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

  • Thread starter nasshi
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In summary, the conversation is about using Einsenstein's Criterion for irreducibility in polynomials. The example given in Dummit & Foote's textbook states that if f(x) is converted into g(x)=f(x+1), and g(x) is irreducible, then f(x) must also be irreducible. The question is asked for clarification on the factors being referred to in the example. The response is that the factors are referring to f(x) if it were factorable, and an example of this technique when a polynomial is reducible is requested. An example is provided, where if g(x) is irreducible and defined as g(x)=f(x^{2}+45x-2), then f(x
  • #1
nasshi
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This is for clarification of a method.

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

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

My question is, "In each of the factors of what?". [itex]f(x)[/itex] if it were factorable? In [itex]g(x)[/itex] since [itex]f(x)[/itex] 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.
 
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  • #2
If [itex]f(z)[/itex] is reducible, then [itex]f(z) = p(z)q(z)[/itex] These are the factors the book is talking about.
 
  • #3
So as an example, if defining [itex]g(x)=f(x^{2}+45x-2)[/itex] and Eisenstein's criterion showed that [itex]g(x)[/itex] is irreducible, then [itex]f(x)[/itex] is irreducible? Or can I only use linear factors such as [itex]g(x)=f(x-2)[/itex]?
 
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  • #4
I think you could prove a more general statement, but linear polynomials are obvious.
 

1. How does using [itex]g(x)=f(x+1)[/itex] help to show irreducibility?

Using this trick allows us to shift the input of the polynomial by 1, which can reveal any hidden factors or symmetries that were not apparent before. If the polynomial remains irreducible after this shift, then it is truly irreducible.

2. Can this trick be used for any polynomial?

Yes, this trick can be used for any polynomial with integer coefficients. It is particularly useful for polynomials that are difficult to factor or have a high degree.

3. How do I apply this trick in a proof of irreducibility?

To apply this trick, simply substitute [itex]x+1[/itex] for [itex]x[/itex] in the polynomial. Then, proceed with your proof as usual, keeping in mind that the polynomial is now in terms of [itex]x+1[/itex] instead of just [itex]x[/itex].

4. Can I use this trick for proving reducibility?

No, this trick is specifically designed for proving irreducibility. If the polynomial is reducible after using this trick, it does not necessarily mean that the original polynomial is also reducible.

5. Are there any other methods for showing irreducibility?

Yes, there are other methods for showing irreducibility, such as using the rational root theorem or Eisenstein's criterion. However, using [itex]g(x)=f(x+1)[/itex] is a common and effective method in many cases.

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