Explaining the trick of using [itex]g(x)=f(x+1)[/itex] to show irreducibility

  • Thread starter Thread starter nasshi
  • Start date Start date
nasshi
Messages
18
Reaction score
0
This is for clarification of a method.

Dummit & Foote, pg 310, Example (3).

f(x)=x^{4}+1 is converted into g(x)=f(x+1) in order to use Einsenstein's Criterion for irreducibility. The example states "It follows that f(x) must also be irreducible, since any factorization of f(x) would provide a factorization of g(x) (just replace x by x+1 in each of the factors)."

My question is, "In each of the factors of what?". f(x) if it were factorable? In g(x) since f(x) was theoretically factorable by their explanation?

Please provide a more detailed explanation if possible. An example of this technique when a polynomial is reducible would be great. I was unable to create one since the wording has confused me.
 
Physics news on Phys.org
If f(z) is reducible, then f(z) = p(z)q(z) These are the factors the book is talking about.
 
So as an example, if defining g(x)=f(x^{2}+45x-2) and Eisenstein's criterion showed that g(x) is irreducible, then f(x) is irreducible? Or can I only use linear factors such as g(x)=f(x-2)?
 
Last edited:
I think you could prove a more general statement, but linear polynomials are obvious.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top