Prove the following polynomials are irreducible over $\mathbb Z_5[x]$: $2x^3+x^2+4x+1$ and $x^4+1$.

So the exponents are already as reduced as they can be in terms of $\mathbb Z_5$. I cannot find any rational roots by factoring or using the rational root test. I don't believe I can't use Eisenstein's criterion on the first polynomial, the second polynomial concludes irreducibility over $\mathbb Q$. Can I then conclude that since no rational roots can be found that they are irreducible over $\mathbb Z_5$?

EDIT: Using N.S.'s advice I will use $a(x)=2x^3+x^2+4x+1$ where $x=0,-1,1,-2,2$ which all give numbers other than zero, thus there are no roots in $\mathbb Z_5[x]$.

  • $\begingroup$ Irreducibility over $\mathbb{Q}$ does not imply irreducibility over $\mathbb{Z}_5$. For example $x^2 + 1$ is irreducible in $\mathbb{Q}$ but factors as $(x - 2)(x+2)$ in $\mathbb{Z}_5$. $\endgroup$ – Tob Ernack Nov 15 '17 at 16:09

Since you work over $\mathbb Z_5$, rational root test is useless.

For the first one, since it has degree 3, it is irreducible if and only if it doesn't have roots. And there are only 5 possible roots in $\mathbb Z_5$, try them.

For the second, note that $$X^4+1=X^4-4$$

  • $\begingroup$ Since $x^4-4= (x^2-2)(x^2+2)$ doesn't that make this polynomial reducible? Should I go further and say $(x^2+2)(x^2-2)$ is irreducible over $\mathbb Z_5[x]$? $\endgroup$ – K Math Nov 15 '17 at 16:23
  • $\begingroup$ @KMath Exactly, it is reducible. Now what you mean by the last statement is that each of $X^2-2$ and $X^2+2$ are irreducible. $\endgroup$ – N. S. Nov 16 '17 at 15:46

$\mathbb{Z}_5[x]$ is not a field, hence it does not make sense to discuss the irreducibility or reducibility of those polynomials over $\mathbb{Z}_5[x]$. On the other hand $\mathbb{F}_5$ is a field. $x^4+1=\Phi_8(x)$ factors over any finite field. In the $\mathbb{F}_5$ case we have $$ x^4+1 = (x^2-2)(x^2+2) $$ and $x^2\pm 2$ is irreducible over $\mathbb{F}_5$ since $\pm 2$ is not a quadratic residue $\!\!\pmod{5}$.
In order to prove that $2x^3+x^2-x+1$ is irreducible over $\mathbb{F}_5$ it is enough to prove it has no linear factors, i.e. no roots in $\mathbb{F}_5$. $0$ is not a root and if $x\in\mathbb{F}_5^*$ we either have $x^2=1$ or $x^2-1$. In the former case $$2x^3+x^2-x+1 = 2x+1-x+1 = x+2 \neq 0, $$ in the latter case $$2x^3+x^2-x+1 = -2x-1-x+1 = -3x \neq 0 $$ and we are done.


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