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

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!

• Where are your doubts? I do not find it so long. You could try "shifted" Eisenstein, or some idea from this duplicate. Jan 3 at 12:39
• My doubts are about making the coefficients of g and h to be whole numbers Jan 3 at 12:40
• This is no problem because it is irreducible over $\Bbb Q$ iff it is irreducible over $\Bbb Z$. By the way, it is irreducible over $\Bbb F_5$, so you are done. So $p=5$ works. Jan 3 at 12:41
• Alright, then my reasoning with Gauss-lemma does work was unsure about that one. Thank you so much. What you pointed out with $p=5$ seems interesting would indeed be much quicker. I will try to figure it out myself Jan 3 at 12:44
• Yes, I forgot the $-1$ solution. you are correct. But that does not help either since I would still get $\beta_1=-\beta_2$ for the $X^3$ part and $-\beta_2-\beta_1=2$ in the $X$ part which is still a contradiction. Jan 3 at 12:48

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 $$\mathbb{Q}$$.

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 $$2$$.

• So, if f factored over $\mathbb{Q}$ it would need to do so as a cubic and a linear - which would be a contradiction to the first case, right? Amazing, so I do not even have to use the rational root theorem in this case, right? Jan 3 at 18:15
• Correct. And yes, this method doesn't use the Rational Root Theorem, though doing so as in Sil's clever answer saves you the trouble of factoring modulo $2$. I've shortened my answer: We don't need to know the factorization of $f$ modulo $2$, only that it has no roots modulo $2$, and to check that it's enough to verify that $f(0), f(1)$ are nonzero modulo $2$. Jan 3 at 21:19
• In any case it's not so bad to check the irreducibility of a quartic modulo $2$: The only irreducible quadratic modulo $2$ is $X^2 + X + 1$, so the only quartic modulo $2$ that has no roots but is nonetheless reducible is $(X^2 + X + 1)^2 = X^4 + X^2 + 1$. A similar trick can be used to check the irreducibility of a quintic modulo $2$, as there are only $2$ irreducible cubics modulo $2$. Jan 3 at 21:24

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.

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 $$P= Q^2$$.

• Nice. I think it would be worthwhile to spend a minute explaining why this polynomial is not a square Jan 3 at 23:26
• @FShrike: Thanks, and a good hint, added that ( $P(-1)=-3$). Jan 4 at 3:15

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.

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 $$\alpha\notin\Bbb F_5$$.

• I think there is a mistake in the last equality of your calculation of $N(\alpha)$; should be $N(\alpha) = 3\alpha^3 + 3\alpha^2 - \alpha + 2$. Jan 3 at 17:35
• Along similar lines, you can use the trace to get another relation: $Tr(\alpha) = \alpha^5 + \alpha = 3\alpha^3 + 3\alpha^2$, from which we can conclude that $-\alpha+2 = N(\alpha) - Tr(\alpha) \in \mathbb{F}_5$, so $\alpha \in \mathbb{F}_5$, contradiction. Jan 3 at 17:40
• @BenjaminWright many thanks. Fixed Jan 3 at 17:55
• This does work, even though the calculations may become a bit hairy. Any irreducible quadratic over $\Bbb{F}_5$ is a factor of $x^{24}-1$ (exactly because its roots are in $\Bbb{F}_{25}$. It seems to me that you are in some sense doing something equivalent. After all, $x^{24}-1=(x^6-1)(x^6-2)(x^6-3)(x^6-4)$, and the norm $\alpha^6$ makes the selection among those sextics :-). My upvote was there already. Jan 3 at 19:48
• @JyrkiLahtonen thanks, I like this method because it’s fairly reliable. It seems to be widely applicable and it’s gratifying to “find a use for” Galois theory ;) I learned the approach of (bash it with norm and trace until it crumbles) from some other MSE post. Do you know if there is some general theorem which guarantees “this” to always work? Jan 3 at 20:28

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.

• I don’t believe this. Although the evaluation of $X^4$ is always $1$, $X^4\neq1$ as a polynomial Jan 3 at 15:46
• You're both right, sorry. I fixed the mistake. For some reason I forgot that "irreducible iff no roots" didn't hold for quartics. Jan 3 at 16:53
• @FShrike What do you mean with "I don't believe this" by the way? Are you talking about my method? Jan 4 at 17:07
• Yes, I did not believe the method in the original revision of this post Jan 4 at 17:09
• For application to an entirely different polynomial, yes Jan 4 at 17:26