Show that $x^{3}-3$ irreducible over $\mathbb{Q}(\sqrt{-3})$ Is there a slick way to show that $x^{3}-3$ is irreducible over $F= \mathbb{Q}(\sqrt{-3})$? What I did seems kind of convoluted (showing directly that there is no root in F).
Thanks
 A: One way may be to using the counting theorem?
The degree of $\sqrt{-3}$ over $\mathbb{Q}$ is 2. 
Since $x^3-3$ is of degree 3, if it was reducible in $\mathbb{Q}(\sqrt{-3})$, it would have a root $\alpha$ in $\mathbb{Q}(\sqrt{-3})$.
But by Eisenstein we also know that $x^3-3$ is irreducible over $\mathbb{Q}$. So assume for contradiction that it was reducible over $\mathbb{Q}(\sqrt{-3})$, and that $\alpha$ existed. Then we would have:
$[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}]=2=[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)][\mathbb{Q}(a):\mathbb{Q}]=[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)] \cdot 3$.
But $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)]$ must be an integer, so we have our contradiction from $2=[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)]\cdot3$.
In order to use this we also must have that $Q(\sqrt{-3})$ is a finite extension of $Q(\alpha)$. But since $Q(\sqrt{-3})$ is a finite extension of $\mathbb{Q}$, we have that $1, \sqrt{-3}$ is a basis for $Q(\sqrt{-3})$ over $\mathbb{Q}$. If $\sqrt{-3}$ should happen to be in $Q(\alpha)$ they must be the same, if not, we then have that $1, \sqrt{-3}$ over $Q(\alpha)$ span $Q(\sqrt{-3})$, but we also must have that $1, \sqrt{-3}$ must be linearly independent when having coefficients in $\mathbb{Q}(\alpha)$, if not $g_1+q_2\sqrt{-3}=0$, where not both coefficients in $Q(\alpha)$ is zero. But then it is easy to see that $\sqrt{-3}$ must be in $\mathbb{Q}(\alpha)$. 
This last fact that $\mathbb{Q}(\sqrt{-3})$ is a finite extension of $\mathbb{Q}(\alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)] $ is the degree of $(\mathbb{Q}(\alpha))(\sqrt{-3})$ over $\mathbb{Q}(\alpha)$. And we know that since $\sqrt{-3}$ is algebraic over $\mathbb{Q}$, it must also be algebraic over $\mathbb{Q}(\alpha)$.
A: There are so many ways of proving this!
Consider $f(X)=X^3-3$,over $\Bbb Q(\sqrt{-3}\,)$. Now, $f$ is irreducible if and only if $f(\sqrt{-3}X)=g(X)=-3\sqrt{-3}X^3-3$ is irreducible, if and only if $-g(X)/3=\sqrt{-3}X^3+1$ is irreducible, if and only if the reverse of the last, namely $X^3+\sqrt{-3}$, is irreducible, and it is so by Eisenstein.
A: Hint: The norm squared any root of $X^3 - 3$ is $3^{2/3}$, irrational, while the norm squared of every element in $\mathbb{Q}(\sqrt{-3})$ is rational, so $X^3 - 3$ has no root in $\mathbb{Q}(\sqrt{-3})$. 
A: Hint: $f$ in $K[x]$ irreducible of degree $m$, $K \subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$. 
$\bf{Added:}$
Let $\alpha$ a root of $f$.   We have 
$$[L(\alpha) \colon K]= [L(\alpha) \colon L]\cdot [L\colon K] \le m \cdot n$$
But   $[L(\alpha) \colon K]$ is divisible by both $[K(\alpha) \colon K]=m$ and  $[L\colon K]=n$ and so by $m\cdot n$. Therefore, we must have equality, $[L(\alpha) \colon L] = m$, and thus $f$ is irreducible in $L[x]$.
A: There's a number of ways of attacking this problem.  Here's but one way of going about this that requires a little less calculation:
Since $\deg(f) \leq 3$, we know $f(x) = x^3 - 3$ is reducible in $\mathbb{Q}(\sqrt{-3}) \iff$ a root of $f$ is contained in $\mathbb{Q}(\sqrt{-3})$.  
Let $\alpha$ be a root of $f$.  If $\alpha \in \mathbb{Q}(\sqrt{-3})$, then we must have $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt{-3})$.
Given that multiplicativity of degrees gives $\Big[ \mathbb{Q}(\sqrt{-3}): \mathbb{Q} \Big] = \Big[ \mathbb{Q}(\sqrt{-3}): \mathbb{Q}(\alpha) \Big] \cdot \Big[ \mathbb{Q}(\alpha): \mathbb{Q} \Big]$, think about the degrees of these extensions over $\mathbb{Q}$ to arrive at a contradiction.
A: Suppose it was reducible.  Then it would have a root in $\mathbb Q(\sqrt{-3})$.  But it is irreducible over $\Bbb Q$, so that would mean $\mathbb Q(\sqrt{-3})$ would contain an element of degree three over $\Bbb Q$.  But it is an extension of degree two so that is impossible.
