How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$? [duplicate]

How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$?

I checked that there are no linear factors since there is no root. How to know that there is no quadratic factor? I checked from Mathematica that there are $32$ irreducible quadratic in $\Bbb Z_5[x]$.

• @OnurcanB. $\Bbb F_5$ is $\Bbb Z/5\Bbb Z$. – Santosh Linkha Jul 21 '17 at 7:17
• @OnurcanB. anything works lol. – Santosh Linkha Jul 21 '17 at 7:20
• This is a special case of this oft recurring slightly more general question. I probably shouldn't be the first to vote to close this as a duplicate because I answered that. – Jyrki Lahtonen Jul 21 '17 at 7:45
• @OnurcanB. People often want to write $\Bbb{F}_5$ instead of $\Bbb{Z}_5$. There are many reasons for this. A) This makes the notation of all finite fields a bit more uniform as, for example, the field of nine elements can then be denoted $\Bbb{F}_9$, whereas $\Bbb{Z}_9$ is something else. B) In some contexts the finite fields appear together with the rings of $p$-adic integers, and those are always denoted $\Bbb{Z}_p$ (and in such a context $\Bbb{Z}/p\Bbb{Z}$ is the only possible way to denote a ring of residue classes of integers). – Jyrki Lahtonen Jul 21 '17 at 7:51
• @Amin235 I'm not sure I understand. As long as what is inside the parens is a power of a prime number, the field is specified uniquely up to isomorphism. In 1) I would be inclined to think that $p$ is a prime, and $q$ can be any positive integer. In 2) I would be inclined to think that $q$ is not necessarily a prime, but can be a prime power instead, and $k$ can be any positive integer. – Jyrki Lahtonen Jul 21 '17 at 8:01

We will show that $P(x):=x^5 -x+r$ is irreducible in $\Bbb Z_5[x]$ for $r=1,2,3,4$.
Since $P(k)\equiv r\pmod{5}$ for $k\in \Bbb Z_5[x]$, $P$ has no linear factors and if it is reducible then $P(x)=Q(x)T(x)$ where $Q$ has degree $2$ and $T$ has degree $3$.
Consider the isomorphism $f: \Bbb Z_5[x]\to \Bbb Z_5[x]$, given by $f(P)(x)=P(x+1)$. Then the polynomial $P(x)=x^5-x+r$ is a fixed point of $f$: $$f(P)=f(x^5-x+r)=(x+1)^5-(x+1)+r=x^5+1-x-1+r=x^5-x+r=P.$$ Therefore we should have $f(Q)=Q$ and $f(T)=T$ ($f$ is degree-invariant). Now we show that $f$ has not any fixed polynomial of degree $2$: $$f(ax^2+bx+c)=a(x+1)^2+b(x+1)+c=ax^2+(2a+b)x+(a+b+c)\\=ax^2+bx+c$$ iff $2a\equiv 0 \pmod{5}$ and $a+b\equiv 0 \pmod{5}$ which is impossible because $a\not\equiv 0 \pmod{5}$.
By long division, if an $x^2 + a x + b$ divided $x^5 - x + 3$, you would need $a^4 - 3 a^2 b + b^2 - 1 \equiv 0$ and $a^3 b - 2 a b^2 + 3 \equiv 0 \mod 5$. You can "complete the square" in $a^3 b - 2 a b^2 + 3 \mod 5$ to obtain $3 a (b + a^2)^2 + 2 a^5 + 3 \equiv 3 a \left((b+a^2)^2 +4 a^4 + 1/a\right)$. For this to be $0$ mod $5$, we need $a^4 - 1/a$ to be a quadratic residue mod $5$ (thus $0$, $1$ or $4$); the possibilities are then $(a,b) = (1,4), (3,3)$ and $(3,4)$. But none of those cases makes $a^4 - 3 a^2 b + b^2 - 1 \equiv 0 \mod 5$.