I'm solving exercise with extensions of fields. I just want to ask if I can write in this way: $\mathbb{Q(3, \sqrt{3}, \sqrt{11}})=\mathbb{Q(\sqrt{3}, \sqrt{11}})$, because $3\in\mathbb{Q}$. Is it my equation correct?

  • 2
    $\begingroup$ Pretty sure it is $\endgroup$
    – Seth
    Mar 24, 2019 at 21:58
  • 3
    $\begingroup$ $3\in \mathbb Q$ so $\mathbb Q(3) = \mathbb Q$ and $\mathbb Q(3, \alpha_1,.... \alpha_n) = \mathbb Q(\alpha_1,... \alpha_n)$. $\endgroup$
    – fleablood
    Mar 24, 2019 at 22:11

1 Answer 1


Abolutely. When you write $\mathbb{Q}[\sqrt{3}]$, you are referring to the field of rational numbers adjoined with $\sqrt{3}$. writing $\mathbb{Q}[3]$ is redundant since, as you pointed out, $\mathbb{Q}[3] = \mathbb{Q}$. If you want to be more nit-picky, you could even say that the minimal polynomial of $\mathbb{Q}[3]$ already has a solution in $\mathbb{Q}$, so its "extension" is $\mathbb{Q}$.


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