Intermediate extensions are those corresponding to basis elements?

Question

An exercise asks us to determine all the intermediate field extensions between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$. One thing i could do would be to calculate the minimal normal extension of this to then apply Galois correspondence, but i hope there's an easier way. The solution provided just claims that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{2})$ are the only ones since $\mathbb{Q}(\sqrt[3]{2}\sqrt{2})$ is the whole field, which sounds like they are assuming that the only possible intermediate extensions are those generated by the basis elements. I don't see how to prove this, and trying to find counterexamples doesn't seem easy either. Any hint?

Answer 1

We don't need Galois theory here.

Let's first observe that if $K=\mathbb{Q} (\sqrt{2},\sqrt[3]{2})$ then $[K:\mathbb{Q}] =6$. Why??

Let $L=\mathbb {Q} (\sqrt {2}),M=\mathbb{Q} (\sqrt[3]{2})$ then we have $[L:\mathbb{Q}] =2,[M:\mathbb{Q}] =3$ because the polynomials $x^2-2,x^3-2$ are irreducible over $\mathbb {Q} $. And because $2$ does not divide $3$ we can't have $\sqrt{2}\in M$ and thus therefore $K=M(\sqrt {2})$ is of degree $2$ over $M$. Therefore $$[K:\mathbb{Q}] =[K:M] [M:\mathbb{Q}] =6$$ Next we need to show that $L, M$ are the only subfields between $K$ and $\mathbb{Q} $.

Any field $F$ with $\mathbb{Q} \subset F\subset K$ must be such that $[F:\mathbb {Q}] $ divides $[K:\mathbb {Q}] =6$ so that $[F:\mathbb {Q}] $ can be either $2$ or $3$.

Let $[F:\mathbb {Q}] =2$. We prove that $\sqrt{2}\in F$. If it were not so then $[F(\sqrt {2}):F]=2$ and hence $[F(\sqrt {2}) :\mathbb{Q}] =4$ but $4$ doesn't divide $6$. So we get a contradiction. It follows that $\sqrt {2}\in F$ and $\mathbb{Q} \subset L\subseteq F$. Since both $F, L$ are of degree $2$ over $\mathbb {Q} $ we have $F=L$.

Next let us assume $[F:\mathbb{Q}] =3$ and prove that $F=M$. To prove this we need to show that $\sqrt[3]{2}\in F$. Suppose it weren't so. Then $x^3-2$ must be irreducible over $F$. Why?? A cubic polynomial is reducible over its field of coefficients if and only if it has a root in that field of coefficients. It follows that $F(\sqrt[3]{2})$ is of degree $3$ over $F$ and hence of degree $9$ over $\mathbb{Q} $ which is absurd.

Thus we must have $F=M$.

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Observe that the whole argument above is based on tower theorem of field extensions and the following property of polynomials of degree $2$ or $3$ that they are reducible over field of coefficients if and only if they have a root in that field.

Answer 2

$\mathbb{Q}(\sqrt{2})$ has degree 2 because the minimal polynomial is $x^2 - 2$. $\mathbb{Q}(\sqrt[3]{2})$ has degree 3 because the minimal polynomial is $x^3 - 2$. Since 2 does not divide 3 and 3 does not divide 2, they cannot have a common subextension bigger than $\mathbb{Q}$ (this is equivalent to saying a group of order 2 and group of order 3 cannot have quotient group of the same size unless the size is 1). The Galois group is a compositum of the 2 groups of these sub extensions and must have order 6 with exacly one factor group of order 3 and one factor group of order 2. [I hope I got it right because it's been a long while...]

PS sorry, I did get it wrong. The splitting field of $X^3-2$ has order 6, not 3, because you need to throw in a non-real cube root of unity. The extension $\mathbb{Q}(\sqrt[3]{2})$ degree 3 and is not normal. The full extension $\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$ has degree 6 and is not normal either. However, it is still always true that the rank of a subextension must divide the rank of a bigger extension, even if the extensions are not normal.