I want to check whether $f(x)=x^3+2x^2+x-1$ is irreducible over $\mathbb Q, \mathbb R, \mathbb Z_2, \mathbb Z_3$?

Definitely, since $f(x)$ is a polynomial of degree 3, therefore, if it is reducible over any of the fields, there exists a zero of $f(x)$ on that field. But I do not think there is any zero of the polynomial in any of the given fields.

Am I right?


Well, in $\Bbb R$ it must have a root, as $\displaystyle\lim_{-\infty}f=-\infty$ and $\displaystyle\lim_{+\infty}f=+\infty$, and also, $x=1$ would do it for $\Bbb Z_3$. Else, correct:)


Since the degree of the polynomial is odd, you know that it reduces over $\mathbb{R}$. This is because there is a real root of the polynomial. To determine whether the root is rational you can try to apply the Rational Root Theorem.

For roots over $\mathbb{Z}_2$ and $\mathbb{Z}_3$ try to look for roots. This is easily done because these fields are small.

  • $\begingroup$ By Rational Root theorem, 1 and -1 are the only possible zeros. Definitely, they are not zeros of $f(x)$. $\endgroup$ – Anupam Feb 24 '14 at 1:42
  • $\begingroup$ @Anupam: That is correct. $\endgroup$ – Thomas Feb 24 '14 at 1:43

HINT: Recall the Mod $p$ Irreducibility Test. I shall state the test for you.

Let $p$ be a prime and suppose that $f(x)\in \mathbb Z[x]$ with $\deg f(x)\ge 1$. Let $\bar f(x)$ be the polynomial in $\mathbb Z_p[x]$ obtained from $f(x)$ by reducing all the coefficients of $f(x)$ modulo $p$. If $\bar f(x)$ is irreducible over $\mathbb Z_p$ and $\deg\bar f(x)=\deg f(x)$, then $f(x)$ is irreducible over $\mathbb Q$.

Another Hint: There is a theorem that states,

Let $f(x)\in \mathbb Z[x]$. If $f(x)$ is reducible over $\mathbb Q$, then it is reducible over $\mathbb Z$.

Hopefully these hints help!


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