# Thread: Irreducible polynomials over Q

1. ## 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!

2. 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 $p(x) \in R[x]$ 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

3. 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.

4. 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

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

multiply by $p!$ to get

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

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

Note that all of the coeffeints are integers so now this is eisenstien in p so it is irreducable.

5. TheEmptySet's solution needs a little bit fixing:

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

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

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

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

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

7. It seems that this is also true if $p$ is any positive integer, not just prime.

8. 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 $p^2$ does not divide the constant term. This is why it is critical that p be prime.

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