# Splitting fields

1. Jan 22, 2007

### calvino

1. The problem statement, all variables and given/known data
This question is from my text.

a)show that the splitting field of (x^4)-5 over Q is Q(5^(1/4), i).
b)show that [Q(5^(1/4), i): Q = 8].
c)what is the order of the galois group of (x^4)-5 over Q?

3. The attempt at a solution

I haven't yet thought about b) and c). For a),

edit2: I have no clue how to show this. I originally thought it would be easy simply by showing the I can rewrite the polynomial into factors t(x-c1)(x-c2), where t is some constant and the cs are the given solutions, stated in the splitting field. I couldn't do it. I still have some practicing to do, but I would love any help (if any). thanks in advance.

Last edited: Jan 22, 2007
2. Jan 22, 2007

### StatusX

A splitting field of f(x) is a field extension in which f(x) reduces to linear factors, ie, splits, and such that it splits in no proper subfield. So to show a given field extension is a splitting field, show it contains all the roots of f(x), and that these roots generate the field extension (do you see how this is equivalent to what I said above?).

3. Jan 23, 2007

### HallsofIvy

Staff Emeritus
What can't you do it? If you are working problems about splitting fields and Galois groups, certainly you learned long ago that (x2- a2)= (x- a)(x+ a)!

$$x^4- 5= (x^2)^2- (\sqrt{5})^2= (x^2- \sqrt{5})(x^2+ \sqrt{5})$$
$$x^2- \sqrt{5}= x^2- (^4\sqrt{5})^2= (x- 5^{\frac{1}{4}})(x+ 5^{\frac{1}{4}})$$
and
$$x^2+ \sqrt{5}= x^2- (^4\sqrt{5}i)^2= (x- i5^{\frac{1}{4}})(x+ i5^{\frac{1}{4}})$$

4. Jan 23, 2007

### calvino

Thank you. I don't know why I tend to forget my basics. Perhaps that is my tragic flaw. Nothing seems to be retained anymore. I should rethink my studying strategies. Thanks again.

EDIT: and sometimes I get flustered by thinking a question is too complex, that I do not know where to start reasoning. Not sure...just lately with algebra...i feel lost. I'll be sure to post better questions next time.

Last edited: Jan 23, 2007
5. Jan 23, 2007

### HallsofIvy

Staff Emeritus
Abstract algebra is a tough course. Hang in there and "practice, practice, practice"!