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I am just learning about field extensions, and I wanted to check my understanding. Is it true that $\mathbb{Q}(\sqrt{3},\sqrt{7})=\mathbb{Q}(\sqrt{3},\sqrt{21})$?

My reasoning is that $\mathbb{Q}(\sqrt{3},\sqrt{7})$ contains $\sqrt{3}$ and $\sqrt{3}\sqrt{7}=\sqrt{21}$, and because $\mathbb{Q}(\sqrt{3},\sqrt{21})$ is the smallest subfield of $\mathbb{R}$ containing these, it is that $\mathbb{Q}(\sqrt{3},\sqrt{21})\subseteq \mathbb{Q}(\sqrt{3},\sqrt{7})$. Also, $\mathbb{Q}(\sqrt{3},\sqrt{21})$ contains $\sqrt{21}$ and $(\sqrt{3})^{-1}=1/\sqrt{3}$, so it contains $\sqrt{21}/\sqrt{3}=\sqrt{7}$ and therefore $\mathbb{Q}(\sqrt{3},\sqrt{7})\subseteq \mathbb{Q}(\sqrt{3},\sqrt{21})$.

Is this reasoning correct? Thank you.

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Yes, I believe your reasoning is correct.

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    $\begingroup$ This really should be a comment - but I understand there is a minimum rep to leave a comment, so let me give you a +1 to help you along. $\endgroup$
    – Deepak
    Feb 10, 2016 at 0:32
  • $\begingroup$ @Deepak. Can you substantiate that with a meta link? Seems like this should be an answer and not a comment to me. math.stackexchange.com/help/privileges/comment. I don't wish to have an extended conversation in the comments but I didn't find it easy to get a ruling by skimming meta. I can ask this on meta if you think it worthwhile. $\endgroup$
    – Mason
    Jul 22, 2018 at 14:45
  • $\begingroup$ @Mason meta.stackexchange.com/questions/214173/… $\endgroup$
    – Deepak
    Jul 22, 2018 at 15:29

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