Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is this problem that I would like to ask for any verification whether my answer is correct.

Edited: Thanks @andybenji.

Show that for any $n\ge1$, there exists an irreducible polynomial $f\in\mathbb{Q}[X]$ of degree $n$.

My answer:
For degree n=0, a non-zero constant is a unit, hence it is irreducible in $\mathbb{Q}[X]$.
For all $n\ge1$, $x^n+2$ satisfies Eisenstein's Criterion with p=2, therefore it is irreducible in $\mathbb{Q}[X]$.

I am particularly doubtful about the case of degree 0. Is it correct that a non-zero constant is irreducible in $\mathbb{Q}[X]$? I saw my friend's note which says there are no irreducible polynomials of degree 0. Which one is correct?


share|cite|improve this question
You've got it. The question was poorly stated. It should be "Show that there are irreducible polynomials of degree $n \geq 1$ in $\mathbb{Q}[x]$." – andybenji Jun 9 '13 at 22:17
@andybenji I guess you mean different thing. My question asked to show for all n, there are irreducible polynomials in Q[X]. Not there are irreducible polynomials for some degree n. Maybe that clarifies your statement. Correct me if I am wrong :) – user71346 Jun 9 '13 at 22:26
@user70346 Alas, my clarification wasn't very clear. A reformulation could be "Show that, for any $n \geq 1$, there exists an irreducible polynomial $f \in \mathbb{Q}[x]$ of degree $n$." – andybenji Jun 9 '13 at 22:29
@andybenji. Thanks for the reformulation. I have edited it in the post. – user71346 Jun 9 '13 at 22:34
The point of the reformulation was that it (intentionally) excludes constant (degree 0) polynomials. – andybenji Jun 9 '13 at 22:35
up vote 1 down vote accepted

As I stated in the comments, the question was unclear, and a possible restatement would be

Show that, for any $n \geq 1$, there exists an irreducible polynomial $f \in \mathbb{Q}[x]$ of degree $n$.

To address the actual question, units are not irreducible. The definition of irreducible states

An element $f \in A$ is called irreducible if $f$ is not zero and not a unit, and for any expression $gh = f$, either $g$ or $h$ is a unit.

So it would, technically, be correct to say there are no irreducible polynomials of degree 0 over a field.

share|cite|improve this answer
Oh yes, you are correct. I have overlooked such a critical thing in the definition. Thanks! – user71346 Jun 9 '13 at 22:36
Yes, it's an important part and helps you to avoid headaches later on :^) – andybenji Jun 9 '13 at 22:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.