Question

Question:Let p(x) and q(x) be irreducible in F(x) and assume that deg p(x) is relatively prime to deg q(x) . Let u be a root of p(x) and v a root of q(x) in some extension field of F , prove that q(x) is irreducible over F(u) .

Short answer

q(x)is irreducible over F(u).

Step-by-step solution

Definition of irreducible polynomial.

Irreducible polynomial is a non constant polynomial that cannot be factored into the product of non constant polynomials.

Showing that D is a field.

Let k be an extension field of F and p(x) and q(x) be irreducible polynomial in F[x] . Then

$degp\left(x\right)·degq\left(x\right)=1$

Let u be the root of p(x) and v be the root of q(x) . That is

$p\left(u\right)=0,q\left(v\right)=0$

Consider $h\left(x\right)\in F\left(u\right)\left[x\right]$be the minimal plolynomial v of F(u) over .

As v is the root of q(x) . This implies

h(x) / q(x) . Then

$$\begin{gathered} degp\left(x\right),degq\left(x\right)=\left[F(u,v):F\right] \\ =\left[F(u,v):F\left(u\right)\right]\left[F\left(u\right):F\right] \\ =degh\left(x\right)·degq\left(x\right) \end{gathered}$$

Thus

$degh\left(x\right)·degq\left(u\right)$

And as$h\left(x\right)/q\left(x\right)$

Therefore

$q\left(x\right)=k·h\left(x\right)$

Now since h(x) is irreducible over F[u]. So,q(x) is also irreducible over F[u] .