# Irreducible polynomial

• Feb 14th 2009, 06:36 AM
HTale
Irreducible polynomial
I need to prove the following:

"Prove that $\displaystyle f(X)$ is irreducible in $\displaystyle \mathbb{Q}[X]$ if and only if $\displaystyle f(\alpha X + \beta)$ is irreducible in $\displaystyle \mathbb{Q}[X]$"

I want to prove "if"; that is, $\displaystyle f(\alpha X + \beta)$ irreducible $\displaystyle \Rightarrow f(X)$ irreducible.

So, I think we have that $\displaystyle f(X)= a(X)b(X)$, with one of $\displaystyle \deg(a), \deg(b) = \deg(f)$.

The hint states that we should argue by contradiction. So, in this instance, we shall assume that $\displaystyle f(\alpha X + \beta)$ is reducible, and hence

$\displaystyle f(\alpha X + \beta) = g(\alpha X + \beta)h(\alpha X + \beta)$

where $\displaystyle \deg(g), \deg(h) < \deg(f)$

Now,

$\displaystyle f(\alpha X + \beta) \equiv [f(\alpha X + \beta) - f(X)] + f(X)$

and so,

$\displaystyle [f(\alpha X + \beta) - f(X)] + f(X) = g(\alpha X + \beta)h(\alpha X + \beta)$

we say, $\displaystyle f(\alpha X + \beta) - f(X)= \tilde{f}(\alpha X + \beta)$, and so

$\displaystyle f(X) = g(\alpha X + \beta)h(\alpha X + \beta) - \tilde{f}(\alpha X + \beta)$

with $\displaystyle \deg(\tilde{f}) < \deg(g)$, by the division algorithm. But, this is a contradiction, and so $\displaystyle \tilde{f}(\alpha X + \beta) = 0$. Thus,

$\displaystyle f(X) = g(\alpha X + \beta)h(\alpha X + \beta) = a(X)b(X)$

which contradicts the fact that $\displaystyle \mathbb{Q}[X]$ is a unique factorisation domain.

I think this is wrong. I'd be really grateful if anyone could pick out the flaws in this proof, and if possible, provide a better (or, correct) proof.

HTale.

• Feb 14th 2009, 07:30 AM
GaloisTheory1
Quote:

Originally Posted by HTale
I need to prove the following:

"Prove that $\displaystyle f(X)$ is irreducible in $\displaystyle \mathbb{Q}[X]$ if and only if $\displaystyle f(\alpha X + \beta)$ is irreducible in $\displaystyle \mathbb{Q}[X]$"

I want to prove "only if"; that is, $\displaystyle f(\alpha X + \beta)$ irreducible $\displaystyle \Rightarrow f(X)$ irreducible.

minor error:

you are are trying to prove the "if" direction.
• Feb 14th 2009, 08:58 AM
HTale
Quote:

Originally Posted by GaloisTheory1

minor error:

you are are trying to prove the "if" direction.

Silly me! Error corrected; thanks for that. Are there any flaws in that proof, as far as you can see?
• Feb 14th 2009, 10:23 AM
ThePerfectHacker
First of all $\displaystyle \alpha \not = 0$! Now define $\displaystyle \phi: \mathbb{Q}[X]\to \mathbb{Q}[X]$ by $\displaystyle \phi (f(X)) = f(\alpha X+\beta)$. Since $\displaystyle \alpha\not = 0$ it means $\displaystyle \deg f = \deg \phi (f)$. Say that $\displaystyle f(X) = f_1(X)f_2(X)$ where $\displaystyle \deg f_1,\deg f_2 < \deg f$ and $\displaystyle f_1,f_2\not \in \mathbb{Q}$. This means, $\displaystyle \phi (f) = \phi (f_1)\phi (f_2)$ but $\displaystyle \deg \phi (f) = \deg (f)$, $\displaystyle \deg \phi (f_1) = \deg f_1$, $\displaystyle \deg \phi (f_2) = \deg f_2$. It follows that $\displaystyle \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f)$ and $\displaystyle \phi(f_1),\phi(f_2) \not \in \mathbb{Q}$, therefore $\displaystyle \phi (f) = f(\alpha X + \beta)$ is not irreducible. We have proved that if $\displaystyle f(X)$ is reducible then $\displaystyle f(\alpha X + \beta)$ is reducible, taking the contrapositive we have proven that if $\displaystyle f(\alpha X + \beta)$ is irreducible then $\displaystyle f(X)$ is irreducible. To prove the converse I am going to give you a hint (Nod). Define $\displaystyle \theta : \mathbb{Q}[X] \to \mathbb{Q}[X]$ by $\displaystyle \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right)$ (again $\displaystyle \alpha\not = 0$). Notice that $\displaystyle \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X)$.
• Feb 14th 2009, 10:58 AM
HTale
Quote:

