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Suppose $f$ is not irreducible. Since $f(x)=x^4+x+4$ does not have any root in $\mathbb{F}_5$, the only possible factorization can be $f(x)=(x^2+ax+b)(x^2+cx+d)$ for $a,b,c,d \in \mathbb{F}_5$.

Now by comparing the coefficients one gets:

$$1. \,\,a+c \equiv 0 $$ $$2. \,\, b+d+ac \equiv0 $$ $$3. \,\,ad+bc \equiv 1 $$ $$4. \,\,bd \equiv 4 $$

From condition $1$ and $4$, one can find possible values of $a,b,c,d$ and show that none of them simultaneously satisfy $2$ and $3$. But this is a rather long process.

I came across this post here but Rabin's test is even more computationally difficult to do by hand. Is there any other way to tackle this without doing tediously long computations especially if one does not have access to computers?

I will appreciate any suggestions.

approximation
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    If the polynomial was reducible, it would have a root in $\Bbb F_{25}$. It has only $25$ elements and it is straightforward to check if any of them is a root. – markvs Mar 03 '22 at 06:58
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    @markvs Thanks for the insight. Could you please explain why it would have a root in $\mathbb{F}_{25}$? – approximation Mar 03 '22 at 07:07
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    @Aqua: If a quadratic polynomial does not have a root in $\Bbb F_p$ ($p$ a prime), it has a root in the quadratic extension which is $\Bbb F_{p^2}$. – markvs Mar 03 '22 at 10:37
  • Great @markvs Is there elementary (high school) way to prove that fact? – nonuser Mar 03 '22 at 10:41
  • What fact? ${}{}$ – markvs Mar 03 '22 at 10:43
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    Depends on the high school students. Usual high school students do not know finite fields at all. If they know elementary field theory then they know the first part of the statement: irreducible polynomial of degree $n$ has a root in an extension of degree $n$. And if they know about finite fields, they should know that all extensions of degree $n$ of $\Bbb F_p$ are isomorphic to $\Bbb F_{p^n}$. – markvs Mar 03 '22 at 10:52
  • I replaced $c$ by $-a$ and solved for $b$ and $d$ getting $a^2-a+1=0$. What is "inelegant" in this? If you want a more general method, working for larger base fields and higher degree polynomials, then you should mention this in the question box, and in this view the chosen example is a bad one. – user26857 Mar 03 '22 at 16:32

4 Answers4

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I don't know if you'd consider this easier than the direct approach, but here's an alternate method that's pretty reasonable to do by hand. First, note that $f$ is squarefree: $f(2) = 2$ is not a square mod $5$, so $f$ can't be the square of a quadratic (and we already know $f$ has no roots).

Since $f$ is monic of degree $4$, the discriminant $D(f)$ is equal to the resultant $R(f, f')$. Since $f'(x) = 4x^3 + 1$, this is equal to the determinant of the Sylvester matrix $$\begin{pmatrix} 1 & 0 & 0 & 1 & 4 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 4 \\ 4 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 & 1 \end{pmatrix}.$$ Applying elementary row operations, we see that $$D(f) = R(f, f') = \det \begin{pmatrix} 1 & 0 & 0 & 1 & 4 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 2 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix} = 2. $$ By the main theorem of [Leonard 1969], since $D(f)$ is not a square mod $5$, the number of irreducible factors of $f$ does not have the same parity as the degree of $f$. Thus, $f$ can't be a product of two irreducible quadratic polynomials.

Reference:

