- #1
bham10246
- 62
- 0
Question: I need to show that [itex]K = \mathbb{Q}(i, 2^{1/4}) [/itex] is a Galois extensions of [itex]\mathbb{Q}[/itex].
If I show that [itex]|Gal(\mathbb{Q}(i, 2^{1/4})/\mathbb{Q})|= [\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}] [/itex], then we're done. Another approach is to find an irreducible polynomial [itex]f(x)\in \mathbb{Q}[x][/itex] such that K is the splitting field for [itex]f[/itex], then we're done.
I first considered [itex] f(x) = x^4 -2 [/itex] but this is a degree 4 polynomial. I'm looking for an irreducible degree 8 polynomial. How do I find such polynomial given K? What's the most efficient way to find such polynomial?
Note that K is a Galois extension over the rational numbers if K is the splitting field for some polynomial f and f does not split completely into linear factors over any proper subfield of K containing [itex]\mathbb{Q}[/itex].
Thank you!
If I show that [itex]|Gal(\mathbb{Q}(i, 2^{1/4})/\mathbb{Q})|= [\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}] [/itex], then we're done. Another approach is to find an irreducible polynomial [itex]f(x)\in \mathbb{Q}[x][/itex] such that K is the splitting field for [itex]f[/itex], then we're done.
I first considered [itex] f(x) = x^4 -2 [/itex] but this is a degree 4 polynomial. I'm looking for an irreducible degree 8 polynomial. How do I find such polynomial given K? What's the most efficient way to find such polynomial?
Note that K is a Galois extension over the rational numbers if K is the splitting field for some polynomial f and f does not split completely into linear factors over any proper subfield of K containing [itex]\mathbb{Q}[/itex].
Thank you!