Is $X^4-3X^2+2X+1$ irreducible over the rationals?

Question

I want to check whether the following polynomials is irreducible over the rationals $f=X^4-3X^2+2X+1$. I think I found that it is irreducible but my solution is really complicated, quite long and I am not sure whether this is correct. I would really appreciate it if someone could point out any mistakes or tell me whether my reasoning is correct. Sadly, I cannot use the classical theorems like Eisenstein (there is no common divisor of the coefficients $-3,2$ and $1$ and neither I have found a good reduction modulo some prime $p$. I have tried $p=2$, which gave me $X^4+X^2+1$ which is reducible and $p=3$ which gave me $X^4+2X+1$ for which $-1$ is a root thus also reducible. Out of options, I resorted to doing it by hand. Since $f$ is normed it is primitive. Thus, $f$ irreducible over $\mathbb{Q}[X]$ if and only if it is irreducible over $\mathbb{Z}[X]$ which follows by Gauss. Then, I checked whether the polynomial had any roots in $\mathbb Z$. It does not because since every root of a polynomial must a divisor of the constant term, here $1$ only $\pm 1$ are possible options for roots. Since $f(\pm 1)\neq 0$ I deduced that f does not have rational roots. Thus, if $f=gh$ for some polynomials $g,h\in \mathbb{Z}[X]$, then both $g$ and $h$ need to be quadratic. So I made the following ansatz \begin{align} f&=gh=(\alpha_1X^2+\beta_1X+\gamma_1)(\alpha_2X^2+\beta_2X+\gamma_2)\\ &=\alpha_1\alpha_2X^4+(\alpha_1\beta_2+\alpha_2\beta_1)X^3+(\beta_1\beta_2+\gamma_1\alpha_2+\gamma_2\alpha_1)X^2+(\beta_1\gamma_2+\beta_2\gamma_1)X+\gamma_2\gamma_1 \end{align}

with $\alpha_i,\beta_i,\gamma_i\in\mathbb{Z}$ for $i=1,2$. Then, I can start doing some simplification by comparing this to the original f. Obviously, $\alpha_1=\alpha_2=\pm1$, from which we can deduce by looking at the term proportional to $X^3$ that $\alpha_1\beta_2+\beta_1\alpha_2=\pm(\beta_1+\beta_2)=0$, it follows $\beta=\beta_1=-\beta_2)$. However, at the same time $\gamma_1=\gamma_2=\pm 1$. Then, by looking at the term proportional to $X$, we see that $\beta_2\gamma_1+\beta_1\gamma_2=\pm(\beta_1+\beta_2)=2$ which is a contradiction. Thus, f must the irreducible. I would really really appreciate any help on this, I have been stuck on this for hours and I have no idea whether what I did was correct. Thank you so much in advance!

Answer 1

Alternative method. Over $\mathbb{Z}_3$ we see $x=-1$ is a root and we find the factorization into irreducibles $f=(x^3+2x^2+x+1)(x+1)$. Now if $f$ factors over $\mathbb{Z}$, it too must factor as product of a linear polynomial and a cubic (product of two quadratics would induce product of quadratics over $\mathbb{Z}_3$). However linear factor over $\mathbb{Z}$ can be ruled out by rational root theorem, and so there is no such factorization over $\mathbb{Z}$, and thus by Gauss lemma also over $\mathbb{Q}$.

Answer 2

As someone commented, for $p=5$ you can indeed check that the $X^4-3X^2+2X+1$ is irreducible in $\mathbb{F}_5$.

However, what you did is absolutely correct! A polynomial is reducible over $\mathbb{Q}$ iff it is so over $\mathbb{Z}$, which you proved it was not.

Another tool I often use to show irreducibility over $\mathbb{Q}$ is the following. We know $\mathbb{R}[X]$ is a UFD. So, if you are given a polynomial over $\mathbb{Q}$, what you can do is factor it into irreducible polynomials over $\mathbb{R}$ (which are linear and quadratic polynomials). Then, if this factorisation does not admit a rational factorisation by means of combining some of these factors, the polynomial is irreducible over $\mathbb{Q}$. Because, if there did exist some rational factorisation, you could reduce it further into a different real factorization, contradicting the uniqueness part in a UFD.

An example: show that $f(X) = X^4+1$ is irreducible over $\mathbb{Q}$. We can factorizate $f$ over $\mathbb{R}$ as $f(X) = (X^2+\sqrt2X+1)(X^2-\sqrt2X+1)$. These are irreducible factors over $\mathbb{R}$ (otherwise $f$ would have real solutions, which it does not). Because any factorisation over $\mathbb{Q}$ must be the product of two quadratics, and we already found one factorisation that is not rational, such a rational factorisation does not exist.

Answer 3

I don’t claim this to be the easiest way but it’s a fun little chase. We need basics from the Galois theory of finite fields; if $x\in\Bbb F_{p^{mk}}$ then its conjugates over $\Bbb F_{p^k}$ are of the form $x^{p^{kj}}$, $0\le j<m$.

Modulo $5$, $X^4-3X^2+2X+1=X^4+2X^2+2X+1\in\Bbb F_5[X]$ has no roots, because $x^4=1$ for nonzero $x\in\Bbb F_5$ and $X^2+X+1$ has no roots by a direct check.

