Proving irreducibility in $\Bbb Q[x]$

Question

How can I prove $f(x) = x^3 - 3x^2 + 9$ is irreducible in $\Bbb Q[x]$?

I tried doing $f(x+1)$ and then applying Eisenstein Theorem but it does not apply to $f(x+1) = x^3 - 3x + 7$

Answer 1

Since it's a degree 3 polynomial irreducibility is equivalent to not having any root in $\mathbb{Q}$ but we know that if a root $r \in \mathbb{Q}$ than $r = \frac{a}{b}$ and for rational root theorem we know that $a|9, b|1$ so the possible solution are $\{ \pm1, \pm 3, \pm9\}$, we can see that

1. $f(1) = 1 - 3 +9 = 7$

2.$f(-1) = -1 - 3 +9 = 5$

3.$f(3) = 27 - 9 +9 = 27$

4.$f(-3) = -27 - 9 +9 = -27$

5.$f(9) = 9^3 - 27 +9 ≠ 0$

6.$f(-9) = -9^3 - 27 +9 ≠ 0$

and we conclude that $f(x)$ is irreducible in $\mathbb{Q}[x]$

Answer 2

Rational root theorem would be your best criterion. However, here is another solution that uses Eisenstein criterion. We need two easy facts to prove $f(x)$ is irreducible. First, we define the reciprocal of a polynomial.

Let $f(x)$ be a polynomial of degree $n$. The reciprocal polynomial is defined as $f^{*}(x)=x^nf(1/x)$. Note that if a polynomial is not $x$ and irreducible then its reciprocal is irreducible (proof is exercise).

Second, for $f(x)$ is a polynomial over $\mathbb{Q}$ and $a \in \mathbb{Q}$. It is true that $f(ax)$ is irreducible over $\mathbb{Q}$ iff $f(x)$ is also irreducible. (proof is easy too).

For your question, $f(3x)=27x^3-27x^2+9$. Taking the reciprocal of $f(3x)$, you get $9x^3-27x+27=9(x^3-3x+3)$ and $x^3-3x+3$ is irreducible by your favorite criterion!

A deeper reason is the following; inf fact, $f(x)$ satisfies Dumas criterion which is a more general criterion than Eisenstein. The criterion is:

Let $p$ be a prime and $f(x)=a_nx^n+\ldots+a_0 \in \mathbb{Z}[x]$. Suppose that for every $i=0,\ldots,n$. we define $e_i$ to be the highest power of the prime $p$ dividing the coefficient $a_i$. if $f(x)$ satisfies the following:

1. $e_n=0$.
2. $\frac{e_i}{n-i}>\frac{r_0}{n}$ for all $i=1,\ldots,n-1$.(neglect the zero coefficients)
3. gcd $(r_0,n)=1$

Then, $f(x)$ is irreducible over $\mathbb{Q}$. Note that Eisenstein criterion is a special case when $r_0=1$ and the second condition automatically follows.

For your example, $f(x)=x^3-3x^2+9$, the prime is $3$ and the degree is 3. $e_3=0$, $e_2=1$ and $e_0=2$ Applying the conditions, we get

1. $e_3=0$
2. $\frac{1}{3-2}>\frac{2}{3}$
3. gcd$(2,3)=1$

So, $f(x)$ is irreducible over $\mathbb{Q}$