3

One of my homework problems wants me explain why $x^2 +2$ is irreducible in $\mathbb{Z}_5$. The possible roots of $x^2+2$ are $x=\pm \sqrt{-2} \notin \mathbb{Z}_5$. Is it enough to say that since $x^2+2$ has no roots in $\mathbb{Z}_5$, it must be irreducible in $\mathbb{Z}_5$? Is this true for any polynomial in any number system?

TheMobiusLoops
  • 1,359

3 Answers3

7

For a second degree polynomial, they are in fact equivalent. For if a second degree polynomial does not have any root, it cannot have any factors of lower degree, since a factor of degree $1$ implies a root.

On the other hand, it's not true for higher degrees. The polynomial $x^4 + 2x^2 + 1$ has no real roots, but is not irreducible.

5

When you say that $\sqrt{-2}\notin\Bbb Z_5$, do you mean that there is no $x\in\Bbb Z_5$ such that $x^2=-2=3$ in $\Bbb Z_5$? If so, it is enough, since $x^2+2$ is only quadratic, but you have to show that it’s actually true; as it stands, it’s just a restatement of the problem. The simplest way to show that it’s true is just to test each element of $\Bbb Z_5$.

If, however, you’re thinking that $\sqrt{-2}$ isn’t even real, therefore it can’t be in $\Bbb Z_5$, the argument is completely incorrect: the same argument would tell you that $x^2+2$ has no solution in $\Bbb Z_{11}$, which is clearly false, since $x=3$ is a solution.

Brian M. Scott
  • 616,228
2

Some fourth-degree polynomials over some fields are reducible without having roots, since some fourth-degree polynomials can be factored as a product of two second-degree polynomials. Similar remarks apply to polynomials of degree more than $4$. Second- and third-degree polynomials are reducible only if they have roots.

So \begin{align} 0^2 & \equiv 0 \not\equiv -2 \pmod 5, \\ (\pm1)^2 & \equiv 1 \not\equiv -2 \pmod 5, \\ (\pm2)^2 & \equiv 4 \not\equiv -2 \pmod 5. \end{align} That takes care of all five members of $\mathbb Z/5$.

Michael Hardy
  • 1