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

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

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

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

we also have

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

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

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

let $r(x)=q(x)-p(x) \in E[x].$ then $\deg r(x) < \deg q(x)$ by $(3).$ we also have $r(a)=0$ and hence $r(x)=0$ by the minimality of $q(x).$ hence $q(x)=p(x)$ 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 $p(x)$ is irreducible over $E$ iff $p(x) = q(x)$. so i need to show that $r(x) = q(x) - p(x) = 0$. so, considering the facts that $r(x), q(x)$ are in $E[x]$ and $r(a)=q(a)=0$, i only need to show that $\deg r(x) < \deg q(x).$ but both $p(x)$ and $q(x)$ are monic and so i need to show that $\deg p(x)= \deg q(x)$ in order to prove $\deg r(x) < \deg q(x)$. finally, the identity $(2)$ is just a natural way of relating $\deg p(x)=[F(a):F]$ and $\deg q(x)=[E(a):E].$

5. ## Re: Polynomial irreducible over an extension

Sorry this reply is late.

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

Thank you for your help.

6. ## Re: Polynomial irreducible over an extension

Originally Posted by AlexP
Sorry this reply is late.

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

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