p(X)$ in $K$, but is a root of $p(X)$ in $K(a)$. – Surb Jun 27 '16 at 13:47
  • I am so sorry. That was my mistake I mean $\alpha$ is root in $K(\alpha)$ – Thế Long Lê Jun 27 '16 at 13:56
  • $f(X) $is an irreducible polynomial with coefficient in $K$ and degree $n$. Suppose that $\alpha$ is a root of it ,$ \alpha$ is not in K; $p(X)=min(K,\alpha)$. I mean why$ n=deg f =[K(\alpha): K] $and is it $f(X)=p(X) $if f is monoic ? – Thế Long Lê Jun 27 '16 at 14:14