Irreducibility of a polynomial by eisenstein and substitution

[icantadd]

Homework Statement

Show 16x^4 = 8x^3 - 16x^2 - 8x + 1 is irreducible.

Homework Equations

Eisenstein's criteria, if there is n s.t. n does not divide the leading coefficient, divides all the other coefficients, and n^2 does not divide the last coefficient then the polynomial is irreducible (over the rationals)

The Attempt at a Solution

I want to say that consider p'(x) = p(\frac{1}{2}x) = x^4 + x^3 - 4x^2 - 4x + 1 is irreducible by eisenstein if we use the standard trick of substituting x+1 -> x, then we get, x^4+5x^3+5x^2-5x-5 where eisenstein is immediate. What I don't know is that if I can then say that since p' is irreducible, then p is.

 

[HallsofIvy]

Homework Statement

Show 16x^4 = 8x^3 - 16x^2 - 8x + 1 is irreducible.

This should be 16x^2+ 8x^4- 16x^2- 8x+ 1[/itex]<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>Homework Equations</h2><br /> <br /> Eisenstein&#039;s criteria, if there is n s.t. n does not divide the leading coefficient, divides all the other coefficients, and n^2 does not divide the last coefficient then the polynomial is irreducible (over the rationals)<br /> <br /> <h2>The Attempt at a Solution</h2><br /> I want to say that consider p&amp;#039;(x) = p(\frac{1}{2}x) = x^4 + x^3 - 4x^2 - 4x + 1 is irreducible by eisenstein if we use the standard trick of substituting x+1 -&gt; x, then we get, x^4+5x^3+5x^2-5x-5 where eisenstein is immediate. What I don&#039;t know is that if I can then say that since p&#039; is irreducible, then p is. </div> </div> </blockquote> p(x) is reducible if and only if p(x)= a(x)b(x) for polynomials a and b of lower degree than p. In that case p(x/2)= a(x/2)b(x/2) so, yes, if p(x) is reducible so is p(x/2). And, then, if p(x/2) is irreducible, so is p(x).

 

[icantadd]

First of all, thank you for your help!

That makes sense. I was stuck because I had a counter example that in general one cannot make substitutions; for example, p(u) = u^2 + 8u + 36 is irreducible over Q, but under the substitution, [t^2 -> u] p(t) splits into quadratic factors. So perhaps, is it possible to claim that any linear substitution over the single variable will be acceptable (i.e. [a*x + b -> x])?

Is there a theory that deals with substitutions in the abstract sense?