Any reduction of this would then be a factorisation into two irreducible quadratics. It follows there would exist $\alpha\in\Bbb F_{25}\setminus\Bbb F_5$ which is a root of this polynomial.

Consider: $$\begin{align}\mathrm{N}(\alpha)&=\alpha^6\\&=(3\alpha^2+3\alpha-1)\alpha^2\\&=3(3\alpha^2+3\alpha-1)+3\alpha^3-\alpha^2\\&=3\alpha^3+3\alpha^2-\alpha+2\end{align}$$Many thanks to Benjamin Wright for pointing out my previous sloppy algebra and for observing a shorter continuation: from here, we similarly compute $\mathrm{Tr}(\alpha)=3\alpha^3+3\alpha^2$ and then $\alpha=2+\mathrm{Tr}(\alpha)-\mathrm{N}(\alpha)\in\Bbb F_5$, a contradiction.

Original line of attack:

Thus $3Y^3+3Y^2-Y+(2-\mathrm{N}(\alpha))\in\Bbb F_5[Y]$ has $\alpha$ as a root and is divisible by $\alpha$’s minimal polynomial $(Y-\alpha)(Y-\alpha^5)$, with a third root $\beta\in\Bbb F_5$ satisfying $\mathrm{N}(\alpha)\beta=\mathrm{N}(\alpha)-2$. Hence $\beta\neq1$.

If $y\neq0,1$, $3(y^3+y^2)=3y^2(y^2-1)/(y-1)=3(1-y^2)/(y-1)=2(y+1)$ in $\Bbb F_5$. Evaluating at $Y=\beta$ thus finds $\mathrm{N}(\alpha)=\beta-1$ or $\mathrm{N}(\alpha)=2$ (if $\beta=0)$. In the former case, we deduce $(\beta-1)\beta=\beta-3,\beta^2+3\beta+3=0$ but this is impossible! (check that there are no roots).

In the latter case $\beta=0,\mathrm{N}(\alpha)=2$ we find the polynomial is then $3Y^3+3Y^2-Y$ but by the previous computation we see $y=-2=3$ is a root of this polynomial in $\Bbb F_5$, different from $\beta$, contradicting $\alpha\notin\Bbb F_5$.

Answer 4

From the plot of $f$ (or with a bit of testing) we see that it has two real roots: $\alpha_1$ in the interval $(-2,-1)$ and $\alpha_2$ in the interval $(-0.5,0)$. Furthermore, $f$ has a positive local minimum at $x=1$, so the other two roots are a complex conjugate pair.

If $f$ were to factor over the rationals it would either have a rational root, or a quadratic factor in $\Bbb{Z}[x]$ (Gauss's lemma), one of which would have to be $$m(x)=(x-\alpha_1)(x-\alpha_2)=x^2-[\alpha_1+\alpha_2]x+\alpha_1\alpha_2.$$ But from the estimates it follows that $\alpha_1\alpha_2\in(0,1)$, so $m(x)\notin\Bbb{Z}[x]$.

The rational root test quickly shows that $f$ doesn't have any rational roots, and its irreducibility follows.

Answer 5

Hint It suffices to analyze the cases $p = 2, 3$ you've already looked at in a little more detail.

Since $f(-1) \equiv 0 \pmod 3$, we have $f(X) = (X + 1) g(X)$ (modulo $3$) for some $g \in \Bbb Z_3$; we can compute $g$ using polynomial long division, and it turns out to be irreducible. So, if $f$ factors over $\Bbb Q$, it must factor as the product of a linear polynomial and a cubic one. The linear factor corresponds to a root of $f$, hence $f$ has a root modulo $2$, but $f$ has no roots modulo $2$.

Answer 6

The given polynomial $P(x)=x(x+2)(x-1)^2 + 1$ takes value $1$ at three integer points $x=-2,0,1$. Let's show that such a fourth degree integral polynomial is irreducible, or is a square. Indeed, assume that $P(x) = Q(x) \cdot R(x)$, integral. Then $Q(x_i)=R(x_i) = \pm 1$, $i=1,2,3$. If one of the $Q$, $R$ is of degree $1$ we get a contradiction ( it has to be constant). Otherwise, they are both of degree $2$, but then they have to be equal ( taking the same value at $3$ points), and so $P = Q^2$.

Now our particular polynomial is not a square, since $P(-1)=-3<0$. Therefore, it is irreducible.

$\bf{Added:}$ Consider $P(x)\in \mathbb{Z}[x]$ of degree $n$, taking the value $1$ at $k> \frac{2}{3}n$ integral points. Then $P(x)$ is irreducible, or is a square.

Indeed, assume a factorization $P(x)= Q(x) \cdot R(x)$, with $\deg Q \ge \deg R>0$. Then we have $Q(x_i) = R((x_i) =\pm 1$, $i=1,k$. If $\deg R\le n-k$, then $k> 2 \deg R$. We conclude that $R$ takes the same value ( $+1$, or $-1$) at at least $\frac{k}{2}>\deg R$ points, so $R$ is constant, contradiction. Otherwise, both $\deg Q$, $\deg R< k$, but then since $Q$, $R$ take the equal values at $k$ points, we conclude $Q=R$, so $P= Q^2$.