Galois extensions

[shoplifter]

Homework Statement

Let $$\beta$$ be a real, positive fourth root of 5. Is $$\mathbb{Q}(\sqrt{-5})$$ Galois over $$\mathbb{Q}$$? How about $$\mathbb{Q}(\beta + i\beta)/\mathbb{Q}$$ or $$\mathbb{Q}(\beta + i\beta)/\mathbb{Q}(\sqrt{-5})$$?

Homework Equations

An extension is Galois when it is normal and separable.

The Attempt at a Solution

Any push whatsoever in the right direction would be much appreciated.

 

[Hurkyl]

Can you tell me anything at all about those extensions?

 

[rochfor1]

Do you know a condition equivalent to being Galois that involves splitting?

 

[shoplifter]

well, a field $$A$$ is galois over $$B$$ if it is the splitting field of some polynomial in $$B[x]$$ which has distinct roots in $$A$$.

so then, $$\mathbb{Q}(\sqrt{-5})$$ is the splitting field of the polynomial $$x^2 + 5$$, which has coefficients in $$\mathbb{Q}$$. this is because $$x^2 + 5 = (x + \sqrt{-5})(x - \sqrt{-5})$$, and hence the roots are also distinct, and so in the first case, we do have a galois extension.

but i can't apply this to the second two ideas, because i can't tell anything about the field $$\mathbb{Q}(\beta + i\beta)$$. it doesn't seem to be the same as $$\mathbb{Q}(i, \beta)$$ or anything of the sort. am i missing something obvious?

thanks so much.

 

[shoplifter]

also, thinking about the polynomial $$(x + \beta + i\beta)(x + \beta - i\beta)(x - \beta + i\beta)(x - \beta - i\beta)$$ doesn't help, because it gives me the information that $$\mathbb{Q}(\beta + i\beta)$$ is galois over $$\mathbb{Q}(\sqrt{5})$$, but nothing else, and i don't see how this helps in the problem.

 

[Hurkyl]

also, thinking about the polynomial $$(x + \beta + i\beta)(x + \beta - i\beta)(x - \beta + i\beta)(x - \beta - i\beta)$$ doesn't help, because it gives me the information that $$\mathbb{Q}(\beta + i\beta)$$ is galois over $$\mathbb{Q}(\sqrt{5})$$, but nothing else, and i don't see how this helps in the problem.

Is b-ib really in Q(b+ib)?

 

[shoplifter]

no, it isn't, but it doesn't follow that it's not galois, right? I've worked for a while, and can't find explicit separable polynomials whose splitting fields are precisely those, but then what does that prove?

i really don't see a way to disprove they're galois extensions. even if they are, i can't find the explicit polynomials, but my intuition points to a disproof. =( but how do i do that?

 

[shoplifter]

oh i got the fact that it's galois over the third one -- from the polynomial x^2 - 2i*b^2.

what about the last one?