Irreducibility of polynomial (need proof evaluation)

[Avatrin]

One thing I have seen several times when trying to show that a polynomial $p(x)$ is irreducible over a field $F$ is that instead of showing that $p(x)$ is irreducible, I am supposed to show that $p(ax + b)$ is irreducible $a,b\in F$. This is supposedly equivalent. That does make sense, and I have a logical explanation for it. However, many times I've seen that my logic is wrong. So, I need my explanation evaluated.

Proposition: $p(x)$ is irreducible iff $p(x+b)$ is irreducible.

In the algebraic closure of F, $p(x)$ can be written as product of linear factors:

$p(x) = \prod_{i \in I} (x-x_i) \textrm{ where } x_i \in F' \textrm{ for all }i$

That means $p(x+b) = \prod_{i \in I} (x+b-x_i)$

So, the roots of $p(x+b)$ are $-b+x_i$ for $i \in I$ , and since $b$ is in the field, these are in the field if $x_i$ are in the field. Since b can be negative, this proof works both ways.

If $p(x)$ is irreducible, one of the linear factors will have an $x_i$ that is not in the field. Therefore, $x_i-b$ will not be in the field either. Thus, $p(x+b)$ is irreducible if $p(x)$ is irreducible.

Proposition: $p(x)$ is irreducible iff $p(ax)$ is irreducible.

$p(x) = \prod_{i = 0}^n (x-x_i)$

That means $p(ax) = \prod_{i = 0}^n (ax-x_i) = a^{n+1}\prod_{i = 0}^n \left(x-\dfrac{x_i}{a}\right)$ . So, the roots of $p(ax)$ are $\dfrac{x_i}{a}$ for $i \in I$ . If one of these $x_i$ 's is not in the field, neither is $\dfrac{x_i}{a}$ . Again, this proof works both ways since fields contain multiplicative inverses.

The idea that $p(x)$ is only irreducible if $p(ax + b)$ is irreducible is a consequence of these two propositions.

 

[micromass]

So, the roots of $p(x+b)$ are $-b+x_i$ for $i \in I$ , and since $b$ is in the field, these are in the field if $x_i$ are in the field.

You seem to think that a polynomial is irreducible if their roots are not in the field. But consider the following polynomial over $\mathbb{Q}$: $(x^2+1)^2$. These roots are not in the field, but the polynomial is reducible anyway. And for the converse, consider the polynomial $x-2$ in $\mathbb{Q}$, these have roots but are irreducible.

 

[Avatrin]

That is true... Let's see... I'll try anew:

Proposition: p(x) is irreducible iff p(x+b) is irreducible.

I can try a contrapositive proof. So, if p(x+b) is reducible, so is p(x). In fact, since b is a element in a field, any proof will work both ways; If p(x) is reducible, so is p(x+b).

So, if $p(x) = \prod q_i(x)$, then $p(x+b) = \prod q_i (x+b)$

Thus, p(x+b) is irreducible iff p(x) is irreducible

The same argument can be used for p(ax), by replacing p(x+b) with p(ax).

By employing contraposition, this became worryingly easy. Is this correct? The thing that is making me really uncertain is that my understanding of fields and substitution tells me that proving if will automatically lead to iff. However, while going from $p(x) = \prod q_i(x)$ to $p(x+b) = \prod q_i (x+b)$, is easy; Going the other direction seems much worse ($p(x+b) = \prod q_i (x)$ to $p(x) = \prod r_i (x)$).

 

[micromass]

That is correct. Notice that the if direction immediately proves the iff because you can do x-b and x/a (like you already noticed in your first post).

 

[mathwonk]

the point is that sending x to ax+b is an isomorphism of the polynomial ring (if a ≠0). try proving that.

 

[Avatrin]

the point is that sending x to ax+b is an isomorphism of the polynomial ring (if a ≠0). try proving that.

But, why does a homomorphism with a one-to-one correspondence prove irreducibility?