# Thread: Polynomial irreducible over an extension

1. ## Polynomial irreducible over an extension

Can someone verify whether or not this proof works for me? Thanks.

Suppose that $\displaystyle p(x) \in F[x]$ and $\displaystyle E$ is a finite extension of $\displaystyle F$. If $\displaystyle p(x)$ is irreducible over $\displaystyle F$ and $\displaystyle \mbox{deg}(p(x))$ and $\displaystyle [E:F]$ are relatively prime, show that $\displaystyle p(x)$ is irreducible over $\displaystyle E$.

Suppose $\displaystyle p(x)$ is reducible over $\displaystyle E$. Then $\displaystyle p(x)$ has a root in $\displaystyle E$, call it $\displaystyle a$ (this part feels true but I can't verify it nicely...basically if it's reducible but not by splitting out $\displaystyle x-a$, the coefficients will be from $\displaystyle F$, but then it would be reducible over $\displaystyle F$). Take the subextension $\displaystyle F(a)$. Then since $\displaystyle F(a) \cong F[x]/\langle p(x) \rangle$ we have $\displaystyle [F(a):F]=\mbox{deg}(p(x))$. Then $\displaystyle [E:F]=[E:F(a)][F(a):F]$ but this implies that $\displaystyle [F(a):F]$ divides $\displaystyle [E:F]$ which it does not since $\displaystyle [F(a):F]=\mbox{deg}(p(x))$ and $\displaystyle \mbox{deg}(p(x))$ and $\displaystyle [E:F]$ are relatively prime.

2. ## Re: Polynomial irreducible over an extension

Originally Posted by AlexP
Can someone verify whether or not this proof works for me? Thanks.

Suppose that $\displaystyle p(x) \in F[x]$ and $\displaystyle E$ is a finite extension of $\displaystyle F$. If $\displaystyle p(x)$ is irreducible over $\displaystyle F$ and $\displaystyle \mbox{deg}(p(x))$ and $\displaystyle [E:F]$ are relatively prime, show that $\displaystyle p(x)$ is irreducible over $\displaystyle E$.

Suppose $\displaystyle p(x)$ is reducible over $\displaystyle E$. Then $\displaystyle p(x)$ has a root in $\displaystyle E$, call it $\displaystyle a$ (this part feels true but I can't verify it nicely...basically if it's reducible but not by splitting out $\displaystyle x-a$, the coefficients will be from $\displaystyle F$, but then it would be reducible over $\displaystyle F$). Take the subextension $\displaystyle F(a)$. Then since $\displaystyle F(a) \cong F[x]/\langle p(x) \rangle$ we have $\displaystyle [F(a):F]=\mbox{deg}(p(x))$. Then $\displaystyle [E:F]=[E:F(a)][F(a):F]$ but this implies that $\displaystyle [F(a):F]$ divides $\displaystyle [E:F]$ which it does not since $\displaystyle [F(a):F]=\mbox{deg}(p(x))$ and $\displaystyle \mbox{deg}(p(x))$ and $\displaystyle [E:F]$ are relatively prime.
no, your proof is not complete because $\displaystyle p(x)$ may not have any root in $\displaystyle E$. you need to pick a root $\displaystyle a$ of $\displaystyle p(x)$ in some extension of $\displaystyle F$ and let $\displaystyle q(x)$ be the minimal polynomial of $\displaystyle a$ over $\displaystyle E$. clearly

$\displaystyle \deg q(x) \leq \deg p(x). \ \ \ \ \ \ \ (1)$

we also have

$\displaystyle [E(a):E)][E:F]=[E(a):F(a)][F(a):F]. \ \ \ \ \ \ (2)$

from $\displaystyle (2)$ and the problem hypothesis it follows that $\displaystyle \deg p(x) \mid \deg q(x)$ and so by $\displaystyle (1)$

$\displaystyle \deg p(x)=\deg q(x). \ \ \ \ \ \ (3)$

let $\displaystyle r(x)=q(x)-p(x) \in E[x].$ then $\displaystyle \deg r(x) < \deg q(x)$ by $\displaystyle (3).$ we also have $\displaystyle r(a)=0$ and hence $\displaystyle r(x)=0$ by the minimality of $\displaystyle q(x).$ hence $\displaystyle q(x)=p(x)$ and we are done.

3. ## Re: Polynomial irreducible over an extension

Thanks. If you don't mind, could you explain your approach/way of thinking about the problem? Often when I don't succeed in finding a proof and see it from someone else, I know I never would have thought of it. I'm really curious about your method.

4. ## Re: Polynomial irreducible over an extension

Originally Posted by AlexP
Thanks. If you don't mind, could you explain your approach/way of thinking about the problem? Often when I don't succeed in finding a proof and see it from someone else, I know I never would have thought of it. I'm really curious about your method.
here is how i found the solution:

i said $\displaystyle p(x)$ is irreducible over $\displaystyle E$ iff $\displaystyle p(x) = q(x)$. so i need to show that$\displaystyle r(x) = q(x) - p(x) = 0$. so, considering the facts that $\displaystyle r(x), q(x)$ are in $\displaystyle E[x]$ and $\displaystyle r(a)=q(a)=0$, i only need to show that $\displaystyle \deg r(x) < \deg q(x).$ but both $\displaystyle p(x)$ and $\displaystyle q(x)$ are monic and so i need to show that $\displaystyle \deg p(x)= \deg q(x)$ in order to prove $\displaystyle \deg r(x) < \deg q(x)$. finally, the identity $\displaystyle (2)$ is just a natural way of relating $\displaystyle \deg p(x)=[F(a):F]$ and $\displaystyle \deg q(x)=[E(a):E].$

5. ## Re: Polynomial irreducible over an extension

Is p(x) monic because otherwise it would be reducible (every nonzero element in a field is a unit)?

no, multiplying a polynomial by a non-zero scalar doesn't change its reducibility or irreducibility. so from the beginning you may assume that $\displaystyle p(x)$ is monic. another way is to let $\displaystyle r(x)=cq(x)-p(x),$ where $\displaystyle c$ is the leading coefficient of $\displaystyle p(x)$.