Originally Posted by ThePerfectHacker
First of all $\displaystyle \alpha \not = 0$! Now define $\displaystyle \phi: \mathbb{Q}[X]\to \mathbb{Q}[X]$ by $\displaystyle \phi (f(X)) = f(\alpha X+\beta)$. Since $\displaystyle \alpha\not = 0$ it means $\displaystyle \deg f = \deg \phi (f)$. Say that $\displaystyle f(X) = f_1(X)f_2(X)$ where $\displaystyle \deg f_1,\deg f_2 < \deg f$ and $\displaystyle f_1,f_2\not \in \mathbb{Q}$. This means, $\displaystyle \phi (f) = \phi (f_1)\phi (f_2)$ but $\displaystyle \deg \phi (f) = \deg (f)$, $\displaystyle \deg \phi (f_1) = \deg f_1$, $\displaystyle \deg \phi (f_2) = \deg f_2$. It follows that $\displaystyle \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f)$ and $\displaystyle \phi(f_1),\phi(f_2) \not \in \mathbb{Q}$, therefore $\displaystyle \phi (f) = f(\alpha X + \beta)$ is not irreducible. We have proved that if $\displaystyle f(X)$ is reducible then $\displaystyle f(\alpha X + \beta)$ is reducible, taking the contrapositive we have proven that if $\displaystyle f(\alpha X + \beta)$ is irreducible then $\displaystyle f(X)$ is irreducible. To prove the converse I am going to give you a hint (Nod). Define $\displaystyle \theta : \mathbb{Q}[X] \to \mathbb{Q}[X]$ by $\displaystyle \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right)$ (again $\displaystyle \alpha\not = 0$). Notice that $\displaystyle \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X)$.

Thanks a lot for your help! When you say $\displaystyle f_1,f_2\not \in \mathbb{Q}$, are you saying that $\displaystyle f_1,f_2\not \in \mathbb{Q}[X]$? I'm not entirely sure what you are trying to say here.
• Feb 14th 2009, 11:17 AM
ThePerfectHacker
Quote:

Originally Posted by HTale
Thanks a lot for your help! When you say $\displaystyle f_1,f_2\not \in \mathbb{Q}$, are you saying that $\displaystyle f_1,f_2\not \in \mathbb{Q}[X]$? I'm not entirely sure what you are trying to say here.

A polynomial $\displaystyle h(X)$ is in $\displaystyle \mathbb{Q}$ if and only if $\displaystyle h(X)$ is konstant.
That is all I meant. (Smile)
• Feb 14th 2009, 11:20 AM
HTale
Quote:

Originally Posted by ThePerfectHacker
A polynomial $\displaystyle h(X)$ is in $\displaystyle \mathbb{Q}$ if and only if $\displaystyle h(X)$ is konstant.
That is all I meant. (Smile)

Aha, in other words, non-unit polynomials. Thank you once again! I will try to attempt the second part of the proof, and post something on here soon
• Feb 14th 2009, 11:37 AM
ThePerfectHacker
Quote:

Originally Posted by HTale
Aha, in other words, non-unit polynomials. Thank you once again! I will try to attempt the second part of the proof, and post something on here soon

