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Let be $K$ the set of real numbers that can be written as $a+b\sqrt2$, with $a$ and $b$ rational numbers. Prove that $K$ is a field.

I have already proved that $0$ and $1$ $\in K$, and that sum and product of two elements $\in K$. I have also already proved that the opposite $\in K$. I don't know how to prove that the reciprocal $\in K$. Can you help me?

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Hint $\ $ [Rationalize the denominator]( of $\rm\ \dfrac{1}{a + b\sqrt{2}}\quad$ – Bill Dubuque Jun 23 '12 at 14:41
up vote 3 down vote accepted

Let $a+b\sqrt{2}\in K$ and $a+b\sqrt{2}\neq 0$. If $b=0$, then $a+b\sqrt{2}=a\neq 0$ and $1/a$ is its multiplicative inverse. Therefore, we assume $b\neq 0$, which implies that $a^2-2b^2\neq 0$. Otherwise, if $a^2-2b^2=0$, then $\sqrt{2}=\frac{a}{b}$ which is rational since $a, b$ are rational, which contradicts to the fact that $\sqrt{2}$ is irrational.

Define $$\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}.$$ First note that it is well-defined because $a^2-2b^2\neq 0$. Moreover, it belongs to $K$ because $\displaystyle\frac{a}{a^2-2b^2},\frac{b}{a^2-2b^2}$ are rational which follows from the fact that $a,b$ are rational numbers. Finally, we have $$(a+b\sqrt{2})\cdot\left(\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}\right)=\frac{(a+b\sqrt{2})(a-b\sqrt{2})}{a^2-2b^2}=\frac{a^2-2b^2}{a^2-2b^2}=1.$$ That is to say, $\displaystyle\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$ is the multiplicative inverse of $a+b\sqrt{2}$.

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One of $a,b$ might be zero. – tomasz Jun 23 '12 at 13:57
Thank you @Paul.. I have quite understood... but so it isn't really important that the multiplicative inverse is written as $a+b\sqrt2$, is it? – Surfer on the fall Jun 23 '12 at 13:57
Ehm @Paul, why $\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$ is rational? $\sqrt2$ isn't rational! – Surfer on the fall Jun 23 '12 at 14:03
@user1294101: I said that $\displaystyle\frac{a}{a^2-2b^2},\frac{b}{a^2-2b^2}$ are irrational, not $\displaystyle\frac{a}{a^2-2b^2}+\frac{b}{a^2-2b^2}\sqrt{2}$. Also, it's important to show the the multiplicative inverse is in the form of $A+B\sqrt{2}$ where $A,B$ are rational, which shows that it is also in the field $K$. – Paul Jun 23 '12 at 14:04
$\displaystyle\frac{a}{a^2-2b^2}+\frac{b}{a^2-2b^2}\sqrt{2}$ is in the form of $A+B\sqrt{2}$ where $A=\frac{a}{a^2-2b^2}$ and $ B=\frac{b}{a^2-2b^2}$. As I said, $\displaystyle\frac{a}{a^2-2b^2},\frac{b}{a^2-2b^2}$ are rational because $a, b$ are rational. Therefore, $\displaystyle\frac{a}{a^2-2b^2}+\frac{b}{a^2-2b^2}\sqrt{2}=A+B\sqrt{2}$ is in $K$. – Paul Jun 23 '12 at 14:11



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And then? Please help me ;) – Surfer on the fall Jun 23 '12 at 14:04
@user1294101: It should be pretty obvious after you multiply out and get a rational denominator; the hint pretty much does all the real work. Call the denominator $d$; then you have $\frac{a}d-\frac{b}d\sqrt2$, where clearly $\frac{a}d,\frac{b}d\in\Bbb Q$. – Brian M. Scott Jun 23 '12 at 14:13
You're right: obvious! Thanks, I'd give you too the accepted answer, but I can't. Sorry ;) – Surfer on the fall Jun 23 '12 at 16:45

In fact, suppose that $\alpha$ is a root of a monic polynomial with rational coefficients, $$m(x) = x^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0,\qquad a_i\in\mathbb{Q}.$$ Then $$\mathbb{Q}[\alpha]=\{p(\alpha)\mid p(x)\in\mathbb{Q}[x]\}$$ is a field. Moreover, $$\mathbb{Q}[\alpha] = \{q(\alpha)\mid q(x)\in\mathbb{Q}[x],\text{ and }q(x)=0\text{ or }\deg(q)\lt n\}.$$

(Your problem is the special case of $\alpha$ being a root of $x^2-2$).

