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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 $\Bbb Z_5[x]$.

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    $\begingroup$ @OnurcanB. $\Bbb F_5$ is $\Bbb Z/5\Bbb Z$. $\endgroup$ – Santosh Linkha Jul 21 '17 at 7:17
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    $\begingroup$ @OnurcanB. anything works lol. $\endgroup$ – Santosh Linkha Jul 21 '17 at 7:20
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jul 21 '17 at 7:45
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    $\begingroup$ @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). $\endgroup$ – Jyrki Lahtonen Jul 21 '17 at 7:51
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    $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Jul 21 '17 at 8:01
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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 $a\not\equiv 0 \pmod{5}$.

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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 $a^4 - 3 a^2 b + b^2 - 1 \equiv 0 \mod 5$.

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The only possibilities that your polynomial can factor is to degree 1 and 4 or two degree 2 and 3. You already know that there is no linear factor, so the first case is omitted. For the second case, do the polynomial division for the 32 irreducibles and if it does not factor the polynomial is irreducible :-)

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  • $\begingroup$ haha .. that's the problem. there are 32 of these, I am required to solve this within 15 min $\endgroup$ – Santosh Linkha Jul 21 '17 at 7:21
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    $\begingroup$ I admit that Robert's idea is better. But, however, Mathematica should be able to divide 32 polynomials in under 15 minutes. :-) $\endgroup$ – Rofl Ukulus Jul 21 '17 at 7:25

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