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Is $\mathbb{Q}(\sqrt{pq})$ a subfield of $\mathbb{R}$?

I have a feeling that the answer is no but I can't prove it. It clearly contains $0$ and $1$, and is closed under addition and multiplication. There is an inverse to addition and the inverse to multiplication for $a+b\sqrt{pq}$ is given by $\frac{a-\sqrt{pq}b}{a^2-pqb^2}$

Please help

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SlyxBrd
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    $\Bbb Q(\alpha)$ is a subfield of $\Bbb R$ for all $\alpha\in \Bbb R$ by definition. It is the smallest field containing $\Bbb Q$ and the element $\alpha$. – Dietrich Burde May 31 '20 at 18:26
  • Thank you. It was asked in a way that suggested the answer was no. – SlyxBrd May 31 '20 at 18:29
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    This depends on how $\Bbb Q(\alpha)$ was defined in your homework. Another possibility for a definition is given in this duplicate. (Just replace $2$ by $pq$, it does not matter what $p$ and $q$ are, it is always a quadratic field extension). – Dietrich Burde May 31 '20 at 18:30
  • And it depends on what p and q are. – Arno Fehm May 31 '20 at 18:31
  • What is $\math Q(\sqrt {pq})$? If it's a field extension of $\mathbb Q$ by including the element $\sqrt{pq}$ then its a field by definition. And as $\sqrt{pq}$ is real and as $\mathbb R$ is a field the every element of $\mathbb Q\sqrt{pq}$ are in $\mathbb R$. SO it's a subfield by definition. – fleablood May 31 '20 at 18:34
  • @fleablood Sometimes, for homework, $\Bbb Q(\sqrt{w})$ is defined as all $a+b\sqrt{w}$ and one has to show inverse etc., see the duplicate. – Dietrich Burde May 31 '20 at 18:35
  • the question was about $\mathbb(Q(\sqrt{15})={a+b\sqrt{15}}$ – SlyxBrd May 31 '20 at 18:35
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    This works exactly like $\Bbb Q(\sqrt{2})$, or $\Bbb Q(\sqrt{5})$ here. – Dietrich Burde May 31 '20 at 18:36
  • Thank you. The previous question was about $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ That's why I thought there might be a problem with something like $a+b\sqrt{6}$ – SlyxBrd May 31 '20 at 18:37
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    Okay, then as $\mathbb Q \subset \mathbb Q(\sqrt{pq}) \subset \mathbb R$ then multiplication, and addition are associative, commutative and distributative and $1$ and $0$ are unique multiplicative and additive identities. So it just remains to show that $a+b\sqrt{pq};a,b\in \mathbb Q$ have additive inverse and if $a+b\sqrt{pq}\ne 0$ then it has a multiplicative inverse. So all is done. – fleablood May 31 '20 at 18:39
  • Actually.... stupid question. Are $p,q$ integers? If $p, q$ are real numbers so that $pq$ is irrational then $\mathbb Q(\sqrt{pq})$ need not be a field. If $\sqrt{pq}$ is algebraic and solves a minimal polynomial of degree more than $2$ it might not be. ANd if $\sqrt{pq}$ is transcendental then it must certainly isn't a field. – fleablood May 31 '20 at 19:08
  • I suppose an equally stupid question. Is $pq \ge 0$. If $pq < 0$ then $\sqrt{pq}$ is imaginary and $\mathbb Q(\sqrt{pq}) \not \subset \mathbb R$. – fleablood May 31 '20 at 19:19

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Okay. Let's do this, slowly and tediously..

$\mathbb Q(\sqrt{pq}) = \{a+b\sqrt{pq}|a,b \in \mathbb Q\}$.

1) $\mathbb Q(\sqrt{pq}) \subset \mathbb R$.

well.... duh.... assume $\sqrt{pq} \in R$ then for any $a,b \in \mathbb Q$ then $a + b\sqrt{pq} \in R$ because $R$ is a field and closed under addition and multipliations.

2) $\mathbb Q(\sqrt{pq})$ is field.

Okay. It inherits multiplication and addition from $\mathbb R$ and so multiplication and addition is a associative, commutative and addition distributions over multiplication (or is that terminology the other way around).

And there is a unique additive identity $0$ and a unique multiplicative identity $1$ and every term has a unique additive inverse and ever non-zero term has a unique multiplicative inverse....... IN $\mathbb R$

So all we have to do is show:

a) $0\in \mathbb Q(\sqrt{pq})$

b) $1 \in \mathbb Q(\sqrt{pq})$

c) For every $a+b\sqrt{pq} \in \mathbb Q(\sqrt{pq})$ the $-(a+b\sqrt{pq})\in \mathbb Q(\sqrt{pq})$

d) For every $a + b\sqrt{pq} \in \mathbb Q(\sqrt{pq}; a+b\sqrt{pq}\ne 0$ then $\frac 1{a+ b\sqrt{pq}} \in \mathbb Q(\sqrt{pq}$.

a)b)c) are trivial $0 = 0 + 0*\sqrt{pq}$, $1 = 1+ 0*\sqrt{pq}$. $-(a+b\sqrt{pq})= -a + (-b)\sqrt{pq}$ and $-a,-b\in \mathbb Q$.

And d) is not trivial but very easy.

We can assume $a - b\sqrt{pq} \ne 0$. (if $a= b\sqrt{pq}$ and $\sqrt{pq}$ is irrational than $b=0$ and $a = 0$ so $a+b\sqrt{pq} =0$)

$\frac {1}{a+b\sqrt{pq}} = \frac {1}{a+b\sqrt{pq}}\frac {a-b\sqrt{pq}}{a-b\sqrt{pq}}=\frac {a -b\sqrt{pq}}{a^2 -b^2pq}=\frac {a}{a-b^2pq}- \frac {b}{a-b^2pq}\sqrt{pq}$

and $\frac {a}{a-b^2pq} \in \mathbb Q$ and $-\frac {b}{a-b^2pq}\in \mathbb Q$.

..... Oh.... I suppose I should have asked. Are $p,q$ integers? If $p,q$ are real numbers where $pq$ is not a rational number... well, we can address that in an addendum...

Okay.... addendum.

If $pq$ are not rational then no, $\mathbb Q(\sqrt{pq})$ need not be a field. For example if $pq$ is transcendental then $\sqrt{pq}$ is too. and Now for rational $a,b;b\ne 0$ then $(a+b\sqrt{pq})(x+y\sqrt{pq})= 1$ will have no rational $x,y$ solutions.

fleablood
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