Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$

Question

Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.

I've tried to use the cyclotomic polynomial as: $$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$

So I have that my polynomial is $$\frac{X^5-1}{X-1}$$ and now i have to prove that is irreducible.

The lineal change of variables are ok*(I don't know why) so I substitute $X$ by $X+1$ then I have: $$\frac{(X+1)^5-1}{X}=\frac{X^5+5X^4+10X^3+10X^2+5X}{X}=X^4+5X^3+10X^2+10X+5$$ And now we can apply the Eisenstein criterion with p=5. So my polynomial is irreducible in $\mathbb{Q}$

Now let's prove that it has two different irreducible factors in $\mathbb{R}$

I've tryed this way: $X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D)$ and solve the system. But solve the system is quite difficult. Is there another way?

Answer 1

Finding the complex roots of the polynomial is easy: if $\varphi=2\pi/5$, the roots are $$ r_1=e^{i\varphi},\quad r_2=e^{2i\varphi},\quad r_3=e^{3i\varphi}=\bar{r}_2\quad r_4=e^{4i\varphi}=\bar{r}_1 $$ and so the factorization over the reals is $$ (X^2-(r_1+\bar{r}_1)X+1)((X^2-(r_2+\bar{r}_2)X+1). $$ What you want to prove is that this factorization is not over $\mathbb{Q}$, thereby deducing that the polynomial is irreducible over $\mathbb{Q}$.

The procedure is standard: let $r$ be any root of the polynomial; then $$ r^2+r+1+\frac{1}{r}+\frac{1}{r^2}=0 $$ and so $$ \left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)-1=0 $$ Since the polynomial $X^2+X-1$ has no rational root, you have proved that $$ r_1+\bar{r}_1=r_1+\frac{1}{r_1} $$ is not rational.

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If $p$ is prime, then $X^{p-1}+X^{p-2}+\dots+X+1$ is irreducible over $\mathbb{Q}$. Write it as $$ \frac{X^p-1}{X-1} $$ and substitute $X=Y+1$. You'll see that Eisenstein applies.

Answer 2

let $$P(x)=x^4+x^3+x^2+x^1+1$$ We know if $x=\frac{a}{b}$ is root of $P(x)$ then $b|1\,$ , $\,a|1$. In the other words $a=\pm 1 $ and $b=\pm 1 $ but $P(1)=5$ and $P(-1)=1$, thus we let $$P(x)=(x^2+ax+b)(x^2+cx+d)$$ as a result \begin{align} & bd=1 \\ & ad+bc=1 \\ & b+d+ac=1 \\ & a+c=1 \\ \end{align} This system has not solution in $Q$ because

$$d(ad+bc)=d\times\,1\to\,ad^2+c=d$$ On the other hand $\,c=1-a$ thus $$ad^2+1-a=d\to\,a(d^2-1)=d-1$$ This implies $d=1$ or $ad+a=1$. If $d=1$ then $\left\{\begin{matrix} a+c=1 \\ ac=-1 \\ \end{matrix}\right.$ that this system has not rational roots . If $\,ad+a=1\,$ then $a=\frac{1}{d+1}=\frac{b}{b+1}$ as a result $$b+d+ac=1\to b+\frac{1}{b}+\frac{b}{b+1}\left(1-\frac{b}{b+1}\right)=1$$ we have $$\frac{(b+1)^2}{b}+\frac{b}{(b+1)^2}=-1$$ This equation has not solution in $\mathbb{R}$

Answer 3

Every polynomial splits completely over the complexes. There are only three possibitilies:

• There are four real roots
• There are two real roots and one pair of complex conjugate roots
• There are two pairs of complex conjugate roots

The roots of the polynomial are fifth roots of unity other than $1$, so there are no real roots. Thus we are in the third case, and the factorization consists of two irreducible real quadratics.

Answer 4

A different route to the factorization over the reals (obviously the end result is same as in Egreg's post, but I give the factors explicitly).

Let $p(x)$ be your polynomial. By a direct calculation we see that $$ (x^2+\frac x2+1)^2=x^4+x^3+\frac94x^2+x+1=p(x)+\frac54 x^2. $$ This calculation is aided by palindromic symmetry of both $p(x)$ and this quadratic. Anyway, this gives us the factorization $$ \begin{aligned} p(x)&=(x^2+\frac x2+1)^2-(\frac{\sqrt5}2\,x)^2\\ &=(x^2+\frac{1-\sqrt5}2\, x+1)(x^2+\frac{1+\sqrt5}2\,x+1) \end{aligned} $$ by the usual $$ a^2-b^2=(a-b)(a+b) $$ formula.

So the Golden ratio (not surprisingly given that the zeros are vertices of a regular pentagon) makes an appearance.