The polynomial is irreducible iff the condition is satisfied

[mathmari]

Hey!

I need some help at the following exercise:

Show that the polynomial $f(x)=x^n+1 \in \mathbb{Q}[x]$ is irreducible if and only if $n=2^k$ for some integer $k \geq 0$.

Could you give me some hints what I could do?? (Wondering)

 

[mathbalarka]

This is a fairly straightforward algebra exercise. Hint : show that

$$x^{2n+1} + 1 = (x + 1)\left ( x^{2n} - x^{2n-1} + \cdots - x + 1 \right)$$

 

[Euge]

Hey!

I need some help at the following exercise:

Show that the polynomial $f(x)=x^n+1 \in \mathbb{Q}[x]$ is irreducible if and only if $n=2^k$ for some integer $k \geq 0$.

Could you give me some hints what I could do?? (Wondering)

Hi mathmari,

Note that $n$ is of the form $2^k \ell$, where $k \ge 0$ and $\ell$ is odd. Thus $x^n + 1 = x^{2^k \ell} + 1 = (x^{2^k})^{\ell} + 1$. Hence, if $\ell > 1$, $x^{2^k} + 1$ is a proper factor of $x^n + 1$. In case $\ell = 1$, show that $(x + 1)^n + 1$ is irreducible by application of Eisenstein's criterion with $p = 2$. Show that this implies $x^n + 1$ is irreducible.

 

[mathbalarka]

Eh, Eisenstein's criterion is a bit too much. Some easy polynomial factorizations, Gauss lemma and a pack of popcorns does the trick.

 

[blue_raver22]

Sure! To prove this statement, we can use the fact that a polynomial $f(x) \in \mathbb{Q}[x]$ is irreducible if and only if it does not have any roots in $\mathbb{Q}$.

Hint 1: Consider the roots of $f(x)$ in an extension field of $\mathbb{Q}$.

Hint 2: Recall that $x^n+1$ can be factored as $(x+1)(x^{n-1}-x^{n-2}+...+x^2-x+1)$.

Hint 3: Use induction on $n$ to show that if $n=2^k$, then $f(x)$ does not have any roots in $\mathbb{Q}$.

Hint 4: For the converse, use the fact that if $f(x)$ is irreducible, then any factor of $f(x)$ must have degree $n$ or $0$.

Hope this helps!