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I know how to show $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\leq[\mathbb{Q}(\sqrt2):\mathbb{Q}][\mathbb{Q}(\sqrt3):\mathbb{Q}]$, but don't know how to show the converse inequality. $[\mathbb{Q}(\sqrt2):\mathbb{Q}]$ and $[\mathbb{Q}(\sqrt3):\mathbb{Q}] $ both divides $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]$, but they are not relatively prime, so I guess I can't the converse inequality this way. I'm considering using the fact that the intersection of $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt 3)$ is $\mathbb{Q}$, but I don't know how to proceed...

user42383
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  • How did you arrive at $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\leq[\mathbb{Q}(\sqrt{2}):\mathbb{Q}][\Bbb{Q}(\sqrt{3}):\Bbb{Q}]$? Write out a most general element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ – Marc Bogaerts Sep 25 '14 at 16:53
  • @Nimda I know how to do that way, but I'm actually trying to prove something more general: if x and y are algebraic over K and the intersection of K(x) and K(y) is K, then [K(x,y):K]=[K(x):K][K(y):K]. – user42383 Sep 25 '14 at 23:41
  • If you know the fact that the intersection of $\mathbb Q\left(\sqrt 2\right)$ and $\mathbb Q\left(\sqrt 3\right)$ is $\mathbb Q$, try this: If $\left[\mathbb Q\left(\sqrt 2, \sqrt 3\right) : \mathbb Q\right]$ was equal to $\left[\mathbb Q\left(\sqrt 2\right) : \mathbb Q\right]$, then the $\mathbb Q$-vector space $\mathbb Q\left(\sqrt 2, \sqrt 3\right)$ would be identical to its subspace $\mathbb Q\left(\sqrt 2\right)$. Can you see why this cannot be? – darij grinberg Sep 26 '14 at 01:22
  • @darijgrinberg Because $\sqrt3$ is not in $\mathbb{Q}(\sqrt2)$? I was wondering how this helps prove the desired result. – user42383 Sep 26 '14 at 13:12

1 Answers1

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The specific question in the title can be settled by elementary methods. Or as a part of more general fact about the field you get by adjoining square roots of coprime integers. See the excellent write up by Bill Dubuque.

In a comment the question was specified to be about a general fact about the degree of the compositum of two algebraic extensions of a field $K$ that intersect trivially. It may be natural to suspect that if $K(x)\cap K(y)=K$, where $K(x), K(y)$ are both algebraic extensions of $K$, then we could have a degree formula like $[K(x,y):K]=[K(x):K][K(y):K]$.

However, this is FALSE in general. An easy to digest counterexample is that of $K=\Bbb{Q}$, $x=\root3\of2$, $y=\omega\root3\of2$, where $\omega=(-1+\sqrt{-3})/2$ is a primitive cubic root of unity. Both $x$ and $y$ are zeros of the irreducible polynomial $p(x)=x^3-2$, so $[K(x):K]=3=[K(y):K]$. We have $K(x)\subset\Bbb{R}$, but $K(y)\cap\Bbb{R}=\Bbb{Q}$, so $K(x)\cap K(y)=K$. Furthermore, $K(x,y)$ is the splitting field of $p(x)$ over $K$, which is known to have degree six, not nine. This follows from $\omega$ being a root of a quadratic. Anyway, the above degree formula does not hold in this case.

I gathered that you are in a context where you do need a more general result. The key concept here is that of linearly disjoint extensions. The degree formula holds, iff $K(x)$ and $K(y)$ are linearly disjoint over $K$. A not so difficult to prove general result here is that if $K(x)/K$ and $K(y)/K$ are both Galois extensions (inside a bigger field, e.g. en algebraic closure of $K$), and $K(x)\cap K(y)=K$, then they are necessarily linearly disjoint.

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Jyrki Lahtonen
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