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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Because a=-b;c=-ab=a^2;ac=a^3=-sqrt 2. $\endgroup$ – fleablood Mar 18 '17 at 22:50
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    $\begingroup$ 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

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.


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