# Irreducible polynomials over Q

• May 18th 2009, 06:05 PM
curiousmuch
Irreducible polynomials over Q
Let p be any prime number. Show that

1 + x + x^2/2! + x^3/3! ... + x^p/p! is irreducible over Q.

Okay for this one, I know that if the polynomial were degree 2 or 3 I could simply check whether it has roots, but I want to generalize to any prime p.

I would love to use Eisenstein Criterion, but because of the polynomial is not in Z[x] I cannot. I have spent a long time trying to convert the polynomial into a form that is usable. I came across an equation that said x^p -1/x-1 = 1+x+...+x^p-2+x^p-1. I have had no success so far.

I would really appreciate the help!
• May 18th 2009, 06:14 PM
TheEmptySet
Quote:

Originally Posted by curiousmuch
Let p be any prime number. Show that

1 + x + x^2/2! + x^3/3! ... + x^p/p! is irreducible over Q.

Okay for this one, I know that if the polynomial were degree 2 or 3 I could simply check whether it has roots, but I want to generalize to any prime p.

I would love to use Eisenstein Criterion, but because of the polynomial is not in Z[x] I cannot. I have spent a long time trying to convert the polynomial into a form that is usable. I came across an equation that said x^p -1/x-1 = 1+x+...+x^p-2+x^p-1. I have had no success so far.

I would really appreciate the help!

Actually you can use Eisenstien, but you need to fix it up first.

Gauss's lemma is the key that you need.

Let R be a Unique factorization domain with field of fractions F and let $\displaystyle p(x) \in R[x]$ if p(x) is reducible in F[x] then p(x) is reducible in R[x].

What you actually want is the contraposative.

if p(x) is irreducible in R[x] then it is irreducable in F[x]

So in your case R is the integers and F is the rationals.

You should be able to finish from here

Happy hunting :)
• May 18th 2009, 06:39 PM
curiousmuch
Wow that is so useful! (we actually have never learned Gauss's lemma in class). Sorry but I'm still a bit confused how to look at the polynomial in Z[x]. Thanks a lot.
• May 18th 2009, 06:45 PM
TheEmptySet
Quote:

Originally Posted by curiousmuch
Wow that is so useful! (we actually have never learned Gauss's lemma in class). Sorry but I'm still a bit confused how to look at the polynomial in Z[x]. Thanks a lot.

so you have

$\displaystyle 1+x+...+\frac{x^p}{p!}=0$

multiply by $\displaystyle p!$ to get

$\displaystyle p! +p!x+\frac{p!}{2}x^2+...+\frac{P!}{(p-1)!}x^{p-1}+\frac{p!}{p!}x^p=0$

$\displaystyle p! +p!x+\frac{p!}{2}x^2+...+px^{p-1}+x^p=0$

Note that all of the coeffeints are integers so now this is eisenstien in p so it is irreducable.
• May 18th 2009, 07:12 PM
NonCommAlg
TheEmptySet's solution needs a little bit fixing:

let $\displaystyle f(x)=1+x + \cdots + \frac{x^{p-1}}{(p-1)!} + \frac{x^p}{p!}.$ suppose $\displaystyle f(x)$ was reducible over $\displaystyle \mathbb{Q}.$ then $\displaystyle p!f(x)$ would also be reducible over $\displaystyle \mathbb{Q}$ too. but by Eisenstein's criterion $\displaystyle p!f(x)$ is irreducible over $\displaystyle \mathbb{Z},$

and therefore irreducible over $\displaystyle \mathbb{Q}$ as well, by Gauss's lemma. contradiction!
• May 18th 2009, 07:14 PM
TheEmptySet
Quote:

Originally Posted by NonCommAlg
TheEmptySet's solution needs a little bit fixing:

let $\displaystyle f(x)=1+x + \cdots + \frac{x^{p-1}}{(p-1)!} + \frac{x^p}{p!}.$ suppose $\displaystyle f(x)$ was reducible over $\displaystyle \mathbb{Q}.$ then $\displaystyle p!f(x)$ would also be reducible over $\displaystyle \mathbb{Q}$ too. but by Eisenstein's criterion $\displaystyle p!f(x)$ is irreducible over $\displaystyle \mathbb{Z},$

and therefore irreducible over $\displaystyle \mathbb{Q}$ as well, by Gauss's lemma. contradiction!

Thanks
• May 23rd 2009, 05:56 PM
ThePerfectHacker
It seems that this is also true if $\displaystyle p$ is any positive integer, not just prime. (Surprised)
• May 23rd 2009, 10:18 PM
Gamma
To use Eisentstein's criterion it needs to be the case that your polynomial is monic (although a stronger condition can be shown to work so long as p does not divide the leading term) and that p divides every other term, but $\displaystyle p^2$ does not divide the constant term. This is why it is critical that p be prime.

If p = 4 for instance you would have $\displaystyle 4!p(x)=24 + 24x + 12x^2 + 4x^3 + x^4$, so look at the $\displaystyle x^3$ coefficient, the only prime dividing 4 is 2, but $\displaystyle 2^2=4|24$ so Eisenstein's criterion fails and this method no longer works.