Why does irr(a,F(b)) divide irr(a,F)?

Question

My textbook asserts, as part of a proof about totally inseparable extensions, the following:

Let F be a field Let a and b be elements of the algebraic closure of F such that they are separable over F irr(a,F(b)) divides irr(a,F)

I do not understand why this is the case.

I see that irr(a,F) consists of factors whose zeros are not in F, and that irr(a,F(b)) consists of factors whose zeros are not in F(a), which is a bigger field. So, each time we adjoin an element to the field under consideration, the irreducible polynomial of a given element over that field lowers in degree. It wouldn't surprise me if this corresponded to the elimination of one or more factors of that polynomial, but I don't see why this has to be the case. Can someone explain it to me?

Thank you in advance.

Answer 1

Hints:

$$\Bbb F\subset \Bbb F(b)\implies irr(a,\Bbb F)\,,\,irr(a,\Bbb F(b))\in\Bbb F(b)[x]$$

But $\;irr(a,\Bbb F(b))\;$ is the monic polynomial of minimal degree in $\;\Bbb F(b)[x]\;$ that has $\;a\;$ as a root and thus it divides any other polynomial in $\;\Bbb F(b)[x]\;$ that has $\;a\;$ as a root...