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Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.

I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:

$f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$

For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2

Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method... so have any idea how we can prove that $f$ is irreducible?

user26857
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Lucas
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  • I think after verifying $f(x+1)$ is not irreducible, it has already shown that $f(x)$ is irreducible. Am I correct? – MonkeyKing May 01 '15 at 15:18
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    Write down a real factorization (degree 2 times degree 2) and note that the coefficients of the factors are not all contained in the rationals. – MooS May 01 '15 at 15:20
  • @MooS why doesn't work with my method? with show if $f(x\pm c)$ is irreducible it means that $f(x)$ is irreducible? – Lucas May 01 '15 at 15:22
  • The problem with your method is, that there is absolutely no reason, why you should find a $c$ with Eisenstein appliable to $f(x \pm c)$ in a reasonable time (if it exists at all). – MooS May 01 '15 at 15:25
  • @MooS by that logic, $x^4-4x^2+4$ is irreducible, since that polynomial can be factored as two real not-all-rational-coefficient polynomials of degree two as $(x^2+2\sqrt{2}x+2)(x^2-2\sqrt{2}x+2)$. However $x^4-4x^2+4=(x^2-2)^2$ is reducible. One would need to verify any factorization into two quadratics fails to be a factorization over $\Bbb Q$, not just a particular one. – anon May 01 '15 at 15:34
  • No, you are not right. If you have one factorization into irreducible real quadratics, you are done, since any factorization over the rationals would then contradict the unique factoriration over the reals. The keyword is irreducible here. Your given factorization is no counterexample, since the quadrics are not irreducible over the reals. – MooS May 01 '15 at 16:09
  • The simple equation $$x^4-4x^2+16=(x^2-\sqrt{12}x+4)(x^2+\sqrt{12}x+4)$$ together with the fact that those quadrics obviously have no real roots is enough to show the irreducibility over $\mathbb Q$. This is by far the shortest approach. No need for nine long answers... – MooS May 01 '15 at 16:14
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    @MooS Your original comment, the one I was responding to, said absolutely nothing about the quadratic factors being irreducible over the reals. All you mentioned to readers in that comment was that the quadratic factors did not have all-rational coefficients. – anon May 01 '15 at 16:17

9 Answers9

13

The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there's no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square: \begin{align} x^4-4x^2+16 &=x^4+8x^2+16-12x^2\\ &=(x^2+4)^2-(\sqrt{12}\,x)^2\\ &=(x^2-\sqrt{12}\,x+4)(x^2+\sqrt{12}\,x+4) \end{align} These two polynomials have negative discriminant (no need to verify it) and so they're irreducible in $\mathbb{R}[x]$. If the given polynomial were reducible over the rationals, the two factorizations in $\mathbb{Q}[x]$ and $\mathbb{R}[x]$ would coincide.

Therefore the given polynomial is irreducible over the rationals.

What's the general rule? Suppose you have $x^4+px^2+q$, with $p,q$ integers and $p^2-4q<0$ (so $q>0$). Write $q=r^2$, with $r>0$ (it need not be integer), and $$ x^4+px^2+q=x^4+2rx^2+r^2-(2r-p)x^2 $$ Note that $2r-p>0$: it's obvious if $p<0$; if $p\ge0$ it's the same as $4q>p^2$, which is true by hypothesis. Then $$ x^4+px^2+q=(x^2-\sqrt{2r-p}\,x+r)(x^2+\sqrt{2r-p}\,x+r) $$ is the decomposition of the polynomial in $\mathbb{R}[x]$. It is in $\mathbb{Q}[x]$ if and only if $\sqrt{q}$ and $\sqrt{2\sqrt{q}-p}$ are integers.

