Prove if $a$ is a root of $x^2-2$ and $b$ is a root of $\displaystyle x^2-\frac{1}{2}$, then $\Bbb{Q}(a)\cong \Bbb{Q}(b)$

Asked Mar 28 '20 at 20:31

Active Mar 28 '20 at 23:50

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This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise E6(c).

If $a$ is a root of $x^2-2$ and $b$ is a root of $\displaystyle x^2-\frac{1}{2}$, then $\Bbb{Q}(a) \cong \Bbb{Q}(b)$.

$a^2-2=0\implies a=\pm\sqrt{2}$, whereas $\displaystyle b^2-\frac{1}{2}=0\implies b=\pm\sqrt{\frac{1}{2}}$.

Note $\pm2b = a$. In general, $F(a) = F(ca)$ for $c\in F$. (Assume $c \ne 0.)$

Hence $\Bbb{Q}(b)=\Bbb{Q}(\pm2b)=\Bbb{Q}(a)$.

Correct?

edited Mar 28 '20 at 23:50

asked Mar 28 '20 at 20:31

hchar

2

That's good. $ $ – Calvin Lin Mar 28 '20 at 20:34
1

No. It may be the case that $2b\ne a$ but $2b=-a$ – Hagen von Eitzen Mar 28 '20 at 20:44
Hint: $\ 2b^2 = 1\iff (2b)^2 = 2.\ $ Generally we can monicize polynomials this way as in the AC-method – Bill Dubuque Mar 28 '20 at 20:52

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