Okay so I just started working on splitting fields today and I wanted to make sure that I understand it well. Here is a specific question I've been working on...
Construct the splitting field for the polynomial $(x ^2 − 3)(x ^2 − 5)$ over $Q(\sqrt{ 2})$. What is the degree of the extension? Why?
here is my work, if there's any mistakes no matter how nitty gritty (or glaringly obvious) I would very much so appreciate having these pointed out and if anyone thinks there's a faster or neater way to show anything I try to show here then I would very much so be grateful for such suggestions as well.
$(x^2-3)(x^2-5)=(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})$
so the roots of the original polynomial are $+-\sqrt{3}$ and $+-\sqrt{5}$
these roots are not elements of $\Bbb Q(\sqrt{2})$ as we shall prove thusly
elements of $\Bbb Q(\sqrt{2})$ are of the form $\{a+b\sqrt{2}|a,b \in \Bbb Q\}$ so if we were to have $\sqrt{3}$ as an element of this then it would imply that:
$3= a^2+ab\sqrt{2}+2b^2$
which has three cases
1) $ab \neq 0 \Rightarrow ab=\frac{3-a^2-2b^2}{\sqrt{2}}$ which is impossible as $ab \in \Bbb Q$ so it cant have an irrational denominator.
2) $a=0 \Rightarrow 3=2b^2 \Rightarrow b=\sqrt{\frac{3}{2}}$ again impossible as $b \in \Bbb Q$
3)$b=0 \Rightarrow a=\sqrt{3}$ again impossible as $a \in \Bbb Q$
so we have shown $ \sqrt{3} \notin \Bbb Q(\sqrt{2}) $ therefore we seek an extension of $\Bbb Q(\sqrt{2})$ which does contain these roots.
By the irreducibility test we know that $x^2-3$ is irreducible in $\Bbb Q(\sqrt{2})[x]$ as we shown that $\sqrt{3}$( the root of that polynomial) is not an element of that field. so we conclude:
$\Bbb Q(\sqrt{3},\sqrt{2}) \cong \Bbb Q(\sqrt{2})[x]/<x^2-3>$ and so $\Bbb Q(\sqrt{3},\sqrt{2}):= \{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}|a.b.c.d. \in \Bbb Q \}$
Next lets show that $\sqrt{5}$ is not an element of $\Bbb Q(\sqrt{3},\sqrt{2})$
suppose it is
$\Rightarrow 5= a^2+2b^2+3c^2+6d^2+2ab\sqrt{2}+2ac\sqrt{3}+(2ad+2bc)\sqrt{6}+2bd\sqrt{12}+2cd\sqrt{18}$
in seeking to get rid of all the square roots we have 4 cases.
1)$a=0, b=0, c=0 \Rightarrow d=\sqrt{\frac{5}{6}}$ impossible as $d \in \Bbb Q$
we obtain similar results for case 2 where a=0, c=0, d= 0 and case 3 where a=0 , b= o, d=0 and case 4 where b=0, c=0, d=0.
so we conclude $\sqrt{5} \notin \Bbb Q(\sqrt{3},\sqrt{2})$ so we also need to adjoin this element on.
by the irreduciblity test we that $x^2-5$ is irreducible in $Q(\sqrt{2}, \sqrt{3})[x]$ as we have shown that its root is not an element of $\Bbb Q(\sqrt{2},\sqrt{3})$
so we conclude
$\Bbb Q(\sqrt{3}, \sqrt{2}, \sqrt{5}) \cong \Bbb Q(\sqrt{2},\sqrt{3})[x]/<x^2-5>$
so our splitting field is $\Bbb Q(\sqrt{2},\sqrt{3},\sqrt{5})$ which is of the form $\{a+b\sqrt{5}+c\sqrt{2}+d\sqrt{10}+e\sqrt{3}+f\sqrt{15}+g\sqrt{6}+g\sqrt{30} |a,b,c,d,e,f,g,h \in \Bbb Q \}$ and by tower law is a degree 4 extension over $\Bbb Q(\sqrt{2})$