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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.

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you need to define the set of c. is it Q, Z, R, C, matrices, polynomes ...? –  Hassan Oct 5 '11 at 3:19
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). –  Arturo Magidin Oct 5 '11 at 5:18
@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. –  lhf Oct 5 '11 at 11:46
@lhf: Good point; of course, certainly $K(c)$ is just not the right thing. –  Arturo Magidin Oct 5 '11 at 15:55

2 Answers 2

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.

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HINT $\ $ Inverses arise by rationalizing denominators. $\:$ Put $\rm\: F = \mathbb Q\:,\ D = \mathbb Q(\sqrt{c}),\ \alpha = \sqrt{c}\ $ in

LEMMA $\ \ $ TFAE $\:$ for $\rm\:F\:$ a subfield of a domain $\rm\:D\:,\:$ and $\rm\: \alpha\in D\:.\:$

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

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

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

Proof $\rm\ \ (1\Rightarrow 2\Rightarrow 3)\ $ trivial. $\ (3\Rightarrow 1)\ \ $ $\rm\:\alpha^2 \in \alpha\ F + F\:$ implies that $\rm\:\alpha\ F + F\:$ is closed under multiplication, so it is a ring. Let $\rm\ 0\ne \beta\in \alpha\ F + F\:,\:$ so $\rm\ \beta = d + e\ \alpha,\ d,e\in F\:.\:$ If $\rm\:e = 0\:$ then $\rm\:0\ne \beta\in F\:$ so $\rm\:\beta\:$ is invertible since $\rm\:F\:$ is a field. Else by $\rm\:(3),\ \alpha^2 = b\ \alpha + c\ $ for some $\rm\:b,c\in F\:,\:$ hence $\rm\:(d + e\ \alpha)\:(d + e\:(b-\alpha)) = d^2 + b\ d\:e - c\ e^2\: =:\: f \in F\:.\:$ Therefore, if $\rm\ f\ne 0\ $ then $\rm\:(d + e\ \alpha)^{-1}\! = (d + e\:(b-\alpha))/f\in F + \alpha\ F\:.\:$ Otherwise $\rm\:0 = f = (d + e\ \alpha)\:(d + e\:(b-\alpha))\:.\: $ By $\rm\:D\:$ a domain, some factor $= 0\:.\:$ Solving for $\rm\:\alpha\:,\:$ using $\rm\:e\ne 0\:,\:$ yields $\rm\:\alpha = -d/e\ \ or\ \ d/e+b\:.\:$ Therefore $\rm\:\alpha\in F\:,\:$ hence $\rm\:\alpha\ F + F = F\:$ is a field. $\quad$ QED

Alternatively, since $\rm\: \alpha\ F + F\:$ is a $2$-dimensional vector space over $\rm\:F\:$ we deduce that $\rm\ \beta^2,\ \beta,\ 1\ $ are $\rm\ F$-linear dependent, therefore $\rm\ a\ \beta^2 + b\ \beta + c = 0\:$ for some $\rm\:a,b,c\in F\:.\:$ If $\rm\ c=0\:$ then $\rm\:\beta\in F\:.\:$ Else $\rm\ (a\ \beta + b)\:\beta = -c\ $ hence $\rm\:\beta^{-1}\! =\: -(a\ \beta + b)/c\:.$ This manipulation of minimal polynomials to obtain inverses also works for higher degree algebraic elements, yielding the generalization

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

See also my posts on rationalizing denominators and related concepts.

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