show that $f(x)=x^3+x-6$ is irreducible over $\mathbb{Q}[i]$ Question: Want to show that $f(x)=x^3+x-6$ is irreducible over $\mathbb{Q}[i]$.
My Thoughts: I am a bit stuck here.  If $f(x)$ would have been something like $x^3-3$, then this is pretty straightforward.  But right now, I can just use the Rational Roots test to claim $f(x)$ has no linear factors over $\mathbb{Q}$.  I see that I can't apply Eisenstein's here either.   Since $x^2+1$ is the polynomial responsible for the extension from $\mathbb{Q}$ to $\mathbb{Q}[i]$, maybe I can play with that a bit?  Any help would be greatly appreciated!
Thank you.
 A: Hint: If $f(\alpha)=0$ then $f(\bar\alpha)=0$, which means that at least one root is in $\Bbb Q$. Now use the rational root test.

Some details to complete the answer:
Complex Conjugation
Let $C(\alpha)=\bar\alpha$ denote complex conjugation, then

For $\alpha,\beta\in\Bbb C$ we have

*

*$C(\alpha\beta)=C(\alpha)C(\beta)$

*$C(\alpha+\beta)=C(\alpha)+C(\beta)$

Proof: Let $\alpha=a_1+a_2i,\beta=b_1+b_2i$

*

*$C(\alpha+\beta)=C(a_1+b_1+(a_2+b_2)i)=a_1+b_1-(a_2+b_2)i=(a_1-a_2i)+(b_1-b_2i)=C(\alpha)+C(\beta)$


*$C(\alpha\beta)=C(a_1b_1-a_2b_2+(a_1b_2+a_2b_1)i)=a_1b_1-a_2b_2-(a_1b_2+a_2b_1)i=(a_1-a_2i)(b_1-b_2i)=C(\alpha)C(\beta)$
This means that $C(f(\alpha))=f(C(\alpha))$, in other words, $\overline{f(\alpha)}=f(\bar\alpha)$. Thus if $f(\alpha)=0$, then $$f(\alpha)=0\\\overline{f(\alpha)}=\overline{0}\\f(\bar\alpha)=0$$ so $\bar\alpha$ is a root as well.
The Rational Root Test
Let $f(x)=a_nx^n+\ldots+a_1x+a_0\in\Bbb Z[x]$, then we have

Suppose $f\in\Bbb Z[x]$, and suppose there is a rational root $r=\frac{p}{q}\in\Bbb Q$ of $f$ with $\gcd(p,q)=1$. Then $p\mid a_0$ and $q\mid a_n$.

Proof: Assuming $r\neq 0$ we have $$f\left(\frac{p}{q}\right)=a_n\left(\frac{p}{q}\right)^n+\ldots+a_1\left(\frac{p}{q}\right)+a_0=0$$
Multiplying with $q^n$ we get the equation $$a_np^n+a_{n-1}p^{n-1}q+\ldots+a_1pq^{n-1}+a_oq^n=0$$
We then observe:

*

*Every term is divisible by $p$, so $p\mid a_0q^n$. As $p,q$ have no common factor we have $p\mid a_0$


*Every term is divisible by $q$, so again $q\mid a_np^n$. As $p,q$ have no common factor we have $q\mid a_n$
A: The following does not use the existence of complex numbers.
Let's start from the fact that given polynomial $f(x) =x^3+x-6\in\mathbb {Q} [x] $ has no rational roots. This means that the polynomial $f(x) $ is irreducible over $\mathbb{Q} [x] $.
Let $a$ be any root of $f(x) $. Then by the irreducibilty of $f(x) $ over $\mathbb {Q} [x] $ we have $[\mathbb {Q} (a) :\mathbb {Q}] =3$. Similarly if $b$ is a root of $g(x) =x^2+1$ then $[\mathbb {Q} (b) :\mathbb {Q}] =2$.
If $b\in\mathbb {Q} (a) $ then $\mathbb {Q} (b) \subseteq \mathbb{Q} (a) $ and hence $[\mathbb{Q} (b) :\mathbb {Q}] =2$ must divide $[\mathbb{Q} (a) :\mathbb {Q}] =3$ which is absurd. Thus $b\notin\mathbb {Q} (a) $. Hence $g(x) $ is irreducible over $\mathbb{Q} (a) $ and $[\mathbb {Q} (a, b) :\mathbb {Q} (a)] =2$ so that $$[\mathbb {Q} (a, b) :\mathbb {Q}] =[\mathbb {Q} (a, b) :\mathbb {Q} (a)] [\mathbb {Q} (a) :\mathbb {Q}] =2\cdot 3=6$$ Now we can observe that $$[\mathbb{Q} (a, b) :\mathbb{Q} (b)] =\frac{[\mathbb {Q} (a, b) :\mathbb {Q}] } {[\mathbb {Q} (b) :\mathbb {Q}] }=\frac{6}{2}=3$$ Let's note that $a$ is a root of $f(x) $ and if $f(x) $ were reducible over $\mathbb {Q} (b) $ then we would have $$[\mathbb {Q} (a, b) :\mathbb {Q} (b)] <\operatorname {deg} (f(x)) =3$$ and this contradiction proves that $f(x) $ is irreducible over $\mathbb{Q} (b) $.
