Prove that $p(x) = x^3 - 3$ is irreducible over $F$. Let $F = \mathbb{Q(i)}$. Prove that $p(x) = x^3 - 3$ is irreducible over $F$.
attempt: Suppose  $F = \mathbb{Q(i)}$. Then let $p(x) = x^3 -3$ ,then $p(x) = x^3 -3$  is irreducible over $\mathbb{Q}$ since it is Eisenstein and since it is a cubic root which has no rational root. 
Hence $[\mathbb{Q}(\alpha): \mathbb{Q}] = 3$ for any root $\alpha$ of $p(x)$.
i.e., the degree of $\alpha$ over $\mathbb{Q }$ is the degree of the extension it generates over $\mathbb{Q }$.
We also know that in $\mathbb{Z(i)}$ , $p(x) = x^3 -3$ is irreducible, where $\mathbb{Z[i]}$ is a UFD. 
And suppose $F = \mathbb{Q(i)}$ is an extension field of $\mathbb{Z(i)}$ containing the root $\alpha$ of $p(x)$ . Then $Z[\alpha] \cong Z[x]/(x^3 - 3)$.
Can someone please help me? I am stuck. I was hoping that because $p(x)$ is irreducible in $Z[i]$ , then that might help with showing $p(x)$ is irreducible in $ F.$
Thank you for any help!
 A: Here is a crude argument. If our cubic is reducible over $\mathbb{Q}(i)$, it has a root in that field. But we can easily write down the roots, and show none is in $\mathbb{Q}(i)$. That basically comes down to the irrationality of $\sqrt[3]{3}$.
Remark: After one has a few tools, "nicer" arguments are available. If $x^3-3$ is reducible over $\mathbb{Q}(i)$, then the field $K$ that contains $i$ and the roots of our cubic is $\mathbb{Q}(i)$ or a quadratic extension of it, and so has degree $2$ or $4$ over the rationals. But $K$  contains $\mathbb{Q}(\sqrt[3]{3})$, which has degree $3$ over the rationals. Since $3$ divides neither $2$ nor $4$, this is impossible.
A: Suppose $p(x)$ is reducible over $Q(i)$. It is cubic and so it must have a root in $Q(i)$. Then the degree of $\min(3^{1/3},Q(i))$ is $1$.
$Q(i,3^{1/3})$ is the field extension of $Q(i)$ containing the root of $p(x)$.
Now $[Q(i,3^{1/3}):Q]=[Q(i)(3^{1/3}):Q(i)][Q(i):Q]$ and also $[Q(i,3^{1/3}):Q]=[Q(3^{1/3})(i):Q(3^{1/3})][Q(3^{1/3}):Q]$.
So we have by equating, $[Q(i)(3^{1/3}):Q(i)][Q(i):Q]=[Q(3^{1/3})(i):Q(3^{1/3})][Q(3^{1/3}):Q]$.
By assumption, LHS is $1\times2=2$ while RHS is divisible by $3$ because $[Q(3^{1/3}:Q]=3$.
So $3$ divides RHS but not LHS, which is a contradiction.
So $[Q(i)(3^{1/3}):Q(i)]\geq2$ showing irreducibility.
A: Yet a third argument: We can still apply Eisenstein over $\mathbb{Z}[i]$, we just need to check that $(3)$ is a prime ideal in $\mathbb{Z}[i]$.
One quick way to do this is to compute $\mathbb{Z}[i]/(3) \cong \mathbb{Z}[X]/(3,X^2+1) \cong (\mathbb{Z}/3\mathbb{Z})[X]/(X^2+1)$, and verify that $X^2+1$ is irreducible in $\mathbb{Z}/3\mathbb{Z}$.