For example, $q=4$ and $p=0$ is a case. For $q=16$ we need $8-p$ to be a square, so $q=16$ and $p=4$ is another case.

egreg
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  • @egreg where I can find this general rule? have a link? is available just for degree 4? – Lucas May 01 '15 at 16:35
  • @Lucas I wrote it without looking anywhere. I don't claim it's original, of course. – egreg May 01 '15 at 16:39
  • @egreg Very good answer.I have stumbled on this by being rushed but I hope that my answer is correct (I sought maximum simplification) – Piquito May 01 '15 at 16:45
  • @LuisGomezSanchez It is correct, but the method of indeterminate coefficient is complicated, particularly when there's a simpler one available. – egreg May 01 '15 at 16:52
  • why we note $q=r^2$ and not something else? – Lucas May 01 '15 at 17:01
  • @egreg not really when there are not termes of 3 and 1 degree: you have immediately ($x^2$ + ax + 4) ($x^2$ -ax + 4) then your irrational value for a. – Piquito May 01 '15 at 17:02
  • @LuisGomezSanchez Because it works. ;-) I want to form an expression of the form $(x^2+r)^2-c^2x^2$, so $r^2=q$. – egreg May 01 '15 at 17:08
  • @egreg when we find $x^2=2\pm i\sqrt{12}$ why we can't say the factored quadratic is: $(x^2-2-i\sqrt{12}) \cdot (x^2-2+ i\sqrt{12})$ ? why we must to have this form: $(x^2-x\sqrt{2r-p}+r)(x^2+x\sqrt{2r-p}+r)$ ? what's the trick? – Lucas May 02 '15 at 07:40
  • @Lucas The polynomial can be completely split in the complex numbers, but only pairing conjugate roots we can get a factorization in the real numbers. Any other would not be conclusive about reducibility or irreducibility over the rational numbers. – egreg May 02 '15 at 07:59
  • if we find all complex roots, is enough to say the roots are not belong in $\mathbb{Q}$ ? why we must to pariring? and if we pairing conjugate roots we don't obtain same thing where you go: $(x^2-2-i\sqrt{12}) \cdot (x^2-2+i\sqrt{12})$ ? – Lucas May 02 '15 at 08:12
  • @Lucas The polynomial $x^4+2x^2+1$ has no real roots, but it's reducible over the rationals; $x^4-4x^2+4$ has real but not rational roots, but it's reducible over the rationals. For the rest, I don't understand: those two factors are not obtained by pairing conjugate roots. – egreg May 02 '15 at 08:15
12

The polynomial has no real roots, because it is equal to $(x^2-2)^2+12$. The remaining possibility is thus that it is a product of two quadratic factors. By Gauss' Lemma these need to have integer coefficients, so we are looking for a possibile factorization like $$ p(x)=x^4-4x^2+16=(x^2+ax+b)(x^2+cx+d) $$ with some integers $a,b,c,d$. Modulo $3$ we have the factorization $$p(x)=(x^2-2)^2+12\equiv(x^2+1)^2.$$ Therefore $a$ and $c$ must both be divisible by three, and $b\equiv d\equiv 1\pmod3$. Modulo $5$ we have $$ p(x)\equiv x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1). $$ This means that $b\equiv d\equiv1\pmod 5$ as well. The Chinese Remainder Theorem (or case-by-case check) then shows that $b\equiv d\equiv 1\pmod{15}$.

Because $bd=16$ the only remaining possibility is that they are $1$ and $16$ in some order. But this is impossible because modulo $2$ we have $$p(x)\equiv x^4,$$ so all of $a,b,c,d$ must be even.

The conclusion is that $p(x)$ is irreducible.

