Cfrgtkky

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

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.

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.

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 $f(sqrt{-3}X)=g(X)=-3sqrt{-3}X^3-3$ 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.

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 $g_1+q_2sqrt{-3}=0$, 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 $mathbb{Q}(alpha)$.

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 $mathbb{Q}(sqrt{-3})$.

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
$$[L(alpha) colon K]= [L(alpha) colon L]cdot [Lcolon K] le m cdot n$$

But $[L(alpha) colon K]$ is divisible by both $[K(alpha) colon K]=m$ and $[Lcolon K]=n$ and so by $mcdot n$. Therefore, we must have equality, $[L(alpha) colon L] = m$, and thus $f$ is irreducible in $L[x]$.