2
$\begingroup$

For which $c$ will the set $K(c) = \{a + b\sqrt{c} : a, b \in \mathbb{Q}\}$ be a field? I know for example, that $K(\frac{2}{3})$ will be one, I am just wondering what the most general result of this type is.

  • 1
    you need to define the set of c. is it Q, Z, R, C, matrices, polynomes ...?2011-10-05
  • 4
    The notation you use is nonstandard. Usually, $K(c)$ denotes the smallest field (in some specified overfield) that contains $K$ and $c$. What you are describing is instead $\mathbb{Q}[\sqrt{c}]$ (note both the $\mathbb{Q}$, and the brackets).2011-10-05
  • 0
    @Arturo, the set is $\mathbb{Q}[\sqrt{c}]$ only if $\sqrt{c}$ is a quadratic algebraic number; otherwise the set is just $\mathbb{Q}+\sqrt{c}\,\mathbb{Q}$, a vector space, not a ring, as Bill notices.2011-10-05
  • 0
    @lhf: Good point; of course, certainly $K(c)$ is just not the right thing.2011-10-05

2 Answers 2