Jyrki Lahtonen
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    This is a long winded answer. Look at my answer for a simple way. – Adhvaitha May 01 '15 at 15:41
  • I thought about doing it your way, @user17762. But I think this is simpler. Tastes vary :-) Both our techniques are important for newbies to master. – Jyrki Lahtonen May 01 '15 at 15:43
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    @user17762 Using only integers instead of complicated radical surds is surely going to be more preferable to a great number of readers. It's true the underlying idea driving the computations in your answer is simple and doesn't require outside knowledge (like Gauss' lemma and modular arithmetic). – anon May 01 '15 at 15:43
  • "The polynomial has no real roots," , but why we need to have real roots? if we don't real roots that's means is irreducible over R[X]* and it include over Q[X], can you explain why we need to have real roots? – Lucas May 01 '15 at 16:49
  • @Lucas: The logic at that point was that because the polynomial has no real roots, it does not have any rational roots either. Therefore its has no linear factors, and we are left to handle the possibility of two quadratic factors. Sorry about not making that clear. – Jyrki Lahtonen May 01 '15 at 16:51
  • @JyrkiLahtonen I don't understand the final part, when you say bd=16 why they aren't 2 and 8 in some order? – Lucas May 06 '15 at 17:53
  • @Lucas: Neither $2$ nor $8$ is congruent to $1$ modulo $15$. – Jyrki Lahtonen May 06 '15 at 17:55
  • @JyrkiLahtonen why is necessary to be modulo 15? what did you led to say modulo 15 and nothing else? – Lucas May 06 '15 at 18:03
  • @Lucas: That's exactly what the earlier parts of my answer were all about! The constant terms must be $\equiv1\pmod3$ and $\equiv1\pmod5$. – Jyrki Lahtonen May 06 '15 at 18:05
  • @JyrkiLahtonen I'm very interested for logic, otherwise you choose modulo 3 and modulo 5 factorization to prove, that free terms are congruent with 1 modulo $3\cdot5$ ? – Lucas May 06 '15 at 18:16
  • why they can't be ≡ 1 (mod 4) ? – Lucas May 06 '15 at 18:23
  • @Lucas: Because modulo $2$ factorization tells us that $b$ and $d$ are even. – Jyrki Lahtonen May 06 '15 at 19:11
  • The idea here is that factorization modulo a prime is unique, because $\Bbb{Z}_p$ is a field. When I'm trying to show that factors like $x^2+ax+b$ and $x^2+cx+d$ cannot exist, I first look at it modulo $p=3$. That gives me that $x^2+ax+b\equiv x^2+1\pmod3$ and $x^2+cx+d\equiv x^2+1\pmod 3$ because $p(x)$ has no other factors modulo $3$. Similarly with $p=5$ I get that one of $x^2+ax+b$ or $x^2+cx+d$ is congruent to $x^2+x+1$ and the other to $x^2-x+1$ modulo $5$. In any case $b\equiv d\equiv1\pmod 5$. Together these imply that $b\equiv d\equiv1\pmod{15}$. – Jyrki Lahtonen May 06 '15 at 19:16
  • Next over $\Bbb{Z}$ (no modulo!) we know that $bd=16$. This forces ${b,d}={1,16}$. Finally this contradicts the modulo $p=2$ factorization: $$x^2+ax+b\equiv x^2+cd+d\equiv x^2\pmod 2.$$ – Jyrki Lahtonen May 06 '15 at 19:18
10

Below is an explicit proof. Note that $x^4-4x^2+16 = (x^2-2)^2 + 12$, which clearly has no real root. Hence, the only possible way to reduce $x^4-4x^2+16$ over $\mathbb{Q}$ is $(x^2+ax+b)(x^2+cx+d)$. However, the roots of $x^4-4x^2+16$ are $x = \pm \sqrt{2 \pm i\sqrt{12}}$, which are all complex numbers. Since complex roots occur in conjugate pairs, $\sqrt{2 \pm i\sqrt{12}}$ must be the roots of one of the factored quadratic. Hence, the factored quadratic must be $$x^2-(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2-2\sqrt3 x+4$$ The other factored quadratic must be $$x^2+(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2+2\sqrt3 x+4$$ Hence, $x^4-4x^2+16$ is irreducible over $\mathbb{Q}$.

