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This question came from the answer here.

The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?

user557
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4 Answers4

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Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.

The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.

Therefore, the smallest field containing $\mathbb{F}_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.

Alon Amit
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  • Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible? – user557 Aug 06 '18 at 20:00
  • Take a factor of that polynomial and consider the splitting field of that factor. – Alon Amit Aug 06 '18 at 20:01
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    I have lost count of the questions I have settled here using this argument. Welcome to the club :-) – Jyrki Lahtonen Aug 06 '18 at 21:12
  • @JyrkiLahtonen proud to be a member! – Alon Amit Aug 06 '18 at 22:08
  • Sorry but I still can't see how considering the splitting field of a factor implies the result. – user557 Aug 06 '18 at 23:08
  • Do you see that if $f(x)=g(x)h(x)$ then any root of $g$ is also a root of $f$? And therefore, if your polynomial factors, any root of such a factor in any field must have order $5$? – Alon Amit Aug 06 '18 at 23:58
  • I can see the first claim and that any root of such a factor gives $1$ when raised to the 5th power (but I'm not sure why it's the minimal power that gives $1$). Even assuming that it has order 5, I don't see why the splitting field must contain $<7^4 $ elements. – user557 Aug 07 '18 at 00:23
  • Because, as I explained in the answer, any field with $7$, $7^2$ or $7^3$ elements cannot have any elements of order $5$. The multiplicative group of these fields has $6$, $48$ or $342$ elements, and none of these numbers is divisible by $5$. If a group contains an element of order $5$, the order of the group must be divisible by $5$ (Cauchy's theorem in group theory). – Alon Amit Aug 07 '18 at 00:25
  • I understood this argument, but I don't understand the very last sentence of your answer; I don't quite see how it follows from your last comment. I see it partially. If the polynomial were reducible, it would either have a quadratic or a linear factor. In the latter case it would have a root which (as you said) would have order $5$, but $(\mathbb F_7){^\times}$ doesn't have elements of order $5$. So the latter case does not occur. Thus the remaining questions are: why cannot it have a quadratic factor and why if it has a root, then no power less than $5$ gives $1$? – user557 Aug 07 '18 at 00:41
  • And of course it has something to do with the order of $2\equiv7$ in the group $\Bbb F_5^\times$. – Lubin Aug 07 '18 at 05:00
  • @user437309 I’m not sure what confuses you since I don’t know what you know. The splitting field of a quadratic polynomial has $7^2$ elements, so it can’t have elements of order $5$. An element satisfying $x^5=1$ has order 5 or order 1. – Alon Amit Aug 07 '18 at 05:23
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    @user437309 If $f(x)$ splits into two irreducible quadratics $g(x)h(x)$, we can construct a finite field $K = \mathbb F_7[x]/(g(x))$. Now we can still compute in $K$: $$x^5-1 = (x-1)f(x) = (x-1)g(x)h(x) \equiv 0 \pmod{g(x)}$$ This means we have constructed a finite field $K = \mathbb F_7[x]/(g(x))$ with $7^2$ elements which also contains an element $x$ with order $5$. But that is not possible since $5\nmid 7^2-1 = 48$. – Yong Hao Ng Aug 07 '18 at 07:19
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    Now I understand the solution as follows. Let $a$ be a root of $x^4+x^3+x^2+x+1$ in a field extension of $\mathbb F_7$. If the polynomial were reducible, then the degree $[\mathbb F_7(a):\mathbb F_7]$ would be strictly less than $4$, and also $a$ would have order $5$ in this extension. The first condition means that $\mathbb F_7(a)=(\mathbb F_7)^k$ where $k=1,2,3$, but none of these three fields contains an element of order $5$. – user557 Aug 07 '18 at 19:26
  • I can't edit the previous comment. Of course I meant "order $5$ in the multiplicative group of this extension" and "none of these three fields contains an element of order $5$ in their multiplicative groups". – user557 Aug 07 '18 at 19:31
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    @user437309 yes, that’s right. – Alon Amit Aug 07 '18 at 19:45
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If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^{-1}$. Thus, the equation $ac+b+d=1$ becomes $$a(1-a)+b+b^{-1}=1\,.\tag{*}$$ The equation $bc+ad=1$ is now $$b(1-a)+ab^{-1}=1\,.$$ Thus, $$a^{-1}\big(1-b(1-a)\big)=b^{-1}=1-b-a(1-a)\,.$$ That is, $$1-b(1-a)=a-ab-a^2(1-a)\,.$$ Consequently, $$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1)\,.\tag{#}$$ Thus, $a\notin\{0,1,4\}$ for the equation above to be possible (noting that $a\neq 0$ and $b\neq 0$). You are left with four choices of $a$, namely, $a\in\{2,3,5,6\}$.

If $a=2$, then, using (#), we get $b=3^{-1}\cdot 5=5\cdot 5=4$, so $b^{-1}=2$, but then $$a(1-a)+b+b^{-1}=2\cdot(-1)+4+2=4\neq 1\,.$$ If $a=3$, then we have by (#) that $b=5^{-1}\cdot20=-3=4$, whence $$a(1-a)+b+b^{-1}=3\cdot(-2)+4+2=0\neq 1\,.$$ If $a=5$, then $b=3^{-1}\cdot (-1)=-5=2$, so $b^{-1}=4$, but then $$a(1-a)+b+b^{-1}=5\cdot(-4)+2+4=0\neq 1\,.$$ Finally, if $a=6$, then $b=4^{-1}\cdot3=2\cdot 3=-1$, whence $b^{-1}=-1$, and $$a(1-a)+b+b^{-1}=6\cdot(-5)+(-1)+(-1)=-4\neq 1\,.$$ Thus, (*) cannot be satisfied.

