1-) Firstly, I have shown that $f(x)$ is irreducible over $\mathbb{Q}$. To show, I first use the theorem that says: $$\text{Suppose there exist } r \in \mathbb{Q} \text{ such that } f(r)=0.\text{ Then } r\in\mathbb{Z} \text{ and }r|a_{0} \text{, where } a_0 \text{ constant coefficient of the polynomial.}$$
$$f(1)\neq0,\;f(-1)\neq0,\;f(2)\neq0,\;f(-2)\neq0,\;f(4)\neq0,\;f(-4)\neq0$$
But, since $\deg(f(x))=4$, we cannot say anything about roots. Clearly, we cannot use Eisenstein Criterion either. Now, we can check whether it has any non-linear factors. Fortunately, it has!
$$x^4+x^2+4=(x^2+x\sqrt{3}+2)(x^2-x\sqrt{3}+2)$$
But, $\sqrt{3} \notin \mathbb{Q}$.Therefore $x^4+x^2+4$ is irreducible over $\mathbb{Q}$.
Here is the question. Can I say $f(x)$ is irreducible after all these
calculations?
2-) I defined $$f_{1}(x)=x^2+x\sqrt{3}+2, \;f_{2}(x)=x^2-x\sqrt{3}+2$$
We can say $f=f_{1}\cdot f_{2}$ is reducible over $\mathbb{Q(\sqrt{3})}$ since $\sqrt{3} \in \mathbb{Q(\sqrt{3})}$. Even $f(x)$ is reducible over extension field $\mathbb{Q(\sqrt3)}$, $f_{1}$ and $f_{2}$ are irreducible over $\mathbb{Q(\sqrt3)}$
For $f_{1}(x)=x^2+x\sqrt{3}+2$ we have $x_{1,2}=\dfrac{-\sqrt{3}\pm \sqrt{-5}}{2}$ and also for $x_{3,4}=\dfrac{\sqrt{3}\pm\sqrt{-5}}{2}$. We know that neither $\sqrt{-5} \notin \mathbb{Q}$ nor $\sqrt{-5} \notin \mathbb{Q({\sqrt{3}})}$. Therefore, we need some new splitting field and it is, clearly, $\mathbb{Q(\sqrt{3},\sqrt{-5})}$.
I think there is no problem. But I have to ask to be sure, Can I extend the field like I did to split the polynomial?
3-) There is extension $\mathbb{Q(\sqrt{3},\sqrt{-5})}$ over $\mathbb{Q(\sqrt{3})}$. There is also extension $\mathbb{Q(\sqrt{3})}$ over $\mathbb{Q}$. For the field extension $\mathbb{Q(\sqrt{3},\sqrt{-5})}$ over $\mathbb{Q}$ we have the structure such as: $$\mathbb{Q(\sqrt{3},\sqrt{-5})}=\lbrace r+s\sqrt{3}+t\sqrt{-5}+z\sqrt{-15}:r,s,t,z\in \mathbb{Q}\rbrace$$ So, for the basis, dimension or degree of minimal polynomial we have $4$ and this basis is $\lbrace1,\sqrt{3},\sqrt{-5},\sqrt{-15}\rbrace$
For, the minimal polynomial of $\mathbb{Q(\sqrt{3},\sqrt{-5})}$ over $\mathbb{Q(\sqrt{3})}$ we have $P_{\sqrt{-5}}(x)=x^2+5$ this polynomial is irreducible since $\sqrt{-5}\notin\mathbb{Q(\sqrt{3})}$. For the minimal polynomial of $\mathbb{Q(\sqrt{3})}$ over $\mathbb{Q}$ we have $P_{\sqrt{3}}(x)=x^2-3$ and it is irreducible over $\mathbb{Q}$ by Eisenstein Criterion.
4-) Clearly $\mathbb{Q(\sqrt{3},\sqrt{-5})}$ is splitting field of $x^4+x^2+2$ over $\mathbb{Q}$ and since it is finite extension, we can say it is a normal extension and it is also separable extension since all the roots of polynomial is simple roots.
My whole solving strategies makes sense?
Thanks in advance!
Best Answer
For Question 1, more justification is needed as was stated in the comments to your question. You've obtained the factorization $x^4 + x^2 + 4 = (x^2 + x\sqrt{3} + 2)(x^2 - x\sqrt{3} + 2)$. You can apply the quadratic formula to each quadratic factor to get a product of four linear factors. Then you want to multiply each pair of linear factors to see whether there is a quadratic factor in $\mathbb{Q}[x]$ dividing that polynomial. It's a good trick to know for quartic polynomials like this where you can find all four roots. Also, I'd like to point out that it is not immediately obvious that you cannot apply Eisenstein's Criterion. A trick you can sometimes do is apply Eisenstein's Criterion to $f(x + a)$ for some $a \in \mathbb{Z}$, and sometimes this will give rise to a polynomial where Eisenstein's Criterion is applicable. You can convince yourself that $f(x+a)$ being irreducible is equivalent to $f(x)$ being irreducible. Sometimes, this method cannot work, and this is one such case as pointed out in a comment to this post.
For Question 2, you seem more or less correct. You should always keep in mind though that the splitting field is found by adjoining the roots to the base field, and in this case, these roots can be found: $( \sqrt{3} \pm \sqrt{-5})/2$ and $( -\sqrt{3} \pm \sqrt{-5})/2$. You want to convince yourself now that
$$ \mathbb{Q}(( \sqrt{3} \pm \sqrt{-5})/2, (- \sqrt{3} \pm \sqrt{-5})/2 ) = \mathbb{Q}(\sqrt{3}, \sqrt{-5} ) $$
I agree with your reasoning in 3
For 4, your justification for the extension being normal sounds good, but the fact that the extension is separable follows from the fact that $\mathbb{Q}(\sqrt{3},\sqrt{-5})$ has characteristic zero.