Determine the irreducibility of $x^4 + x + 1$ in $\Bbb{Q}[x]$.
Well, suppose there is a rational solution $\frac{c}{d}$ to this polynomial. Then $c | \pm 1$ and $d | \pm 1$, implying the only possible factorization in $\Bbb{Q}[x]$ would be $(x - 1)(x^3 +\beta x + \rho)$. But $1$ or $-1$ isn't a solution, as $f(1) \neq 0$, so this polynomial is irreducible.
Determine the irreducibility of $x^3 - \frac{x^2}{2} - \frac{1}{2}$.
Let $f(x) = x^3 - \frac{x^2}{2} - \frac{1}{2}$. Then $2f(x) = 2x^3 - x^2 - 1$.
Well, suppose there is a rational solution $\frac{c}{d}$ to this polynomial. Then $c | \pm 1$ and $d | \pm 2$. Thus our possible solutions are from the set $$\pm \{1, \frac{1}{2}\}$$
But plugging these in, we get $f(1) = 0$, $f(-1) \neq 0$, $f(\frac{1}{2}) = \frac{1}{4} - \frac{1}{4} - 1 \neq 0$ and $f(-\frac{1}{2}) \neq 0$ as well. Thus $2x^3 - x^2 - 1 = (x - 1)(x^2 + \beta x + \rho)$. (I stop here as I do not need to calculate the work; see question.)
Question: Are my solutions correct?
Question: Another solution to the second problem (and perhaps the first) involves using this proposition:
Proposition: Suppose that $f(x)$ is in $\mathbf{Z}[x]$ and $p$ is a prime. Let $f_p(x)$ be the associated polynomial in $\mathbf{Z}_p[x]$ obtained by reducing the coefficients of $f(x) \mod p$. If deg$(f(x)) =$ deg$f_p(x))$ and if $f_p(x)$ is irreducible in $\mathbf{Z}_p[x]$, then $f(x)$ is irreducible in $\mathbf{Z}[x]$.
I don't really grasp the proposition and would like any more concrete examples and hints for a solution using this proposition. The authors of these exercises mention that "Since it is a cubic polynomial, we can check to see if it has a nontrivial factorization by checking to see if it has a root in $\mathbf{Z}_3$". BUT WHY does it necessarily have to be $\mathbf{Z}_3$??
I wholeheartedly appreciate it.
There is a general result due to Gauss, that if a polynomial with integer coefficients factors properly over the rationals, then it factors properly over the integers.
To prove the irreducibility of $x^4+x+1$ over the integers, we note that since the polynomial has no rational zeros, any factorization would be as a product of quadratics with integer coefficients. Without loss of generality we may assume that the lead coefficient of each quadratic is $1$. So the factorization can be taken to be of shape $(x^2+ax+b)(x^2+cx+d)$. A little exploration shows this is impossible. The coefficient of $x^3$ is $a+c$, so $c=-a$. It is clear that $b=1,d=1$ or $b=-1,d=-1$. The coefficient of $x$ in the product is therefore $0$ (false). So we have shown that $x^4+x+1$ is irreducible over the integers, and hence the rationals.
We redo the above calculation, showing $x^4+x+1$ is irreducible over $\mathbb{Z}_2[x]$. There is no zero in $\mathbb{Z}_2$. So any factorization must be as a product of quadratics, say $(x^2+ax+b)(x^2+cx+d)$.
It is clear that $a=c$ and $b=d=1$. It follows that the coefficient of $x$ in the product is $0$. But that is not true.
What does that say about reducibility over the integers? If the polynomial $x^4+x+1$ split over the integers, then taking the coefficients modulo $2$, the splitting over the integers would give a splitting over $\mathbb{Z}_2[x]$. We have shown there is no such splitting.
Remark: Unfortunately, in this case the direct proof of irreducibility over the integers is too easy, so using $\mathbb{Z}_2$ offers no clear advantage.