# An elementary question regarding Galois theory

1. Aug 11, 2007

### bham10246

Question: I need to show that $K = \mathbb{Q}(i, 2^{1/4})$ is a Galois extensions of $\mathbb{Q}$.

If I show that $|Gal(\mathbb{Q}(i, 2^{1/4})/\mathbb{Q})|= [\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}]$, then we're done. Another approach is to find an irreducible polynomial $f(x)\in \mathbb{Q}[x]$ such that K is the splitting field for $f$, then we're done.

I first considered $f(x) = x^4 -2$ 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 $\mathbb{Q}$.

Thank you!

2. Aug 11, 2007

### BSMSMSTMSPHD

$$[\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}] = 8$$. Here's why:

$$[\mathbb{Q}(2^{1/4}):\mathbb{Q}] = 4$$ (using the polynomial you have which is irreducible via Eisenstein).

Then, since $$i \notin \mathbb{Q}(2^{1/4})$$, $$[\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}(2^{1/4})] = 2$$ (using the polynomial $$x^2 + 1 \in \mathbb{Q}(2^{1/4})[X]$$).

Therefore, $$[\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}] = [\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}(2^{1/4})] \cdot [\mathbb{Q}(2^{1/4}):\mathbb{Q}] = 8$$

Last edited: Aug 11, 2007
3. Aug 11, 2007

### Kummer

What is wrong with $$f(x) = x^4 - 2$$? The polynomial does not need to have degree equal to 8, as long as $$\mathbb{Q}(i,2^{1/4})$$ is a splitting field over $$\{ x^4 - 2\}$$. Indeed it is.

That means, $$|\mbox{Gal}(\mathbb{Q}(i,2^{1/4})/\mathbb{Q})| = \{ \mathbb{Q}(i,2^{1/4}) : \mathbb{Q} \} = [\mathbb{Q}(i,2^{1/4}:\mathbb{Q}]=8$$, since it is a splitting field (as shown above) and furthermore it is a seperable extension (since the field charachteristic is zero). Thus, this is a Galois group.

(A useful result to know is that given $$\alpha_1 , ... ,\alpha_n$$ algebraic elements. Then the smallest field which contains them all is $$F(\alpha_1, ... ,\alpha_n)$$).

Last edited: Aug 11, 2007
4. Aug 11, 2007

### Kummer

I think you mean this.

Definition: Given a field $$F$$ and the algebraic extension $$E$$ we say $$E$$ is a "splitting field" of $$\{ f_i(x)\in F[x] \}$$ over $$F$$ so that $$E$$ is the minimal field having all their zeros.

Definition: Given a field $$F$$ and the algebraic extension $$E$$ we say $$E$$ is a "splitting field" of $$F$$ iff it is a splitting field of some set of polynomial in $$F$$.

Now if we want $$|\mbox{Gal}(E/F)|=\{ E: F\}$$ is is necessary and sufficient for $$E$$ to be a splitting field over $$F$$. Meaning, it makes no difference what set of polynomials it is.

5. Aug 11, 2007

### learningphysics

My apologies Kummer. You're right... I was thinking something wrong...

6. Aug 11, 2007

### mathwonk

the same field can be generated by different sets of elements. you have two elements generating your field. if you can find one element generating that same fiekld, then that element will be a root of an irreducible polynomial of degree 8 over Q.

The Galois group G of this extension has degree 8 and acts on the field with orbits of order dividing 8. If an orbit has order n then its elements are roots of an irreducible polynomial of degree n and generates a subfield of degree dividing n!

thus orbits of order either 4 or 8 could generate the full field, and will correspond to irreducible polynomials of degree either 4 or 8.

The difference is that in the case of a generating orbit of degree 4, adjoining just one root will not give all the others, you will have to adjoin two roots to get the full field. In the case of an irreducible polynomial of degree 8 for elements of an orbit, any one which is adjoined will also give the others.

e.g. the galois group of a polynomial like x^4 -7X^2 + 7 cannot have maximal order 4! = 24, since adjoining any root gives also its negative, another root, so the degree of the splitting field is either 4 or 8. I.e. it is irreducible so adjoining one root gives a degree 4 extension, and also a second root. then you may have to make another quadratic extension.

7. Aug 12, 2007

### bham10246

Thanks. For some reason, I was thinking that $|Gal(\mathbb{Q}(i, 2^{1/4})/\mathbb{Q})|$ must equal the degree of the irreducible polynomial, where $\deg_\mathbb{Q} (f)= [\mathbb{Q}(i, 2^{1/4}):\mathbb{Q}]$.....

8. Aug 12, 2007

### mathwonk

every finite extension of Q is "simple", which means it CAN be generated by a single element. But not every choice of a single element will generate.

if c is a root of an irreducible Q polynomial f , the the field Q(c) has degree equal to the degree of f. This is the basic rule for the degree of a simple extension.

NOW YOUR FIELD IS SIMPLE, but no one root of your polynomial does the trick. If it takes more than one root of f to generate the field, then the degree of the fiel is alrger than the degree of f.

The worst case scenario is when eaCH TIME YOU ADD IN ONE more ROOT OF F, thats all you get. I.e. you never get any more roots for free, until the enxt to akst one.

thus the relation between irreducible polynomials and splitting fields is that the degree of the splitting field can be any factor of n! where n is the degree of the polynomial.

do you see why the degree cannot be n! if the polynomial is not irreducible?

9. Aug 12, 2007

### Hurkyl

Staff Emeritus
Sometimes, you can make a problem easier by generalizing!

The first two things that leap to mind are:

Conjecture: If L is a Galois extension of K, and M is a Galois extension of L, then M is a Galois extension of K.

Conjecture: If L and M are two Galois extensions of K, then their compositum LM is a Galois extension of K.

Your problem is an easy consequence of either statement, and I suspect they are both true. Can you prove either?

10. Aug 12, 2007

### mathwonk

the first conjecture seems false. take the extension of Q generated by a square root of 2. then add in a real square root of that square root. both are galois since all degree 2 extensions are galois. but the final extension is generated by the 4th root of 2 which is not galois becauase there are complex 4th roots of 2.

indeed the conjecture if true would imply all extensions of degree 2^n are galois, clearly false in view of the present discussion.

the second conjecture is easy using the criterion that a galois extension is one generated by the roots of any polynomial over hte base field.

but the question is easy directly, since his field is generated by the roots of X^4-2.

11. Aug 12, 2007

### Hurkyl

Staff Emeritus
Ah, now I feel better about not seeing how to prove the first one. I was hoping not to spoil the second one for the OP though.

12. Aug 12, 2007

### mathwonk

well i did REPEAT HIS OWN CHARAcTERIZATION OF A GALOIS EXTENSION FROM HIS ORIGINAL POST. sorry if that spoiled it.

Last edited: Aug 12, 2007