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Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.1

Solution: Let $F=\{x+y\sqrt{2}\mid x,y\in\mb Q\}$. Then we must show six things:

1. $0$ is in $F$
2. $1$ is in $F$
3. If $x$ and $y$ are in $F$ then so is $x+y$
4. If $x$ is in $F$ then so is $-x$
5. If $x$ and $y$ are in $F$ then so is $xy$
6. If $x\not=0$ is in $F$ then so is $x^{-1}$

For 1, take $x=y=0$.

For 2, take $x=1$, $y=0$.

For 3, suppose $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$. Then $x+y=(a+c)+(b+d)\sqrt{2}\in F$.

For 4, suppose $x=a+b\sqrt{2}$. Then $-x=(-a)+(-b)\sqrt{2}\in F$.

For 5, suppose $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$. Then $$xy=(a+b\sqrt{2})(c+d\sqrt{2})=(ac+ 2bd) + (ad+bc)\sqrt{2}\in F.$$For 6, suppose $x=a+b\sqrt{2}$ where at least one of $a$ or $b$ is not zero. Let $n=a^2-2b^2$. Since $\sqrt{2}$ is not rational, we have $n\ne 0$. Let $y=a/n + (-b/n)\sqrt{2}\in F$. Then $$xy=\frac1n(a+b\sqrt{2})(a-b\sqrt{2})=\frac1n(a^2-2b^2)=1.$$ Thus $y=x^{-1}$ and $y\in F$.