After you finish posting your proof for the problem you been assigned here is a good application of this theorem. Let $\displaystyle p$ be a prime number and consider the polynomial $\displaystyle X^{p-1} + X^{p-2} + ... + X + 1$. This polynomial is always irreducible. See if you can prove this by $\displaystyle X\mapsto X+1$ (i.e. $\displaystyle \alpha = 1 \text{ and }\beta = 1$).
• Feb 14th 2009, 03:09 PM
HTale
Quote:

Originally Posted by ThePerfectHacker
After you finish posting your proof for the problem you been assigned here is a good application of this theorem. Let $\displaystyle p$ be a prime number and consider the polynomial $\displaystyle X^{p-1} + X^{p-2} + ... + X + 1$. This polynomial is always irreducible. See if you can prove this by $\displaystyle X\mapsto X+1$ (i.e. $\displaystyle \alpha = 1 \text{ and }\beta = 1$).

I think in this instance, we have that $\displaystyle f(X) = \frac{X^p - 1}{X-1}$, and so if we set $\displaystyle X\mapsto X+1$, we get

$\displaystyle \frac{1}{X}((X+1)^p -1) = \sum^{p-1}_{r=0} {p \choose r}X^{p-r-1}$

and I think you apply Eisensteins criterion to finish the proof.

That's just a sketch of what I'd do. If it's correct, then I'll fill in the details and post the other proof and this to this thread.

HTale
• Feb 14th 2009, 06:45 PM
ThePerfectHacker
Quote:

Originally Posted by HTale
I think in this instance, we have that $\displaystyle f(X) = \frac{X^p - 1}{X-1}$, and so if we set $\displaystyle X\mapsto X+1$, we get

$\displaystyle \frac{1}{X}((X+1)^p -1) = \sum^{p-1}_{r=0} {p \choose r}X^{p-r-1}$

and I think you apply Eisensteins criterion to finish the proof.

That's just a sketch of what I'd do. If it's correct, then I'll fill in the details and post the other proof and this to this thread.

HTale

Yes (Clapping). The only issue I will take with your proof is that you wrote $\displaystyle f(X) = \frac{X^p - 1}{X-1}$. This is not a geometric sequence, these are just polynomials (I mean we can add and multiply polynomials we cannot divide them). What I would do to be more formal is say that $\displaystyle f(X)(X-1) = X^p - 1$ - this can be easily seen by multiplying out the LHS. Now if you let $\displaystyle X\mapsto X+1$ then you get $\displaystyle Xf(X+1) = (X+1)^p - 1 = X\sum_{k=0}^{p-1}{{p}\choose {k+1}}X^k$. We can "cancel" the $\displaystyle X$ (I say "cancel" because we are not really dividing since there is no such thing as division of polynomials, however, it is as if we are canceling) to get $\displaystyle f(X+1) = \sum_{k=1}^{p-1} {{p}\choose {k+1}}X^k$. Now apply Eisenstein criterion by using the fact that $\displaystyle p$ divides $\displaystyle {p\choose {\ell}}$ if $\displaystyle 1 < \ell < p$.
• Feb 15th 2009, 03:29 AM
HTale
One other question. If I set, say, $\displaystyle \alpha = 1$, so that our polynomials are monic, could I conclude that this was the case for $\displaystyle \mathbb{Z}[X]$? I guess the maps you could express in a proof would be well defined.
• Feb 15th 2009, 07:59 AM
ThePerfectHacker
Quote:

Originally Posted by HTale
One other question. If I set, say, $\displaystyle \alpha = 1$, so that our polynomials are monic, could I conclude that this was the case for $\displaystyle \mathbb{Z}[X]$? I guess the maps you could express in a proof would be well defined.

If the polynomials have integer coefficients then you can define $\displaystyle X\mapsto \pm X + \beta$ to get a transformed polynomial. This transformed polynomial is irreducible if and only if the original polynomial was irreducible by following the exact same proof except $\displaystyle \mathbb{Q}$ is replaced by $\displaystyle \mathbb{Z}$. We need to avoid the case $\displaystyle |\alpha| > 1$ because we cannot define an inverse transform because we are in the integers.