When is $\mathbb Q(\sqrt c)$ a field?

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.

This is a field if and only if $\sqrt{c}$ satisfies a quadratic polynomial with coefficients in $\mathbb{Q}$. This certainly holds for all rational $c$ (the polynomial being $x^2-c$), but there are also slightly less expected examples. E.g. $c=3+2\sqrt{2}$ also works, because $3+2\sqrt{2}=(1+\sqrt{2})^2$, so $\sqrt{c}$ satisfies the polynomial $(x-1)^2-2$.

Now, where does this criterion come from? If you adjoin to $\mathbb{Q}$ a root of an irreducible polynomial of degree $n$, and then close the set under addition and multiplication, then the resulting set is a vector space over $\mathbb{Q}$ of dimension $n$ (think about it!). This shows that the above criterion is necessary, since your object is at most two-dimensional over $\mathbb{Q}$. It is also sufficient. Indeed, if $\sqrt{c}$ satisfies a quadratic, then by replacing $\sqrt{c}$ by some suitable $\sqrt{c}+d$, $d\in K(c)$, you may without loss of generality assume that the quadratic is of the form $x^2-\alpha$, $\alpha\in\mathbb{Q}$, and that your elements are all of the form $a+b\sqrt{\alpha}, a,b\in \mathbb{Q}$. An element of the form $a+b\sqrt{\alpha}$ has inverse $\frac{1}{a^2-\alpha b^2}(a-b\sqrt{\alpha})$, and that's also in your set, so you have inverses, and thus a field.

Inverses arise by RTD = rationalizing the denominator. $ $ Put $\, F = \mathbb Q,\ D \!=\! \mathbb Q(\sqrt{c}),\ \alpha = \sqrt{c}\ $ in

Lemma $\ \, $ TFAE $\,$ for $\,F\,$ a subfield of a domain $\,D\,\,$ and $\, \alpha\in D$

$(1)\quad \alpha\, F + F\,$ is a field.

$(2)\quad \alpha\, F + F\,$ is a ring.

$(3)\quad \alpha^2 \in\,\alpha\ F + F.$

Proof $\ \ (1\Rightarrow 2\Rightarrow 3)\ $ trivial. $\ (3\Rightarrow 1)\ $ $\,\alpha^2 \in \alpha\ F + F\,$ implies that $\,\alpha\ F + F\,$ is closed under multiplication, so it is a ring. Let $\ 0\ne \beta\in \alpha\ F + F,\,$ so $\ \beta = d + e\ \alpha,\ d,e\in F\,.\,$ If $\,e = 0\,$ then $\,0\ne \beta\in F\,$ so $\,\beta\,$ is invertible since $\,F\,$ is a field. Else by $\,(3),\ \alpha^2 = \color{#c00}b\ \alpha + c\ $ for some $\,b,c\in F,\,$ so we can invert $ \beta = d + e\ \alpha$ by multiplying it by its conjugate (as in RTD), using $\,\color{#c00}{\bar\alpha = b-\alpha},\,$ i.e. $\beta\bar\beta =(d + e\ \alpha)\,(d + e\,(\color{#c00}{b-\alpha})) = d^2 + b\ d\,e - c\ e^2 =:\, f \in F\,.\,$ Therefore, if $\ f\ne 0\ $ then $\,\beta^{-1}\!=(d + e\ \alpha)^{-1}\! = (d + e\,(b-\alpha))/f\in F + \alpha\ F\,.\,$ Otherwise $\,\beta\bar\beta =f = 0\, $ so by $\,D\,$ a domain one of the factors $\beta,\bar\beta$ is zero, i.e. $\,d + e\ \alpha=0\,$ or $\, d + e\,(\color{0a0}{b-\alpha}) = 0.\,$ Solving for $\,\alpha,\,$ using $\,e\ne 0,\,$ yields $\,\alpha = -d/e\ \ or\ \ d/e+b\,.\,$ Hence$\,\alpha\in F,\,$ so $\,\alpha\ F + F = F\,$ is a field. $ $ QED

Alternatively, since $\, \alpha\ F + F\,$ is a $2$-dimensional vector space over $\,F\,$ we deduce that $\ \beta^2,\ \beta,\ 1\ $ are $\ F$-linear dependent, therefore $\ a\ \beta^2 + b\ \beta + c = 0\,$ for some $\,a,b,c\in F\,.\,$ If $\ c=0\,$ then $\,\beta\in F\,.\,$ Else $\ (a\ \beta + b)\,\beta = -c\ $ hence $\,\beta^{-1}\! =\, -(a\ \beta + b)/c\,.$

Generally this manipulation of minimal polynomials to obtain inverses also works for higher degree algebraic elements, yielding the generalization

$\qquad$ If $\, E < D\,$ is an integral extension of domains then $\,E\,$ is a field $\iff$ $\,D\,$ is a field.

See also my posts on rationalizing denominators and related concepts.