Prove the irreducibility of a polynomial

Question

Let $p(x) = x^2 + ax + b$ a polynomial with $a,b\in\Bbb Z$ odd integers. Prove that $p$ is irreducible at $\Bbb Z[X]$, and at $\Bbb Q[X]$.

I know the Eisenstein's criterion but I'm not sure about how to apply it.

Answer 1

Consider the case over $\mathbb Z$. $p(x)$ is reducible iff it has roots in $\mathbb Z$. If $p(x)$ has roots $c_1, c_2 \in \mathbb Z$, then we have $c_1 c_2 = b$ and $c_1 + c_2 = -a$. If $b$ is odd, then both $c_1$ and $c_2$ are odd. This contradicts with $a$ being odd. Thus, $p(x)$ must be irreducible.

The case over $\mathbb Q$ is equivalent by Gauss's lemma, as pointed by Arturo Magidin in the comments.

Answer 2

To follow up on Claudius's comment, if $p(x)=r(x)s(x)$ in $\Bbb Z[x]$, then $p(x) \equiv r(x)s(x) \pmod{2}$, so $p(x)$ would also be reducible in $\Bbb Z / 2 \Bbb Z[x]$. But $p(x) \equiv x^2+x+1 \pmod{2}$ and it's easy to check that $x^2+x+1$ has no roots in $\Bbb Z /2 \Bbb Z$, so it must be irreducible, which also forces $p(x)$ to be irreducible over $\Bbb Z[x]$.

Answer 3

What happens when

$p(x) = x^2 + ax +b \in \Bbb Z[x] \tag 1$

is reducible over $\Bbb Z$? Well, the factors must be both monic and linear in $x$; thus

$\exists \alpha, \beta \in \Bbb Z, \; p(x) = (x - \alpha)(x - \beta); \tag 2$

we may then write

$p(x) = x^2 + ax + b$ $= (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha \beta, \tag 3$

from which we deduce that

$\alpha + \beta = -a, \; \alpha \beta = b; \tag 4$

since $b$ is odd, both $\alpha$ and $\beta$ must be odd; but the sum of two odds is even, which contradicts the assumption of odd $a$; thus such $p(x)$ are irreducible over $\Bbb Z$.

If $p(x)$ were reducible over $\Bbb Q$, then as above $p(x)$ has two monic first-degree factors, but now in $\Bbb Q[x]$:

$p(x) = (x - \alpha)(x - \beta), \; \alpha, \beta \in \Bbb Q; \tag 5$

we thus may write

$\alpha = \dfrac{r}{s}, \; r, s \in \Bbb Q, \; \gcd(r, s) = 1; \tag 6$

then by the rational root theorem,

$s \mid 1 \Longrightarrow s = \pm 1, \tag 7$

and thus

$\alpha = \dfrac{r}{s} \in \Bbb Z; \tag 8$

whence

$\beta = - a - \alpha \in \Bbb Z \tag 9$

as well, so $p(x)$ is reducible over $\Bbb Z$; but this contradicts what has been shown in the first paragraph; hence, $p(x)$ is $\Bbb Q$-irreducible as well.

Note Added in Edit, Thursday 19 September 2019 7:54 PM PST: The one assertion made above which perhaps deserves further explanation is that $p(x)$ factors into monic factors of the first degree. This is easy to see over $\Bbb Z$, since if

$p(x) = x^2 + ax + b$ $= (\alpha_1 x + \beta_1)(\alpha_2 x + \beta_2) \tag{10}$

then

$\alpha_1 \alpha_2 = 1 \tag{11}$

with

$\alpha_1, \alpha_2 \in \Bbb Z; \tag{12}$

then evidently we must have

$\alpha_1 = \alpha_2 = \pm 1; \tag{13}$

thus we may take the factors in (10) to be monic. If, on the other hand,

$\alpha_1 x + \beta_1, \alpha_2x + \beta_2 \in \Bbb Q[x], \tag{13}$

(11) remains in force, and we write (10) in the form

$p(x) = x^2 + ax + b$ $= \alpha_1 \alpha_2 \left (x + \dfrac{\beta_1}{\alpha_1} \right ) \left (x + \dfrac{\beta_2}{\alpha_2} \right)$ $= \left (x + \dfrac{\beta_1}{\alpha_1} \right ) \left (x + \dfrac{\beta_2}{\alpha_2} \right), \tag{14}$

and we see that $p(x)$ splits into linear monic factors in $\Bbb Q[x]$. End of Note.