Polynomial is irreducible over $\mathbb{Q}$

I've been scratching my head lately over this problem in my textbook:

Prove that the polynomial $$x^4 + x^3 +x^2 +x +1$$ is irreducible over $$\mathbb{Q}$$.

I've done some research and found this link, but they talk about Eisenstein's criterion, which we haven't covered in our math class just yet. Is there a general strategy where we can show whether a polynomial is irreducible over a field?

My textbook really didn't go into depth on the topic of irreducibility of polynomials, but this Wiki link somewhat helps. Perhaps the rational root theorem may be helpful here, but how would I go about starting this proof?

EDIT

Please see the first comment for my initial strategy at showing the required result.

• Here's my strategy: so if I apply the rational root theorem, gather my roots and test each one into the function $x^4 + x^3 +x^2 + x +1 = 0$, then I know I have rational roots. But if none of my choices work, then the original polynomial is irreducible in $\mathbb{Q}$. Mar 11, 2017 at 3:57
• Showing a polynomial has no rational roots only allows you to conclude it is irreducible in $\Bbb Q[x]$ when it is a quadratic, or of odd degree.
– Dave
Mar 11, 2017 at 4:00
• So if my strategy doesn't suffice, what other approach should I consider? Mar 11, 2017 at 4:02
• Otherwise simply showing 4 irrational roots solve it, would be enough by fundamental theorem of algebra. Mar 11, 2017 at 4:17
• @Dave "Showing a polynomial has no rational roots only allows you to conclude it is irreducible in $\mathbb Q[x]$ when it is a quadratic, or of odd degree." The polynomials $x^3-2$ and $x^2-2$ both have no roots in $\mathbb Q$, and hence neither does their product, which is obviously reducible in $\mathbb Q[x]$ despite being of odd degree. Do you mean "Showing a polynomial has no rational roots only allows you to conclude it is irreducible in $\mathbb Q[x]$ when it is a quadratic or a cubic? Mar 11, 2017 at 5:25

A couple of strategies:

1. This may be the only strategy you can use at this point (without Eisenstein's Criterion). If you are able to show that your polynomial (let's say $p(x)$) has no rational roots (by the rational root test for instance), then $p(x)$ has no linear factors in $\Bbb Q[x]$. Thus, since $p(x)$ is quartic, if it is to be reducible, then it must factor as two quadratics. This is because a quartic is either factored as: 4 linear, 1 linear and 1 cubic, 2 linear and 1 quadratic, or 2 quadratics, if it is reducible. So you suppose that $p(x)=(x^2+ax+b)(x^2+cx+d)$ for some $a,b,c,d\in\Bbb Z$. Then you compare the coefficients of the terms of $p(x)$ with the expanded form of that polynomial involving $a,b,c,d$ to see if there is a solution. If there cannot be $a,b,c,d$ satisfying these requirements, then $p(x)$ must be irreducible in $\Bbb Q[x]$.

2. You can use Eisenstein's Criterion here (although if you haven't learned about it this may be a problem). However, it will not work immediately. You can use the result (may require a proof) that $f(x)$ is irreducible if and only if $f(x+a)$ is irreducible for some $a\neq 0$ in the field you are working with. Thus, you can show that $f(x+a)$ is irreducible for some $a\in\Bbb Z$ by Eisenstein's Criterion, which will allow you to conclude that $f(x)$ is irreducible in $\Bbb Q[x]$.

3. Another way to show this polynomial is irreducible in $\Bbb Q[x]$, which is a more interesting approach (although not expected) is the following. Using the aforementioned theorem of $f(x)$ being irreducible if and only if $f(x+a)$ is irreducible, one can show that $f(x)=x^{p-1}+x^{p-2}...+x+1$ is irreducible in $\Bbb Q[x]$ for any positive prime $p$. One also needs to use Eisenstein's Criterion here. In this case, for $p=5$, we see that the polynomial in question is irreducible in $\Bbb Q[x]$.