Batominovski
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Yet an other solution. It is based on the fact that $$ f = x^4+x^3+x^2+x+1\in\Bbb F_7[x] $$ is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)

Assume there is a factorization $$ f=(x^2+ax+b)(x^2+cx+d)\ ,\qquad a,b,c,d\in\Bbb F_7\ .$$ Let $u\ne 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.

First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product $$(x^2+ax+1)(x^2+(1-a)x+1)$$ is reciprocal, and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $a\leftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.

Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials $$ (x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1)\ . $$ The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $\Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)

Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $b\ne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.

Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:

sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True

One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $\Bbb F_7$. Then there is a root in $\Bbb F_{7^2}=\Bbb F_{49}$, so the polynomials $f$ and $x^{49}-x$ have a common divisor. The gcd is but... we type

sage: gcd( x^49 - x, f ) 
1
sage: gcd( x^(7^4) - x, f ) 
x^4 + x^3 + x^2 + x + 1

For this, one can also compute easily $(x^{49}-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^{49}-x,f)=(x^{49}-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.

dan_fulea
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HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.

We have $$x^4+x^3+x^2+x+1=-\dfrac 14(-2x^2+(\sqrt5-1)x-2)(2x^2+(\sqrt5+1)x+2)$$ Since $\mathbb F_7^2=\{1,2,4,0\}$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.

Besides that a linear factor is not possible is checked easily.

Piquito
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  • Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $\mathbb F_7$. This is not possible. – Piquito Aug 06 '18 at 22:38
  • I don't know who downvoted this answer, I just upvoted it. – user557 Aug 06 '18 at 23:05
  • @user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk. – Piquito Aug 07 '18 at 00:07
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    Hmm. Over $\Bbb{C}$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $\zeta^k$ with $\zeta=e^{2\pi i/5}$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-\zeta)(x-\zeta^4)$ and $(x-\zeta^2)(x-\zeta^3)$, don't exist in $\Bbb{F}_7$ because they involve $\sqrt5$. But how do you rule out a factor like $(x-\zeta)(x-\zeta^2)$ showing up? True, Galois theory shows that $\Bbb{Q}(\sqrt5)$ is the only quadratic subfield of $\Bbb{Q}(\zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep. – Jyrki Lahtonen Aug 07 '18 at 05:34
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    (cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $\Bbb{F}_7$: If $\zeta$ is a root of some polynomial with coefficients in $\Bbb{F}_7$ then so is $F(\zeta)=\zeta^7=\zeta^2$, and therefore also $F(\zeta^2)=\zeta^{14}=\zeta^4$, and therefore also $F(\zeta^4)=\zeta^3$. That's four zeros, so any polynomial with $\zeta$ as a root has degree four. – Jyrki Lahtonen Aug 07 '18 at 05:37
  • I didn't downvote because I like the approach of using parallels from slightly different theories. However, I didn't upvote either because this is not quite rigorous. Whenever $5$ is a quadratic residue modulo a prime $p$, you get a factorization of $x^4+x^3+x^2+x+1$ modulo $p$ from this. But, reduction of polynomials from $\Bbb{Z}[x]$ modulo a prime gives factorization information in only one direction. A "famous" example is $x^4+1$, irreducible over $\Bbb{Z}$ but reducible modulo all primes. – Jyrki Lahtonen Aug 07 '18 at 05:42
  • The complex zeros of $x^4+1$ involve $i$, $\sqrt2$ and hence also $\sqrt{-2}$. Its factorization modulo a prime $p$ depends on which of $-1,2$ and $-2$ are quadratic residues modulo $p$. At least one of them is. All of them are iff $p\equiv1\pmod8$. You should not just check whether $$x^4+1=(x^2+\sqrt2 x+1)(x^2-\sqrt2 x+1)$$ "survives" or not. Which is roughly equivalent to what you did here. That only checks whether $2$ is a quadratic residue or not. – Jyrki Lahtonen Aug 07 '18 at 05:44
  • With the quartic in this thread the Galois group is cyclic (as opposed to Klein four you get with $x^4+1$), and that does make a difference (and a justification for this solution). But, you need to spell it out. – Jyrki Lahtonen Aug 07 '18 at 05:48
  • Thank you for your comment, at the same time very illustrative and very kindly. My perspective is much simpler: If $a, b, c, d$ are elements of $\mathbb F_7$ and $f (x) = (x ^ 2 + ax + b) (x ^ 2 + cx + d) = (x ^ 2 + \alpha x + β) (x ^ 2 + γ x + δ)$ where the Greek letters are not in $\mathbb F_7$ (but in a field extension of $\mathbb F_7$) and the quadratic factors are irreducible, then two different factorizations of $f (x)$ would be in a single factorization domain. I would really appreciate it if you would say, without – Piquito Aug 07 '18 at 13:58
  • further explanation, if this reasoning is wrong, in which case I would erase my answer. Thank you very much.(sorry for the bad English). – Piquito Aug 07 '18 at 13:59