- #1

Avatrin

- 245

- 6

Proposition: [itex]p(x)[/itex] is irreducible iff [itex]p(x+b)[/itex] is irreducible.

In the algebraic closure of F, [itex]p(x)[/itex] can be written as product of linear factors:

[itex]p(x) = \prod_{i \in I} (x-x_i) \textrm{ where } x_i \in F' \textrm{ for all }i[/itex]

That means [itex]p(x+b) = \prod_{i \in I} (x+b-x_i)[/itex]

So, the roots of [itex]p(x+b)[/itex] are [itex]-b+x_i[/itex] for [itex]i \in I[/itex] , and since [itex]b[/itex] is in the field, these are in the field if [itex]x_i[/itex] are in the field. Since b can be negative, this proof works both ways.

If [itex]p(x)[/itex] is irreducible, one of the linear factors will have an [itex]x_i[/itex] that is not in the field. Therefore, [itex]x_i-b[/itex] will not be in the field either. Thus, [itex]p(x+b)[/itex] is irreducible if [itex]p(x)[/itex] is irreducible.

Proposition: [itex]p(x)[/itex] is irreducible iff [itex]p(ax)[/itex] is irreducible.

[itex]p(x) = \prod_{i = 0}^n (x-x_i)[/itex]

That means [itex]p(ax) = \prod_{i = 0}^n (ax-x_i) = a^{n+1}\prod_{i = 0}^n \left(x-\dfrac{x_i}{a}\right)[/itex] . So, the roots of [itex]p(ax)[/itex] are [itex]\dfrac{x_i}{a}[/itex] for [itex]i \in I[/itex] . If one of these [itex]x_i[/itex] 's is not in the field, neither is [itex]\dfrac{x_i}{a}[/itex] . Again, this proof works both ways since fields contain multiplicative inverses.

The idea that [itex]p(x)[/itex] is only irreducible if [itex]p(ax + b)[/itex] is irreducible is a consequence of these two propositions.