Proof for strategy 3:

I will leave it to you to prove $f(x)$ is irreducible if and only if $f(x+a)$ is irreducible. Define $\phi_p(x)\in\Bbb F[x]$ as $\phi_p(x)=\frac{x^p-1}{x-1}$ for a positive prime $p$. Then we know that $\phi(x)=x^{p-1}+x^{p-2}+...+x+1$. Evaluating at $x+1$, we have: \begin{align}\phi_p(x+1)&=\frac{(x+1)^p-1}{x+1-1}\\&=\frac{1}{x}\sum_{n=0}^p\binom{p}{n}x^n-1\\&=x^{p-1}+\binom{p}{1}x^{p-2}+...+\binom{p}{p-2}x+\binom{p}{p-1}\end{align} Note that $p\mid\binom{p}{k}$ for every $k\in\{1,2,...,p-1\}$, but $p\nmid 1$ and $p^2\nmid p=\binom{p}{p-1}$. Thus, by Eisenstein's Criterion and the fact that $f(x)$ irreducible iff $f(x+a)$ irreducible, we have that $x^{p-1}+x^{p-2}+...+x+1$ is irreducible in $\Bbb Q[x]$.

In this particular case it is in fact quite easy to determine the actual irreducible factorization over $\mathbb{R}$, which results in a couple of quadratic factors with complex roots and coefficients that are not (all) rational, therefore the original polynomial is irreducible over $\mathbb{Q}\,$.

\begin{align} x^4+x^3+x^2+x+1 &= x^2\left(x^2+\frac{1}{x^2}+x+\frac{1}{x}+1\right) \\[3px] & = x^2\left(\left(x+\frac{1}{x}\right)^2 + \left(x+\frac{1}{x}\right) -1\right) \\[3px] & = x^2\left(x+\frac{1}{x}-\frac{-1 + \sqrt{5}}{2}\right)\left(x+\frac{1}{x}-\frac{-1 - \sqrt{5}}{2}\right)\\[3px] & =\left(x^2 - \frac{-1 + \sqrt{5}}{2} \, x+1\right)\left(x^2 - \frac{-1 - \sqrt{5}}{2} \, x+1\right) \end{align}

• Could you also do that with 6th order? Aug 27, 2021 at 17:21
• @Unknown123 Guess you could, technically, since you get a cubic in $x+1/x$ which is solvable by radicals, and must have a real root, so it is reducible over $\mathbb R$. But the calculations would be cumbersome to the point of making such approach impractical.
– dxiv
Aug 27, 2021 at 17:36

After confirming no rational roots, the possible factorization is quadratic times quadratic. The result of Gauss on content says that, if an integer polynomial factors over the rationals, it also factors over the integers. So, we go to trial and error: $$( x^2 + ax + 1) (x^2 + bx + 1)$$ OR $$( x^2 + ax - 1) (x^2 + bx - 1)$$ Neither one works. Confirming that neither one works for integers $a,b$ is not difficult. You should do it.

Eisenstein.

I suppose you have shown that the polynomial has no rational root. Now if the polynomial is reducible, then its factors must be two irreducible quadratic polynomials. Recognise that the polynomial is $\frac{x^5-1}{x-1}$. The roots of $x^4+x^3+x^2+x+1$ would be the roots of unity other than 1. Find the pair of irreducible quadratic polynomials and see that this factorisation is not possible in rational numbers.

There is no general strategy for testing whether a polynomial is irreducible except when the field is algebraically closed. It is also more difficult to check whether a polynomial is irreducible in $\Bbb Q$ than in $\Bbb R$.

Note that in this case you can use Cohn's irreducibility criterion for base $b=2$, because $f(2)=31$ is a prime and the irreducibility follows.

