Irreducibility of a cubic polynomial

Question

Let $f(x)=x^3+2x^2+x-1$. Then over which of the following fields $k$ is $f$ irreducible?

$k=\mathbb{Q}$

$k=\mathbb{R}$

$k=\mathbb{F}_2$

$k=\mathbb{F}_3$

My Attempt: (2) $f$ is cubic and hence it has a real solution, $\implies$ is not irreducible over $k=\mathbb{R}$

(1) Considering $f'$ and $f''$ and Descartes' Rule of Signs, I concluded that it has only one real root which is positive. Eisenstein Criterion fails. By Rational Root test, if the real root of $f$ is rational, say $\frac{p}{q}$, then $p$ is a factor of $-1$ and $q$ is a factor of $1\implies \frac{p}{q}=1$ (since real root is positive), and since $f(1)\ne0$ we can conclude that it has no rational roots and hence it is irreducible over $\mathbb{Q}$.

(3) Over $\mathbb{F}_2$, $f$ is irreducible (since $x,x^2+1,x^2+x+1$ does not divide $f$)

Is it correct so far? And how do I justify for $\mathbb{F}_3$?

Also, though (2) (looks like) it works, I feel I am overkilling it, is there any other way to justify that $f$ is irreducible over $\mathbb{Q}$?

For (2), I would have used the Rational root theorem at the first step, and then you have to check -1 and 1. — wythagoras, May 30 '15 at 07:29
For the sake of precision, your approach to (1) needs to use the fact it has no negative roots. The way you phrased things, it's unclear if you checked that. — , May 30 '15 at 07:31
@wythagoras, either way, its just a seconds business when not written! — Jesse P Francis, May 30 '15 at 07:38
For $\Bbb F_2$ and $\Bbb F_3$, you can just check whether $f$ has roots, since it's a third-degree polynomial. — Arthur, May 30 '15 at 07:39

score 1 · Accepted Answer · answered May 30 '15 at 07:32

1

For $\mathbb{Q}$, just go straight to the rational root test: any rational root would have to be $1$ or $-1$, but those aren't roots, since $f(1)=3$ and $f(-1)=-1$.

Similarly, for $\mathbb{F}_2$ and $\mathbb{F}_3$, you can check directly whether there are roots, which suffices for cubic polynomials. Alternatively, for $\mathbb{F}_3$, observe that $$x^3+2x^2+x-1=x^3-x^2+x-1=(x-1)(x^2+x+1)$$

answered May 30 '15 at 07:32

Zev Chonoles

129,973

Thank you! And for $\mathbb{Q}$, is there any other way to prove irreducibility than Rational Root Test? – Jesse P Francis May 30 '15 at 11:21

Irreducibility of a cubic polynomial

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