Why do we have that for example $\Bbb Q(\sqrt5, \sqrt{11}) = \Bbb Q(\sqrt5)(\sqrt{11})$?

Question

Why do we have that for example $\Bbb Q(\sqrt5, \sqrt{11}) = \Bbb Q(\sqrt5)(\sqrt{11})$?

I'm trying to understand what this means. The lhs of the equality is supposed to be the smallest field containing $\Bbb Q$ and $\sqrt5$ as well as $\sqrt{11}$.

The rhs is from what I can understand the smallest field containing $\Bbb Q(\sqrt{5})$ and $\sqrt{11}$?

Is there some simple way to see why this equality holds? And is it true for every $a,b \in \Bbb R$ that $\Bbb Q(\sqrt a, \sqrt{b}) = \Bbb Q(\sqrt a)(\sqrt{b})$?

Answer 1

This is indeed true much more generally. But let's first get a handle on what these things mean.

Like you said, $\mathbb{Q}(\sqrt{5})$ is the "smallest field" containing $\mathbb{Q}$ as well as $\sqrt{5}$. To be more precise, $\mathbb{Q}(\sqrt{5})$ is the smallest subfield of $\mathbb{R}$ which contains both $\mathbb{Q}$ and $\sqrt{5}$.

More generally, when you have a subfield $F$ of a field $K$, and elements $\alpha_1, \dots, \alpha_n \in K$, then $F(\alpha_1, \dots, \alpha_n)$ denotes the smallest subfield of $K$ which contains both $F$ and the elements $\alpha_1, \dots, \alpha_n$. It's important to keep in mind what this larger field $K$ is, even though it is usually only implicitly specified through context!

So, in your case:

• $\mathbb{Q}(\sqrt{5})$ is the smallest subfield of $\mathbb{R}$ which contains both $\mathbb{Q}$ and $\sqrt{5}$.
• $\mathbb{Q}(\sqrt{5})(\sqrt{11})$ is the smallest subfield of $\mathbb{R}$ which contains both $\mathbb{Q}(\sqrt{5})$ and $\sqrt{11}$. Combining this with the previous step, this means that is the smallest subfield of $\mathbb{R}$ which contains $\mathbb{Q}$, $\sqrt{5}$, and $\sqrt{11}$.
• $\mathbb{Q}(\sqrt{5}, \sqrt{11})$ is, by definition, the smallest subfield of $\mathbb{R}$ which contains $\mathbb{Q}$, $\sqrt{5}$, and $\sqrt{11}$.
• Thus, $\mathbb{Q}(\sqrt{5})(\sqrt{11}) = \mathbb{Q}(\sqrt{5}, \sqrt{11})$.

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This argument is perhaps a bit lacking in rigor, and if you'd like you remedy that, you should start by writing down what exactly it means that $\mathbb{Q}(\sqrt{5})$ is "the smallest subfield of $\mathbb{R}$ which contains both $\mathbb{Q}$ and $\sqrt{5}$". Let me know if you try this and get stuck!

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Challenge yourself: can you see why more generally we have

$F(\alpha_1)(\alpha_2)\dots(\alpha_n) = F(\alpha_1, \dots, \alpha_n)$

for any field $F$ and any elements $\alpha_1, \dots, \alpha_n$ in a superfield $K$ of $F$?

Answer 2

Yes, the intuition about $\mathbb Q({\root 3 \of 2})$ being "the smallest field extension of $\mathbb Q$ containing ${\root 3 \of 2}$ is definitely correct. But there turn out to be glitches in this, so some refinement is needed.

At least eventually, you'll not want to think of field extensions of $\mathbb Q$ as being subfields of $\mathbb R$ or $\mathbb C$, because fields like $\mathbb Q({\root 3 \of 2})$ have three different imbeddings into $\mathbb C$, exactly one of which has image inside $\mathbb R$, with the image of ${\root 3 \of 2}$ being the real cube root.

It turns out to be better, again, at least eventually, to follow Kronecker, and say that $\mathbb Q({\root 3 \of 2})=\mathbb Q[x]/\langle x^3-2\rangle$, somewhat abstractly. And, yes, this "abstract" field has 3 different imbeddings to $\mathbb C$, etc.