Consturct a splitting field over of . Find its dimensionality over .
Let is the primitive 5th root of unity in , whose minimal polynomial is .
has roots in . Then a splitting field of is . And .
To see this, , because is irreducible by Eisenstein criterion. Furthermore is irreducible in . Suppose not, since , must split into two quadratic polynomial, namely . So . While satisfies , which has roots and . So . This is impossible because does not divide 5.
We can consider this problem in another way. , is irreducible in . Otherwise can be imbedded in . This is impossible because .
Find a primitive element of a splitting field of over .
The Galois group of over are permutations of roots. Every permutation is uniquely determined by and . Let , the only fixes is . So is the splitting field of and is a primitive element.

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I don’t understand why you can conclude that $Q(\sqrt[5]{2})$ can be embedded in $Q(\omega)$ when $x^5 – 2$ is not irreducible over $Q(\omega)$. I know that if one root is in $Q(\omega)$ then so are the rest, but how do you show that $x^5 – 2$ cannot be factored into an irreducible cubic and an irreducible quadratic over $Q(\omega)$? Is there a standard trick that one uses in such circumstances?
I think all the roots of $x^52$ are $\sqrt[5]{2}w^i$, $i=0,1,2,3,4$. The multiplication of any two of roots are not in $Q(\omega)$. So you can not factorize to irreducible quadratic and cubic
Okay, that makes sense. Thank you.