0
$\begingroup$

Let $S=\{\sqrt p \in \mathbb R | p $ is a primer number$\}$.

How can I show that $\mathbb Q(S)|\mathbb Q$ is an infinite field extension?

$\endgroup$

marked as duplicate by Jonas Meyer, Adam Hughes, voldemort, Tunk-Fey, user147263 Aug 26 '14 at 3:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

It follows directly from the infinity of primes and the fact that $\left[\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right]=2^n$ for $p_1,\dots,p_n$ distinct prime numbers. This last fact is the theme of question 113689.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.