You may check through linear algebra that your polynomial is the minimal polynomial of $\alpha=e^{\frac{2\pi i}{5}}$ over $\mathbb{Q}$. $\alpha^5-1=0$ and $\alpha-1\neq 0$, hence $\alpha$ is a root of $$\Phi_5(x)=\frac{x^5-1}{x-1} = x^4+x^3+x^2+x+1.$$ $\mathbb{Q}[x]/(\Phi_5(x))$ is a vector space over $\mathbb{Q}$ with dimension $4$ and a base given by $1,x,x^2,x^3$.
Since the following matrix $$\begin{pmatrix}0 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}$$ representing the multiplication by $x$ in the previous ring, has full rank, $\Phi_5(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $\Phi_5(x)$ is an irreducible polynomial.

• I don't understand this at all - the matrix is just a companion matrix. How does this argument use the polynomial in a way that would not also apply to x^2+2x+1? Oct 11, 2021 at 19:37
• You cannot compare the two situations since your polynomial is not coprime with its derivative, but in general the kernel of a companion matrix gives plenty of informations about the involved polynomial. Oct 12, 2021 at 5:12
• OK then, "f = x^2 + 3x + 2. -1 is a root of f. Q[x]/f is a vector space with dim 2 over Q. companion matrix has full rank 2, so f is the min poly of 1 over Q and hence is irreducible" Surely your argument needs more details Oct 12, 2021 at 12:54
• Also the kernel of a companion matrix is 0 unless X divides the poly, so I'm not sure what you mean by 'gives plenty of info about the poly' Oct 12, 2021 at 12:58

There exists a simple general method for monic quartic polynomials in $\mathbb{Z}[x]$. Suppose you have verified that the polynomial

$$p(x) = x^4 + a x^3 + b x^2 + c x + d$$

has no rational roots. Then we want to see if it has quadratic factors, and if so we want to factor it. Also we don't want to use very elaborate trial and error methods. The first test that needs to be performed is to see whether p(x) is the square of a quadratic polynomial, that's easy to see by computing the GCD of $p(x)$ with its derivative. If a nontrivial GCD, then you're done. If not, then we proceed as follows. We reduce $p(x)$ modulo $x^2 - p x - q$, this yields:

$$\left(p^3 + a p^2+ 2 p q + a q+ b p+c\right) x + p^2 q + q^2 +a p q+ b q+d$$

Then if $p(x)$ has a quadratic factor, this has to be identically zero. We thus need to equate the coefficient of $x$ and the constant term to zero and solve the two equations for $p$ and $q$. It's then convenient to start with eliminating the highest powers of $p$ in favor of lower powers until $p$ has been completely eliminated in favor of $q$. We then end up with an equation for $q$, and an expression for $p$ in terms of $q$. Then since $q$ had to divide $d$, you only have a few cases to check. If none of them work then $p(x)$ is irreducible. If $p(x)$ is reducible you'll find the factorization, unless both factor have the same value for $q$. In the latter case the single solution for $q$ obviously cannot tell you what both values for $p$ are (and they are different because $p(x)$ was verified to be square-free). If this exceptional case does not occur, then we have:

$$p = \frac{a q^2+c q}{d-q^2}\tag{1}$$

and

$$q^6 +b q^5 +(a c-d)q^4 + \left(a^2 d-2 b d+c^2\right)q^3 +\left(a c d-d^2\right)q^2 +b d^2 q+d^3 = 0$$

So, $q$ must be an integer that divides $d$ that satisfies this equation. If two such values for $q$ are found then you have found the factorization, the corresponding values for $p$ follow from Eq. (1). If only one solution for $q$ is found, then Eq. (1) will be singular, the two values for $p$ are then solutions of the equation:

$$a p^2 + a^2 p+a b-2 c = 0$$

Interesting that nobody solved it the following way: $$x^4 + x^3 + x^2 + x + 1 => (x+1)^4 + (x+1)^3 + (x+1)^2 + (x+1) + 1 = x^4 + 5x^3 + 10x^2 + 10x + 5$$. Now using Eisenstein's criterion, p = 5, it is irreducible.

• hi JimmySwisher, welcome to the site! please note that OP explicitly stated that they have not covered Eisenstein's criterion, and so your post does not answer the question they posed. Apr 16, 2021 at 16:05