Polynomial x$^3$ - 2 is not the product of two nonconstant polynomials $f$ = $gh$

Question

This is an example out of Artin's Algebra book on page 334, chapter 11.

He states that x$^3$ - $2$ is in the kernel of the homomorphism $\phi$: $\mathbb{Q}$[x] -> $\mathbb{C}$ via the substitution map that sends x -> 2$^{\frac{2}{3}}$.

Artin's goal is to show that $f$ = x$^3$ - $2$ is the monic polynomial of lowest degree in the kernel, and since $\mathbb{Q}$ is a field, we have that the principle ideal, ($f$), is the kernel of $\phi$.

His reasoning is: "The polynomial x$^3$ - $2$ is in the kernel, and because 2$^{\frac{2}{3}}$ is not a rational number, it is not the product $f$ = $gh$ of two nonconstant polynomials with rational coefficients. So it is the lowest degree polynomial in the kernal, and therefore it generates the kernel."

My question is: Why does the fact that 2$^{\frac{2}{3}}$ is irrational allow us to conclude x$^3$ - 2 is not the product of two constant polynomials $g$ and $h$?

Answer 1

If $x^3-2$ is reducible over $\Bbb Q$ then it must have a linear factor in $\Bbb Q[x]$ implying a rational root but with $2^{\frac{2}{3}}$ irrational so is the only real root $2^{\frac{1}{3}}$. Therefore, $\; x^3-2$ is irreducible over $\Bbb Q$.

Answer 2

Let $\,b:=2^\frac13\,$ and $\,1\ne\,e\in\Bbb C\,$ be the such that $\,e^3=1.\,$ Then polynomial $\,x^3-2\in\Bbb Q[e],\,$ while

$$ x^3-2\,=\,(x-b)\cdot(x-e\cdot b)\cdot(x-e^2\cdot b) $$

None of the linear factors on the right belongs to $\,\Bbb Q[e]\,$ hence there is no decomposition of $\,x^3-2\,$ in $\,\Bbb Q[e].$

Great!

Answer 3

Here we can use some classic one-variable calculus.

$$f(x)= x^3-2$$ Must have at least one real root in $x\in[1,2]$, since we have $$f(1)=-1<0\\f(2)=6>0$$ Together with Bolzano's theorem (really cool mathematician by the way, you should check him out!).

We can furthermore prove that this is the only root, since derivative $$f'(x) = 3x^2$$ exists and is non-negative everywhere. In fact the root is $x = \sqrt[3]{2}$, so the only factorization we can do over real numbers ($\mathbb R$) would be with the factor $(x-\sqrt[3]{2})$ but that number is irrational and therefore not in $\mathbb Q$ which was requested.