Proof. First, note that that if $q_1(x) \equiv q_2(x)\pmod{m(x)}$, then there exists a polynomial $t(x)$ such that $m(x)t(x) = q_1(x)-q_2(x)$. Evaluating at $\alpha$, we have that $q_1(\alpha) - q_2(\alpha) = m(\alpha)t(\alpha) = 0$. Hence $q_1(\alpha)=q_2(\alpha)$. Thus, for every polynomial $p(x)\in\mathbb{Q}[x]$, there is a polynomial $q(x)$ with $q=0$ or $\deg(q)\lt \deg(m)$ (namely, the remainder when dividing $p(x)$ by $m(x)$) such that $q(\alpha)=p(\alpha)$. Thus, we get the second equality asserted above.

It suffices to show that if $p(x)\in\mathbb{Q}$ is such that $p(\alpha)\neq 0$, then there exists $t(x)\in\mathbb{Q}(x)$ such that $p(\alpha)t(\alpha) = 1$ (existence of multiplicative inverses). In order to do that, note that if $m(x)$ is reducible, then we may replace it with an irreducible factor of $m(x)$ that vanishes at $\alpha$ (which must exist if $m(\alpha)=0$. If $m(x)$ is irreducible, then since $\mathbb{Q}[x]$ is a UFD, it follows that for every $p(x)\in\mathbb{Q}[x]$, either $\gcd(m(x),p(x))=1$, or else $m(x)|p(x)$. And since $m(x)|p(x)$ implies that $p(\alpha)=0$, then for every $p(x)\in\mathbb{Q}[x]$ we have that either $p(\alpha)=0$, or else $\gcd(m(x),p(x))=1$.

Now, let $p(x)\in\mathbb{Q}[x]$ be such that $p(\alpha)\neq 0$. Then $\gcd(p(x),m(x))=1$, and since $\mathbb{Q}[x]$ is a PID, there exist polynomials $t(x)$ and $s(x)$ such that $$p(x)t(x) + m(x)s(x) = 1.$$ Evaluating at $\alpha$ we get $1 = p(\alpha)t(\alpha) + m(\alpha)s(\alpha) = p(\alpha)t(\alpha)$.

Thus, every nonzero element of $\mathbb{Q}[\alpha]$ has a multiplicative inverse in $\mathbb{Q}[\alpha]$, proving that it is indeed a field.

In your situation, we have that $\alpha=\sqrt{2}$ satisfies $x^2-2$. Thus, we know that $\mathbb{Q}[\sqrt{2}]$ is a field by the above. Since $x^2-2$ is irreducible, then the second part of the original assertion tells us that in fact $$\begin{align*} \mathbb{Q}[\sqrt{2}] &= \{q(\sqrt{2})\mid q(x)\in\mathbb{Q}[x],\text{ and }q=0\text{ or }\deg(q)\lt 2\}\\ &= \{ a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}.\end{align*}$$

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Very great proof! I'm taking a little time to understand it! – Surfer on the fall Jun 23 '12 at 20:27
@user1294101: It is a standard proof that if $\alpha$ is algebraic over $F$, then $F(\alpha)=F[\alpha]$. – Arturo Magidin Jun 23 '12 at 20:28
Why so complicated? Every finite-dimensional integral domain over a field $k$ is a field by elementary linear algebra: For $x \neq 0$ the endomorphism that multiplies with $x$ is injective, hence surjective. So $x$ is invertible. – Martin Brandenburg Jun 23 '12 at 20:43
@MartinBrandenburg That way is efficient, but it usually provokes an irate "How on earth would anybody know to take that approach?!" from students. (I can appreciate the reaction.) Arturo's entry-level approach is pretty good medicine! – rschwieb Jun 24 '12 at 0:57

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