Equivalence of Irreducibility Tests for Polynomials Under Variable Substitution

[Simfish]

So if we have, say, a polynomial f(x) = a_n x^n + ... + a_0 and, say, use the substitution x = y + a, then do ALL irreducibility tests work the same? And do all OTHER tests also work the same? Is the polynomial FUNDAMENTALLY the same? And what theorem is there to prove this?

 

[John Creighto]

So if we have, say, a polynomial f(x) = a_n x^n + ... + a_0 and, say, use the substitution x = y + a, then do ALL irreducibility tests work the same? And do all OTHER tests also work the same? Is the polynomial FUNDAMENTALLY the same? And what theorem is there to prove this?

It is the same in terms of x. Of course the polynomial in terms of y is a translation of the polynomial which was in terms of x.

 

[Simfish]

Okay. It's a translation. Would the translation work even if you try x = a^n y^n + ... + a_0? And are there theorems to prove this?

 

[morphism]

Are you asking if the irreducibility of f(x) implies that of f(x+a)? Well suppose f(x+a) factors nontrivially into g(x)h(x), then what can we say about f(x)=f(x+a-a)?

If you want a more 'high level' explanation of this, and if your coefficients come from a field F, then consider the isomorphism F[x] -> F[x] given by x->x+a. This induces an isomorphism F[x]/<f(x)> =~ F[x]/<f(x+a)>. f(x) is irreducible iff F[x]/<f(x)> is a field iff f(x+a) is irreducible.