Is $\mathbb Q(\sqrt p) = \{a + b\sqrt p\mid a,b \in\mathbb Q\}$ a field for every prime p?

Will it be a field for any natural number p or for just prime numbers p?

  • 5
    $\begingroup$ It's a field for any $p$, but when $p$ is a perfect square, it is just $\mathbb Q$. $\endgroup$ Aug 21, 2017 at 16:52
  • 1
    $\begingroup$ HInt: try to compute reciprocals by rationalizing the denominator. $\endgroup$ Aug 21, 2017 at 16:53
  • $\begingroup$ Are you familiar with quotient rings and the First Isomorphism Theorem? $\endgroup$ Aug 21, 2017 at 17:01

2 Answers 2


The general result is the following:

For any $\alpha\in\mathbf Q$, $\mathbf Q(\sqrt \alpha)$ is a field. This field equal to $\mathbf Q$ if $\alpha$ is a square in $\mathbf Q$.

If $\alpha$ is not a square, $\mathbf Q(\sqrt \alpha)$ is a quadratic extension of $\mathbf Q$, and there exists a unique square-free integer $d$ such that $\mathbf Q(\sqrt \alpha)=\mathbf Q(\sqrt d) $.


Yes, if $p$ is a prime $X^2-p$ is irreducible and $\mathbb{Q}(\sqrt p)=\mathbb{Q}[X]/(X^2-p)$. The same argument shows that it is a field if $p$ is not a square. if $p$ is a square it is also a field since $\mathbb{Q}(\sqrt p)=\mathbb{Q}$.


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