# Substitution of variables

1. Feb 28, 2008

### 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?

Last edited: Feb 28, 2008
2. Feb 28, 2008

### John Creighto

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.

3. Feb 28, 2008

### 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?

4. Feb 28, 2008

### 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.

Last edited: Feb 28, 2008