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

  • Thread starter nasshi
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  • #1
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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.
 

Answers and Replies

  • #2
6,054
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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
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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
6,054
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I think you could prove a more general statement, but linear polynomials are obvious.
 

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