# Show that there are irreducible polynomials of every degree in $\mathbb{Q}[X]$

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 $n$.

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?

Thanks!

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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

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.

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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