Adhvaitha
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    +1 for this is a correct solution. The method is available whenever we can find the zeros. It is easier to come up with than the method I used. The downside is that this method is not available if we cannot find the zeros. – Jyrki Lahtonen May 01 '15 at 15:46
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    @JyrkiLahtonen Agreed. Though there is a good old proverb that "One needs the right tool for each task". – Adhvaitha May 01 '15 at 16:04
  • "The method is available whenever we can find the zeros" , we can apply this method because $f( \pm \sqrt{2\pm i\sqrt{12}} )=0$ ? but this are complex roots, not real, so in $\mathbb{R}$ we can't find the zero – Lucas May 02 '15 at 06:36
  • For instance, why the first factorization form isn't: $x^2-(\sqrt{2+i\sqrt{12}}-\sqrt{2-i\sqrt{12}})x-\sqrt{2+i\sqrt{12}}\sqrt{2-i \sqrt{12}}$ ? or another form... ? – Lucas May 02 '15 at 08:01
  • why you get the roots with the same sign and not pairing a positive root with a negative root, how is in my example ? – Lucas May 02 '15 at 08:54
  • @Lucas Read the answer completely before commenting. As I have mentioned, "Since complex roots occur in conjugate pairs...". Whenever you have $x^2+px+q =0$, where $p,q \in \mathbb{R}$, the roots are conjugate pairs of each other. So your choice doesn't work. – Adhvaitha May 02 '15 at 12:16
  • $\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}}=\sqrt{2(1+i\sqrt{3})}+\sqrt{2(1-i\sqrt{3})}$ how can I continue to obtain $2\sqrt{3}$ ? – Lucas May 02 '15 at 13:03
  • @Lucas $$\sqrt{2+i\sqrt{12}} = \sqrt{2+2i\sqrt3} = \sqrt2 \sqrt{1+i\sqrt3} = 2 \sqrt{1/2+i\sqrt{3}/2} = 2\sqrt{e^{i\pi/3}} = 2 e^{i\pi/6}$$ – Adhvaitha May 02 '15 at 13:05
5

You've seen that $f(x)$ has no roots, so you want to exclude factorizations of the form $$f(x) = (x^2 + ax + b)(x^2 + cx + d)$$ Since $f(x) = f(-x)$, the above implies $$f(x) = (x^2 - ax + b)(x^2 - cx + d)$$ Here $a,b,c$, and $d$ are integers by Gauss's Lemma.

So a given root $r$ of $x^2 - ax + b$ is a root of $x^2 + ax + b$ or $x^2 + cx + d$.

If $r$ is a root of $x^2 + ax + b$, it is the root of the difference $x^2 + ax + b - (x^2 - ax + b) = 2ax$, which implies $a = 0$ since zero is not a root of $f(x)$.

If $r$ is a root of $x^2 + cx + d$ it is similarly a root of the difference $(c + a)x + (d - b)$, and since $f(x)$ has no rational roots $c = -a$ and $d = b$.

So either $a = 0$ or $c = -a$ and $d = b$. Since the argument is entirely symmetric in the two factors, we also either have $c = 0$ or $c = -a$ and $d = b$. Hence we have two possibilities: Either $a = c = 0$ or $c = -a$ and $d = b$.

In the first case we have $$x^4 - 4x^2 + 16 = (x^2 + b)(x^2 + d)$$ But the roots of $y^2 - 4y + 16$ are irrational (they're not even real) so this can't happen.

In the second case we have $$x^4 - 4x^2 + 16 = (x^2 + ax + b)(x^2 - ax + b) = x^4 + (2b - a^2) x^2 + b^2$$ Hence $b = \pm 4$ and therefore either $8 - a^2 = -4$ or $-8 - a^2 = -4$, neither of which has rational solutions.

Hence $f(x)$ is irreducible.

Zarrax
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Suppose that $f$ is reducible, observe that $f$ is monic, hence $f$ can be written as the product of two polynomials $g$ and $h$ of degree at least one and with integer coefficients.
Now observe that $f$ has no integer root, because with rational root theorem an integer root to $f$ must be a divisor of $f(0)=16$ and you can check it's cases.
So $f$ has no linear factor, hence $g$ and $h$ are two monic polynomials with degree two, with letting $$f(x)=g(x)h(x)=(x^2+ax+b)(x^2+cx+d)=x^4-4x^2+16$$ We get $$ a=-c,b+d+ac=-4,ad+bc=0,bd=16 $$ Now if $a\neq0$, then $b=d$,$b+d+ac=-4$, so $b^2=16$, $2b-a^2=\pm8-a^2=-4$, which contradicts.
Now if $a=-c=0$, then $bd=16$, and $b+d=-4$ which again contradicts.

k1.M
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As Jyrki Lahtonen observed in his answer, $p(x)=x^4-4x^2+16=(x^2-2)^2+12$ has no real roots, so the only possible factorization is of the form

$$(x^2+ax+b)(x^2+cx+d)$$

which expands to

$$x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$$

We conclude that $c=-a$ (from the $x^3$ term), hence* $d=b$ (from the $x$ term), which means $2b-a^2=-4$ and $b^2=16$. Plugging the two possibilities for $b$ into $a^2=2b+4$ gives $a^2=12$ or $-4$, neither of which corresponds to an integer value for $a$.

