Prove that there exists no polynomial in $\mathbb{Q}[x]$ of degree 1 or 2 which divides $x^3-2$. I assume I need to look at the ideal $\left<x^3-2\right>$, but I do not know how to proceed. 
Hints only please!
 A: Hint. A famous irreducibility criterion is applicable here.
A: Here’s another approach: If this polynomial was to factor, it would have to have a root in $\mathbb Q$.
If you can figure out why this is true, and then prove that the conclusion is false,  you’re done.
A: Oooh, I love this question! It's always really fun showing things are irreducible! I'll try not to spoil all the fun!
Okay okay, so the goal is to prove $x^3-2$ is irreducible in $\mathbb{Q}[x]$. Yes, I'm aware that's not what the question directly says, but it's what it's indirectly saying. Because, hypothetically, if we could factor it, we would have for some $f,g,h \in \mathbb{Q}[x]$:
$$x^3-2 = f(x)g(x)h(x)$$
where only these cases are possible:


*

*$\deg f = 2, \deg g=1,$ and $\deg h = 0$ (that is to say, $h \in \mathbb{Q} \backslash \{0\}$ and is a "constant"), and

*$\deg f = \deg g = \deg h =1$.


So showing that nothing of degree $1$ or degree $2$ divides $x^3-2$ essentially shows it's irreducible in $\mathbb{Q}[x]$! 
Alright, I'm going to have to spoil half of the problem because it's impossible not to -- it's just too straightforward. But don't worry, the other, more worthwhile one is all yours. For the degree $1$ case, let's assume that $x^3-2$ has a factor of degree $1$. If it has such a factor, then it must have a root, correct?
To elaborate in case you don't see why, note that if $f\in \mathbb{Q}[x]$ is a polynomial of degree $1$ then it must be of the form
$$a_0+a_1x$$
for $a_0, a_1 \in \mathbb{Q}$. Now specifically because $\mathbb{Q}$ is a field, the element $a_1^{-1} \in \mathbb{Q}$ and more importantly we have $-a_1^{-1}a_0 = -\frac{a_0}{a_1} \in \mathbb{Q}$. Therefore the polynomial $f(x)$ has a zero:
\begin{align*}
f\left(-\frac{a_0}{a_1}\right) &= a_0+a_1 \left( -\frac{a_0}{a_1} \right) \\
&= a_0+(-a_0) \\
&= 0
\end{align*}
So now let's go back to the problem! As shown above, if $x^3-2$ had a factor of degree $1$, then it must have a root in $\mathbb{Q}$. That is, an $x \in \mathbb{Q}$ that satisfies $x^3-2=0$ but that $x$ would also have to satisfy
$$x^3=2$$
as well -- but hey! Wait a minute! $\sqrt[3]{2}$ clearly isn't in $\mathbb{Q}$! And that's the end of that.
Now the other part, showing now factors of degree $2$, follows literally directly from this. I'll lay it out one by one below in spoilers, but try it first!
Assumption:

 Assume $x^3-2$ has a factor of degree $2$. Let that be some polynomial $f \in \mathbb{Q}[t]$.

First Implication:

 We then have $f|(x^3-2) \implies x^3-2 = fg$, where $g \in \mathbb{Q}[t]$.

Final Implications:

 $\deg(x^3-2) = \deg(fg) \implies 3 = \deg f + \deg g \implies \deg g =1$, but we had shown that $x^3-2$ has no factors of degree $1$, thus this is a contradiction and $x^3-2$ has no factors of degree $2$ either.

