1
$\begingroup$

I think it can be shown that $x^3-\sqrt{2}$ is irreducible by arguing that $x=\sqrt[6]{2}$, which is the only real solution, is not in $\mathbb{Q}(\sqrt{2})$, so the polynomial has no solutions over this field, which implies that the polynomial cannot be expressed as a product of polynomials of degree 1 or 2.

However, I'm offered to prove the irreducibility by showing that the polynomial cannot be expressed as a product of polynomials of $\deg 1$ and $\deg 2$. I'm wondering if this is not too complicated to do so as opposed to my version of the solution? In any case, I don't know how to show it the way I'm offered. Maybe I'm having a lapse in my knowledge of the theory. Would appreciate one's insight.

$\endgroup$
7
  • $\begingroup$ The two methods are more or less the same. For your solution, make sure to note also that if a polynomial $f(x)$ has real coefficients, than neither or both (but not exactly one) of the complex roots of $x^3-\sqrt2$ are roots of $f(x)$. $\endgroup$ – Greg Martin Mar 18 '17 at 6:42
  • 1
    $\begingroup$ Well, we could just try it. (x+a)(x^2+bx+c)=x^3+(a+b)x^2+(ab+c)x+ac so a+b=0, ab+c= 0, and ac=-sqrt (2). Soa^3=- sqrt 2 which isn't possible I think. $\endgroup$ – fleablood Mar 18 '17 at 6:43
  • 4
    $\begingroup$ Your argument is ok. An alternative that comes to mind would be to use the facts that $\Bbb{Z}[\sqrt2]$ is a PID, $\sqrt2$ is a prime there, and apply Eisenstein's criterion. $\endgroup$ – Jyrki Lahtonen Mar 18 '17 at 6:44
  • 1
    $\begingroup$ Because a=-b;c=-ab=a^2;ac=a^3=-sqrt 2. $\endgroup$ – fleablood Mar 18 '17 at 22:50
  • 1
    $\begingroup$ If it can't be factored by degree 1 and 2, it won't be factorable by the degree 1 either. $\endgroup$ – fleablood Mar 18 '17 at 22:53
1
$\begingroup$

Let $K=\mathbb{Q}(\sqrt{2})$, and let $f = x^3 - \sqrt{2}$.

Since $f \in K[x]$ is cubic, to show $f$ is irreducible in $K[x]$, it suffices to show $f$ doesn't have a root in $K$.

Suppose otherwise. Thus, suppose $r^3=\sqrt{2}$, for some $r \in K$.

Since $r \in K$, we can write $r = a + b\sqrt{2}$, for some $a,b \in \mathbb{Q}$.

Then $r^3 = \sqrt{2}$ implies $\,r$ is not rational, hence $b \ne 0$. Then

\begin{align*} &r^3 = \sqrt{2}\\[4pt] \implies\; &(a + b\sqrt{2})^3 = \sqrt{2}\\[4pt] \implies\; &\left(a(a^2+6b^2)\right) + \left(b(3a^2+2b^2)\right)\sqrt{2} = \sqrt{2}\\[4pt] \implies\; &\left(a(a^2+6b^2)\right) + \left(b(3a^2+2b^2)-1\right)\sqrt{2} = 0\\[4pt] \implies\; &a(a^2+6b^2)=0\;\;\text{and}\;\;b(3a^2+2b^2)-1=0\\[4pt] \end{align*}

Then $\,a(a^2+6b^2)=0 \implies a = 0\;\;$(since $b \ne 0$).

But then, $\,b(3a^2+2b^2)-1=0 \implies 2b^3 - 1 = 0$,

$\qquad$contradiction, since by the rational root test, the polynomial $2x^3 - 1$ has no rational roots.

It follows that $f$ is irreducible in $K[x]$, as was to be shown.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.