*Zarrax in comments astutely observes my "hence" is mistaken. It ignores the possibility $a=c=0$. To complete the proof that $p(x)$ is irreducible, we need to note that if $a=c=0$ then, letting $u=x^2$, we would have a factorization

$$(u+b)(u+d)=u^2-4u+16=(u-2)^2+12$$

My thanks to Zarrax for pointing out the error.

Barry Cipra
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I think you should try using a method other than Eisenstein's criterion. By Gauss' lemma it suffices to show that $f(x)$ does not factor over the integers, so you need only show the following:

  • There is no integer root: ie $f(x) \ne 0 $ for $x$ an integer dividing $16$.
  • There is no quadratic factorization: You cannot write $f(x) = (x^2+ax+b)(x^2+cx+d)$.

The first of these is straightforward to verify, and deriving a contradiction from the second is not unreasonable with this particular $f(x)$.

Rolf Hoyer
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After substitution $$ g(x)=f(2x-1)=16x^4-32x^3+8x^2+8x+13 $$ it becomes irreducible by Murty's criterion (see Theorem 1) for $n=4$, since $g(4)=2221$ is a prime. The criterion is not so well spread, so here it is:

Let $f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_1x+a_0$ be a polynomial of degree $m$ in $\mathbb{Z}[x]$ and set $$H=\max_{0\leq i\leq m-1} |a_i/a_m|.$$ If $f(n)$ is prime for some integer $n\geq H+2$, then $f(x)$ is irreducible in $\mathbb{Z}[x]$.

Just a note, other substitutions are possible as well. For example, reverting coefficients - basically a substitution $x^4f(1/x)$ - yields again irreducibility by the criterion above, this time with $g(5)=9901$ a prime.

Sil
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$$x^4-4x^2+16=(x^2-(2+\sqrt{-12}))(x^2-(2-\sqrt{-12}))$$

No rational roots and no factorization into quadratics over the rationals. The polynomial is irreducible over the rationals

edit for those who commented that this is not enough. I factorized over $\mathbb{C}[X]$ and thus proved that there are no rational solution i.e no degree $1$ factors. The only factorization possible is therefore into two quadratics. $\mathbb{C}[X]$ is a UFD and therefore we have

$$x^4-4x^2+16=(x-\sqrt{2+\sqrt{-12}})(x+\sqrt{2+\sqrt{-12}})(x-\sqrt{2-\sqrt{-12}})(x+\sqrt{2-\sqrt{-12}})$$

And this is unique. So combining the degree 1 factors in pairs is the only way to factorize in quadratics and there are three different ways to combine and none is rational

marwalix
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    That is not sufficient to prove it is irreducible over $,\Bbb Q,,$ i.e. you need to say more. – Bill Dubuque May 01 '15 at 15:33
  • "- There is no quadratic factorization: You cannot write $f(x) = (x^2+ax+b)(x^2+cx+d)$" – Lucas May 01 '15 at 15:40
  • Set $y=x^2$ and solve the quadratic $y^2-4y+16=0$. – marwalix May 01 '15 at 15:40
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    marwalix: Showing that one particular factorization into two quadratics fails to be a factorization over $\Bbb Q$ is not the same thing as showing that there isn't any factorization into quadratics over $\Bbb Q$. See my comment above for an example. – anon May 01 '15 at 15:40
  • Observe that \begin{align} x^4 - 4x^2 + 16 & = x^4 - 4x^2 + 4 + 12\ & = (x^2 - 2)^2 - (-12)\ & = (x^2 - 2)^2 - (2i\sqrt{3})^2\ & = (x^2 - 2 + 2i\sqrt{3})(x^2 - 2 - 2i\sqrt{3})\end{align} – N. F. Taussig May 01 '15 at 15:41
  • @N.F.Taussig why marwalix's method doesn't work? – Lucas May 01 '15 at 15:45
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    I just explained why marwalix's method doesn't work in my comment above. The fix is in user17762's answer: try all possible quadratic factors and see that none of them have rational coefficients. – anon May 01 '15 at 15:46
  • (The edit is correct.) – anon May 01 '15 at 16:19