2

Let $K = \mathbb{Q}(i2^{1/3}, 3^{1/4})$. Is this a Galois extension?

People seem pretty convinced this is Galois. I present an issue I am having though.

So the monic polynomial p(x) in $\mathbb{Q}[x]$ of least degree whose root field over $\mathbb{Q}$ is $K$

($x^2$+1)($x^6$+4)($x^4$-3) = $x^{12}+x^{10}+4x^6+4x^4-3x^8-3x^6-12x^2-12$

Is the fact that this is degree 12 contradict that $K$ is Galois since the extension K has degree 24?

user148884
  • 151
  • 2
    Continuation of http://math.stackexchange.com/questions/829812/let-k-mathbbqi21-3-31-4-is-this-a-galois-extension. – lhf Jun 12 '14 at 23:53

2 Answers2

4

No, it does not contradict the fact. Here is an example of a Galois extension $K$ such that $[K:F] > \deg(p(x))$:

Let $K = \mathbb{Q}[\sqrt[4]{2}, i]$. You can check that this is a splitting field of the irreducible polynomial $f(x) = (x^4 - 2)$. Note that $\deg(f) = 4$. However, $[K:\mathbb{Q}] = 8$. The important thing is that $K$ is a Galois extension since it is the splitting field of a polynomial over the rationals.

To conclude, the following statements are equivalent to saying that an extension $K$ over a field $F$ is Galois (think if and only if):

  • $K$ is a separable extension, and it is a splitting field of some polynomial $F[x]$. Since $\mathbb{Q}$ and $\mathbb{R}$ are perfect fields, all finite extensions of them are separable. Therefore, we need only check that extensions over $\mathbb{Q}$ or $\mathbb{R}$ are splitting fields.
  • $[K:F] = |Aut(K/F)|$

Returning to the example above, we can check the second condition as well. If we take two "base" $\mathbb{Q}$-automorphisms defined as follows: $\phi(\sqrt[4]{2}) = i\sqrt[4]{2}$ and $\psi(i) = -i$, then we can construct an automorphism group of order $8$ isomorphic to $D_4$ by composing $\psi$ with powers of $\phi$.

Kaj Hansen
  • 33,011
  • 1
    Thank you this clears it up. I agree that K is galois – user148884 Jun 12 '14 at 23:12
  • 2
    I'm glad I could help! I'm slowly editing my response to try to add more detail and clarity, so check back in like half an hour and there might be a little more. – Kaj Hansen Jun 12 '14 at 23:17
2

It might be worth mentioning that if $f(x)$ is a typical degree $n$ irred. polynomial over $\mathbb Q$, then the splitting field of $f$ will have Galois group $S_n$, and in particular be of degree $n!$. So it is not surprising that the degree of the field extension is greater than the degree of the polynomial of which it is a splitting field; indeed, such behaviour is typical.

[Added in response to comments: Here, by "typical" I mean "random". See the links in the comments below for more details about this.]

One theoretical way to think about it is as follows: if $L/K$ is Galois with Galois group $G$, then the minimal degree of a poly. for which it is a splitting field is equal to the minimal degree of a faithful permutation rep'n of $G$.
(If $L$ is the splitting field of a polynomial $f \in K[x]$, then $G$ acts faithfully on the roots of $f$. Conversely, every permutation rep'n of $G$ arises in this way.)

Any group $G$ has a faithful perm. rep'n of degree $|G|$, namely its regular rep'n on itself, but this it typically not minimal.

In your example, the Galois group is an index $2$ subgroup of the direct product $D_6 \times D_8$. Thus it has a faithful permutation rep'n of degree $7$: act on $3$ elements via the projection to $D_6$ and the isomorphism $D_6 \cong S_3$, and act on $4$ elements via the projection to $D_8$ and the embedding $D_8 \subset S_4$.

Concretely, this corresponds to the degree $7$ polynomial $(x^4 - 3)(x^3 - 2)$.

(So incidentally, your claim about $12$ being the minimal degree is false.)

Matt E
  • 123,735
  • I'm not sure I'm following your first paragraph. There are certain irreducible cubics with Galois groups isomorphic to $\mathbb{Z}_3$ (those whose discriminant is a perfect square). The strongest result I can think of would instead be if you are given a degree $n$ irreducible, then $[K:F]$ divides $n!$. – Kaj Hansen Jun 13 '14 at 03:26
  • @KajHansen so is my polynomial wrong? What is the degree supposed to be? I am assuming Matt is getting it from knowing that $2^{1/3}$ is in there, and we can get $i$ from its minimum polynomial's roots, so that is why it is degree 7? – user148884 Jun 13 '14 at 07:49
  • You're fine @user148884. As long as you can show that the extension is a splitting field for some polynomial, then you're good to go. My question is regarding Matt's statement in his first couple sentences. He says "if $f(x)$ is a typical degree $n$ irred. polynomial over Q, then the splitting field of f will have Galois group $S_n$, and in particular be of degree $n!$". My claim is that this is false. For example, the splitting field for $f(x) = x^4 - 2$ has degree $8$ with Galois group $G \cong D_4 \ncong S_4$. – Kaj Hansen Jun 13 '14 at 09:52
  • (As an aside, I hope I'm not missing something obvious and making a fool of myself). – Kaj Hansen Jun 13 '14 at 09:53
  • @KajHansen: Dear Kaj, I think you're missing the word "typical". Assuming the inverse Galois problem has a positive answer, we can get any finite group as a Galois group. But a randomly chosen degree $n$ polynomial will have Galois group $S_n$. See e.g. here, here, or here. Regards, – Matt E Jun 13 '14 at 12:13
  • @user148884: Dear user, Your polynomial does have the right splitting field. I am just quibbling with your claim that it is the least degree polynomial with this property. (The point is that there are infinitely many polynomials, with infinitely many possible degrees, having a given splitting field. You have certainly written down one, but not one of least degree. My answer is supposed to give an indication of how you could determine the least possible degree, and how it does (and doesn't) relate to the size of $G$.) Regards, – Matt E Jun 13 '14 at 12:15
  • @MattE, thanks for clearing things up. Sorry for the trouble. I should note for user14884's benefit that, while oftentimes it is necessary to find the minimal polynomial, this is not one of those times. In order to demonstrate that $K$ is a Galois extension, it is only necessary that you show that it is a splitting field of some polynomial. – Kaj Hansen Jun 13 '14 at 18:38