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Given:

$( \{a+b \sqrt{n}\,|\,a,b\in \mathbb{Z} \},+,\times)$ and $( \{a+bi\sqrt{n}\,|\,a,b\in \mathbb{Z} \},+,\times)$ with $n\in\mathbb{N}^+$.

Show that in the given example, if we switch the ring of the integers for the field of rational numbers (e.g if we take $a,b\in\mathbb{Q}$) we will obtain fields.

I proceded from the definition of a field:

A field is a ring $(K,+,\times)$ that has an inverse multiplicative for every $n\in K\neq 0$ so that $x\times y=1$.

I thought that the demonstration consist in finding a $y$ such that $(a+b \sqrt{n})\times y=1$, solving for $y$, I'll obtain:

$$y=\frac{1}{a+b \sqrt{n}}$$

Which is a rational number for some $n$'s, is that correct? I'm starting to doubt it because I guess for some $n$'s (e.g. $n=2$), we'll have a real number.

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Red Banana
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2 Answers2

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You don't need $y$ itself to be rational (that usually isn't possible). You need $y$ to be of the form $$c+d\sqrt n$$ for some rationals $c,d$. In the case that $n$ isn't a perfect square, try rationalizing the denominator of $$y=\frac1{a+b\sqrt n}$$ to find this form of $y$. If $n$ is a perfect square, then the structure should be far more familiar. (What is it?)

Cameron Buie
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  • I don't understand: How is $c+d\sqrt n$ the multiplicative inverse of $a+b\sqrt n$? – Red Banana May 09 '13 at 22:48
  • Assuming that $n$ isn't a perfect square, set $(a+b\sqrt n)(c+d\sqrt n)=1.$ solve the resulting system for $c$ and $d$. Show that both are rational. – Cameron Buie May 09 '13 at 22:54
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First of all, there is no need to do this separately: just take $n\in\mathbb{Z}$ and the two cases can be treated in the same way.

The simple case is when $n$ is a perfect square: $n=m^2$. Then the set you want to verify to be a field is just $\mathbb{Q}$.

So we can assume that $n$ is not a square. Therefore

$$ a^2 - nb^2\ne0 $$

for all $a,b\in\mathbb{Q}$. Otherwise

$$ n=\biggl(\frac{a}{b}\biggr)^2 $$

which is impossible. (Proof?)

Next, consider

$$ y=\frac{1}{a+b\sqrt{n}} $$

that you know is defined in $\mathbb{C}$; we want to reduce it to the form $y=c+d\sqrt{n}$ where $c,d\in\mathbb{Q}$. Just observe that

$$ y=\frac{a-b\sqrt{n}}{a^2-nb^2} $$

and that the denominator is non zero. Therefore

$$ c=\frac{a}{a^2-nb^2},\qquad d=\frac{-b}{a^2-nb^2} $$

are the rational numbers you wanted.

Also you should check closure for addition and multiplication, which are easy.

egreg
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