Field extensions and irreducibility

Question

I'm having trouble trying to show that the function f=x^3 + x + 3 is irreducible in the rationals. I tried using Eisensteins criterion but it didn't work as it doesnt satisfy all conditions.

the second part of the question is to show that f has one real root which i showed using Rolle's Theorem from Analysis

the third part is showing that denoting the real root as x Q(x)=Q(x^2) where Q(a) [represents the field extension over rationals] My attempt at this question is to show that every member of Q(x) is contained in Q(x^2), by using closure under field operations, but I'm struggling to show that every member of Q(x^2) is inside Q(x).

Any help would be greatly appreciated.

Answer 1

Here is what I think for the first question:

Suppose $f(x) = x^{3} + x + 3$ is not irreducible over $\mathbb{Q}$. Then we can write $f(x)$ as the product $(x + a)(x^{2} + bx + c)$ for some rational numbers $a, b, c$.

But then, multiplying this out gives the expression $x^{3} + bx^{2} + cx + ax^{2} + abx + ac = x^{3} + (b + a)x^{2} + (ab + c)x + ac$

And we know this must equal $x^{3} + x + 1$, so we know that:

1. $b + a = 0$
2. $ab + c = 1$
3. $ac = 3$

But if a polynomial is irreducible over $\mathbb{Z}$, then it is irreducible over $\mathbb{Q}$, so we really only need to find a contradiction with the above equations if $a,b,c$ are integers.

Equation 3 means either $a = 1$, $c = 3$ or $a = 3$, $c = 1$, or $a = -1$, $c = -3$, or $a = -3, c = -1$. But equation 2 implies $b = -a$, so:

If $a = 1$, $c = 3$, then $b = -1$, so from equation 2 we get $ab + c = 1(-1) + 3 = -1 + 3 = 2 \neq 1$. So equation 2 is not satisfied.

If $a = 3$, $c = 1$, then $b = -3$, and by equation 2 we get $3(-3) + 1 = -9 + 1 = -8 \neq 1$, so again equation 2 is not satisfied.

If $a = -1$, $c = -3$, then $b = 1$ and so $(-1)1 - 3 = -4 \neq 1$, which means equation 2 is yet again not satisfied.

Finally, if $a = -3$, $c = -1$, then $b = 3$, and so $(-3)3 - 1 = -10 \neq 1$, which means equation 2 is yet again not satisfied.

So in every case, we can't find a solution of integers that satisfies the three equations, and thus the polynomial is not reducible over $\mathbb{Z}$. Consequently, it is not reducible over $\mathbb{Q}$.

Answer 2

Being of degree $3$, it's reducible over $\mathbb{Q}$ if and only if it has a root in $\mathbb{Q}$ ( that is, a factor of degree $1$).

$\bf{Fact}$: if a rational number is a root of a monic polynomial with integer coefficients that numbers is an integer.

$\bf{Fact}$: An integral root of a monic polynomial with integer coefficients is a divisor of the free term.

There are $4$ divisors of the free term in your case $\pm 1$, $\pm 3$ and none is a root of $x^3 + x+3$, hence the polynomial having not rational roots is irreducible.

ALternatively: a (primitive) polynomial with integer coefficients is irreducible over $\mathbb{Q}$ if and only if it's irreducible over $\mathbb{Z}$. Now your polynomial is irreducible over $\mathbb{Z}$ since it is irreducible $\mod 2$ ( no root $\mod 2$, odd value for any integer, and $\deg 3$).

Second part, note that $f'>0$ so $f$ is strictly increasing.

For the third part: We have the extensions $\mathbb{Q} \subset \mathbb{Q}(x^2) \subset \mathbb{Q}(x)$. The degree of the intermediate extension $\mathbb{Q} \subset \mathbb{Q}(x^2)$ is not $1$ (since $x^2 \not \in \mathbb{Q}$) and divides the degree of $\mathbb{Q} \subset \mathbb{Q}(x)$, which is $3$, so it must be $3$ also, and so $\mathbb{Q}(x^2) = \mathbb{Q}(x)$.

Or, a hands-on approach: $x^2 + x + 3 =0$ so $x^4 + x^2 + 3 x = 0$ and so $x = -\frac{1}{3}(x^2)^2 - \frac{1}{3} x^2$.