Daniel Hast
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    Good answer (+1). But isn't this just the fact that the Galois group of an irreducible polynomial of degree $n$ is a subgroup of $A_n$ if and only if its discriminant is a square. If it were the product of two quadratics, the Galois (as a permutation of the roots) would be generated by a product of two disjoint 2-cycles - an even permutation. – Jyrki Lahtonen Mar 03 '22 at 09:56
  • @JyrkiLahtonen, that's a theorem for polynomials over $\mathbb{Q}$, not $\mathbb{F}_5$, isn't it? Over a finite field, all irreducible polynomials have cyclic Galois group (generated by the Frobenius automorphism). Also, the Galois group of a product of two irreducible quadratics is generated by two disjoint transpositions (not by their product), so we still have odd permutations. – Daniel Hast Mar 03 '22 at 14:01
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    See Theorem 4.13. and its Corollary in Jacobson's Basic Algebra I. The result is stated and proven for all fields of characteristic $\neq2$. Actually even irreducibility is not required. Only separability. But the theorem is about the Galois group viewed as a group of permutations of the roots. So when applied to the product of two irreducible quadratics, we look at the Galois group as a subgroup of $S_4$, when the non-trivial element becomes a product of two disjoint 2-cycles. The Frobenius automorphism (=the only non-identity automorphism) permutes both pairs. – Jyrki Lahtonen Mar 03 '22 at 14:54
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    (cont'd) As you pointed out, the Galois group of an extension of a finite field is always cyclic. Hence it cannot contain two disjoint transpositions no matter what - only their product. – Jyrki Lahtonen Mar 03 '22 at 14:58
  • Oh, I see how your argument works now. Yes, I suppose it could also be proven in that way. – Daniel Hast Mar 03 '22 at 18:12
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    That "main theorem" of Leonard is due to Stickelberger, as he says in the paper, and a nice way to think about it is that it gives a formula for the Moebius function on $\mathbf F_p[x]$ for odd primes $p$: for $f(x)$ in $\mathbf F_p[x]$, $\mu(f) = (\frac{{\rm disc} f}{p})$. More generally, if $q$ is an odd prime power then for $f(x)$ in $\mathbf F_q[x]$, $\mu(f) = \chi({\rm disc}(f))$, where $\chi$ is the quadratic character on $\mathbf F_q$, extended to $0$ by $\chi(0) = 0$. I used this point of view in my answer to https://math.stackexchange.com/questions/28281/reducibility-over-mathbbz-2. – KCd Mar 08 '22 at 15:10
  • The point is that when $\mu(f) = -1$, $f$ has an odd number of (monic) irreducible factors, so a quartic with no linear factors and Moebius value $-1$ has only one (monic) irreducible factor. Thus it is irreducible. – KCd Mar 08 '22 at 15:10
  • Resultants can be computed by an algorithmic process that is very similar to Euclid's algorithm for computing a gcd, so you do not need to set up that huge matrix. – KCd Mar 08 '22 at 15:17
  • Here are properties of resultants that can let us compute it in a manner similar to Euclid's algorithm: $R(h_1,h_2)$ is bimultiplicative, $R(h_1,h_2) = (-1)^{\deg h_1\deg h_2}R(h_2,h_1)$, $R(c,h) = c^{\deg h}$ for nonzero constant $c$ and nonzero $h$, and if $h_1 \equiv h_2 \bmod M$ then $R(M,h_1) = R(M,h_2)$ if the polynomial $M$ is monic. Then in $\mathbf F_5[x]$ with $f = x^4 + x+4$, $R(f,f') = R(f',f) = R(-x^3+1,f) = R(-1,f)R(x^3-1,f) = R(x^3-1,2x-1) = -R(2x-1,x^3-1) = -R(2,x^3-1)R(x+2,x^3-1) = -2^3R(x+2,x^3-1) = (-3)R(x+2,1) = -3 = 2$. And ${\rm disc}(f) = R(f,f')$ since $4\mid\deg f$. – KCd Mar 08 '22 at 16:25
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Checking whether a polynomial is irreducbile is a lot like checking whether a number is prime. You need to check there are no degree-two factors of your polynomial. There are ten irreducible polynomials of degree 2 (which you can list easily), so you can check divisibility by those.

On the other hand Rabin's test which you mention is not that bad here. Since $f$ is square-free (check $\gcd(f, f')=1$) and has no roots we know $f \mid x^{5^4}-x$, so we just have to check $\gcd(f, x^{5^2}-x) = 1$. Moreover we can compute $x^{5^2}$ modulo $f$ via $x^{5^2} = (x^5)^5$, which makes this easy. Modulo $f$ we have $$x^{5^2} \equiv (-x^2+x)^5 \equiv -x^{10}+x^5 \equiv -(-x^2+x)^2 -x^2+x\equiv 2x^3 + 3x^2 + 2x + 4.$$ Call this $g$ and now check $\gcd(f, g-x) = 1$ by the usual algorithm.

Sean Eberhard
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  1. $a+c=0$ implies $c=-a$
  2. $b+d+ac=0$ implies $b+d=a^2$
  3. $ad+bc=1$ implies $b-d = a^{-1}$

Hence $b = 3(a^2+a^{-1})$ and $c = 3(a^2-a^{-1})$

  1. $bd=4$ implies $3(a^2+a^{-1}).3(a^2-a^{-1}) = 4$

Hence $1=(a^4-a^{—2})$

Note that $a\neq 0$ so $a^4=1$. Hence $a^{-2}$ equals 0, this is impossible.

Xoff
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This answer uses a little bit of Galois theory.

To simplify the computation a bit, let's replace $x$ by $-x$, and show that $x^4-x+4$ is irreducible. Suppose $a$ is a zero of $x^4-x+4$, it suffices to show $a\not\in\mathbb F_{5^2}$.

Note that $\text{Gal}(\mathbb F_{25}/\mathbb F_5)=\{1, \sigma\}$ where $$\sigma(a) = a^5=aa^4=a(a-4)=a(a+1)=a^2+a$$ Hence $$a + \sigma(a) = a^2+2a \in\mathbb F_5$$ $$a\sigma(a) = a(a^2+a)=a^3+a^2\in\mathbb F_5$$

Now $(a^2+2a)^2-4(a^3+a^2) = a^4 = a+1\in\mathbb F_5$, $a\in\mathbb F_5$, but it's easy to check no element in $\mathbb F_5$ satisfies the polynomial.

Just a user
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