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I know that for the cubic polynomial $x^3-5$ over $\mathbb{Q}$ the splitting field is $\mathbb{Q}(\sqrt[3]{5},e^{\frac{2\pi i}{3}})$, but I cannot convince myself why the dimension of this splitting field is 6 and not 9.

The basis of the extension field for $\mathbb{Q}(\sqrt[3]{5})$ over $\mathbb{Q}$ gives linearly independent vectors $\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}$ such that any field element in $\mathbb{Q}(\sqrt[3]{5})$ is a linear combination of these three vectors, so $$\mathbb{Q}(\sqrt[3]{5})=\{a+b\sqrt[3]{5}+c\sqrt[3]{25}\mid a,b,c \in \mathbb{Q}\},$$ and similarly the splitting field $$\mathbb{Q}(\sqrt[3]{5},e^{\frac{2\pi i}{3}})=\{A+Be^{\frac{2\pi i}{3}}+Ce^{\frac{4\pi i}{3}}\mid A,B,C \in \mathbb{Q}(\sqrt[3]{5})\}.$$ Therefore, wouldn't the basis for the splitting field $\mathbb{Q}(\sqrt[3]{5},e^{\frac{2\pi i}{3}})$ consist of 9 elements such that $$\mathbb{Q}(\sqrt[3]{5},e^{\frac{2\pi i}{3}})=\left\{{{r+s\sqrt[3]{5}+t\sqrt[3]{25}+ue^{\frac{2\pi i}{3}}+v\sqrt[3]{5}e^{\frac{2\pi i}{3}}}\atop {+\,w\sqrt[3]{25}e^{\frac{2\pi i}{3}}+xe^{\frac{4\pi i}{3}}+y\sqrt[3]{5}e^{\frac{4\pi i}{3}}+z\sqrt[3]{25}e^{\frac{4\pi i}{3}}}}\;\middle\vert\;r,s,t,u,v,w,x,y,z\in\mathbb{Q}\right\}$$ so the dimension of this splitting field would be 9, or am I missing something and three of these I am counting twice and really there are only six?

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up vote 4 down vote accepted

You have some (linearly dependent) redundancies in your set of generators. In general, the degree of a compositum is not the product of the degrees.

EDIT: And as Jacob points out below, even this isn't the full extent of your boo-boo because $[\mathbb{Q}(e^{\frac{2\pi i}{3}}):\mathbb{Q}]=2$--it's minimal polynomial is $x^2+x+1$ the $2$nd Cyclotomic polynomial (look it up!). In fact, this fact is true here. The key is that $[\mathbb{Q}(\sqrt[3]{5}:\mathbb{Q}]=3$ and $[\mathbb{Q}(e^{\frac{2\pi i }{3}}):\mathbb{Q}]=2$ are coprime. Indeed, because of this their intersection $\mathbb{Q}(\sqrt[3]{5})\cap\mathbb{Q}(e^{\frac{2\pi i }{3}})$ is $\mathbb{Q}$ (why? hint: it must have degree dividing $(2,3)$). The fact then follows because of (among other reasons) $\text{Gal}(LK/\mathbb{Q})$ is the subgroup of $\text{Gal}(L/\mathbb{Q})\times\text{Gal}(K/\mathbb{Q})$ consisting of pairs $(\sigma,\tau)$ such that $\sigma_{L\cap K}=\tau_{L\cap K}$. Since in our case the intersection is $\mathbb{Q}$ and every element of any Galois group (over $\mathbb{Q}$) fixes $\mathbb{Q}$ we see that, in fact the Galois group of our compositum is the product of the Galois groups of the individual fields.

There are several ways to actually find the degree of the extension. One of the more tricky and computation free ways is to note that if $K$ is the splitting field for $x^3-5$ then $K/\mathbb{Q}$ is Galois and so $[K:\mathbb{Q}]=|\text{Gal}(K/\mathbb{Q})|$. Now, a very useful theorem involving the Galois group is the following

Theorem: Let $f(x)\in\mathbb{Q}[x]$ be irreducible and have degree $n$. Then, if $K_f$ denotes a splitting field for $f$ one has that $[K_f:\mathbb{Q}]\mid n!$

To see why this is true, note that $f$ has $n$ distinct roots $\alpha_1,\cdots,\alpha_n$ and that $K_f=\mathbb{Q}(\alpha_1,\cdots,\alpha_n)$. Then, since an element of $\text{Gal}(K_f/\mathbb{Q})$ permutes the roots of $f$ and is determined by its action on them we get a faithful action of $\text{Gal}(K_f/\mathbb{Q})$ on $\{\alpha_1,\cdots,\alpha_n\}$. This gives us an embedding $\text{Gal}(K_f/\mathbb{Q})\hookrightarrow S_n$ and thus $[K_f:\mathbb{Q}]=|\text{Gal}(K_f/\mathbb{Q})|\mid n!$.

Now, with this theorem on hand let's prove that if $f(x)=x^3-3$ then $[K_f:\mathbb{Q}]=6$. Well, for starters you know that $K_f\supseteq \mathbb{Q}(\sqrt[3]{5})$ and since $\sqrt[3]{5}$ has minimal polynomial $x^3-5$ we have that $[\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}]=3$ and thus $3\mid [K_f:\mathbb{Q}]$. Now, since $[K_f:\mathbb{Q}]\mid 6$ this implies that either $[K_f:\mathbb{Q}]=3$ or $[K_f:\mathbb{Q}]=6$. If the latter were true then evidently we'd have that $K_f=\mathbb{Q}(\sqrt[3]{5})$, but I think this is clearly untrue since $K_f$ contains non-real numbers (i.e. $e^{\frac{2\pi i }{3}}$) whereas $\mathbb{Q}(\sqrt[3]{5})$ doesn't. Thus, we must have that $[K_f:\mathbb{Q}]=6$ as desired.

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@OP If you look at Alex's answer above you will see where your problem likes. If $L,M$ are finite extensions of $K$ it is not in general true that $[LM:K] = [L:K][M:K]$. – user38268 Feb 4 '13 at 5:59
@BenjaLim Actually that's only one of his issues. His main issue is that he doesn't understand that the third roots of unity generate a degree $2$ and not $3$ extension. In this particular instance we do have what's not true in general. – JSchlather Feb 4 '13 at 6:26

Hint: Are you sure that $1$, $e^{2\pi i/3}$, and $e^{4\pi i/3}$ are linearly independent over $\mathbb{Q}$? What are the roots of $x^2+x+1$?

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