Show that $\{1, \alpha, \alpha^2, \alpha^3\}$ is an integral basis in $k$, ( i.e. show that $\mathcal{O}_k = \Bbb{Z}[\alpha]$.) Let $k = \Bbb{Q}(\alpha)$ where $\alpha$ is a root of the polynomial $m(X) = X^4 − 3.$
I have already shown $m$ is irreducible and worked out the the discriminant of $\{1, \alpha, \alpha^2, \alpha^3\}$. The discriminant is $-6912$ which has 2 square divisors $2$ and $3$.
Show that $\{1, \alpha, \alpha^2, \alpha^3\}$ is an integral basis in $k$, ( i.e. show that $\mathcal{O}_k = \Bbb{Z}[\alpha]$.)
Any help would be much appreciated!
 A: I can recommend the lecture notes for Algebraic Number Theory by Stevenhagen (http://websites.math.leidenuniv.nl/algebra/ant.pdf). Theorem $3.1$ of these notes handles this problem very quickly.
We need to show that the number ring $\mathbb{Z}[\alpha]$ is regular over $2$ and $3$. Following the theorem: $X^4-3\equiv (X-1)^4\bmod 2$, and the remainder of $m(X)$ upon division by $X-1$ is $m(1)=-2\notin 4\mathbb{Z}[X]$, so $\mathbb{Z}[\alpha]$ is regular over $2$.
For $3$ we have $X^4-3\equiv X^4 \bmod 3$, for which the remainder of $m(X)$ upon division by $X$ is just $m(0)=-3\notin 9\mathbb{Z}[X]$, so $\mathbb{Z}[\alpha]$ is regular over $3$ as well, which proves that $\mathbb{Z}[\alpha]=\mathcal{O}_k$.
No heavy lifting procedures involved!
A: As you've noted we have that $\left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right |^2 \big | \text{ disc } \mathbb{Z}[\alpha]$, so if $\left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right | \not = 1$, then it's divisible by $2$ or $3$. 
Now if we can prove that $f$ is an Eisenstein polynomial for $2$ and $3$ we are done, according to the claim here. Now obviously $f$ satisfies the Eisenstein criterion for $3$ and thus by the lemma we have that $3 \nmid \left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right |$.
On the other side $f(x+1)$ satsifies the Eisenstein criterion for $2$ and so $2 \nmid \left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha-1]} \right |$. However we have $\mathbb{Z}[\alpha-1] = \mathbb{Z}[\alpha]$ and hence we have that $2 \nmid \left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right |$. Thus we conclude that $\left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right | = 1$, i.e. $\mathcal O_K = \mathbb{Z}[\alpha]$
