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How can I show that $\Bbb Q(i)$ and $\Bbb Q(\sqrt2)$ as fields are not isomorphic. Is there an element of order $4$ in $\Bbb Q(\sqrt2)$?

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    $\begingroup$ Or, is there an element that squares to $1+1$ in $\Bbb Q(i)$? $\endgroup$ – anon Jan 22 '13 at 2:23
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    $\begingroup$ Or, if you prefer, show that no element of $\mathbb{Q}(\sqrt{2})$ squares to give $-1$. $\endgroup$ – Zach L. Jan 22 '13 at 2:30
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    $\begingroup$ Or, only one of the fields admits a linear ordering that respects the field operations. $\endgroup$ – Andrés E. Caicedo Jan 22 '13 at 2:37
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No. The reason is very simple:

Assume by contradiction that $a+b\sqrt{2}$ has order four. Than $a+b\sqrt{2}$ has order 4 also in the bigger field $\mathbb C$. But the only elements of order four in $\mathbb C$ are $\pm i$ and $a+b\sqrt{2}$ is real...

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Any field homomorphism must send 1 to 1. But 1 generates $\mathbb{Z}$, thus we conclude any field homomorphism fixes $\mathbb{Z}$. What does this say about the action of the homomorphism on $\mathbb{Q}$?

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You can answer your own question about elements of order $4$ by direct computation. A typical element of $\Bbb Q(\sqrt2)$ has the form $p+q\sqrt2$ for $p,q\in\Bbb Q$, and

$$\begin{align*} \left(p+q\sqrt2\right)^4&=\left(p^2+2q^2+2pq\sqrt2\right)^2\\ &=\left(p^2+2q^2\right)^2+2\left(2pq\sqrt2\right)^2+2\left(p^2+2q^2\right)(2pq)\sqrt2\\ &=\left(p^2+2q^2\right)^2+16p^2q^2+4pq\left(p^2+2q^2\right)\sqrt2\;. \end{align*}$$

How set this equal to $1$ and see what you can discover about $p$ and $q$. (This is not the only way to answer the main question, but it certainly works.)

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A little different approach. Assume there's a fields isomorphism $\,f:\Bbb Q(\sqrt 2)\to\Bbb Q(i)\,$ , then

$\exists\,w:=a+b\sqrt 2\in\Bbb (\sqrt 2)\,\,s.t.\,\,f(w)=i\,$ , and from here

$$-1=i^2=f(w^2)=f(a^2+2b^2+2ab\sqrt 2\sqrt 2)=$$

$$=f(a)^2+2f(b)^2+2f(ab\sqrt 2)=a^2+2b^2+2abf(\sqrt 2)$$

since $\,f\,$ fixes all the elements of $\,\Bbb Q\,$ . But then we get that

$$f(\sqrt 2)=-\frac{a^2+2b^2+1}{2ab}\in\Bbb Q\Longrightarrow f(\Bbb Q(\sqrt 2))\subset \Bbb Q\subsetneq \Bbb Q(i)...\text{contradiction!}$$

I'll let you check what happens in the particular case that $\,ab=0\,$.

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  • $\begingroup$ A similar analysis works much more generally - see my answer. $\endgroup$ – Math Gems Jan 22 '13 at 4:40
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Hint $\ $ By the lemma below, if $\sqrt{a},\sqrt{b}\not\in \Bbb Q\,$ then $\,\Bbb Q(\sqrt{a})\cong \Bbb Q(\sqrt{b})\iff \sqrt{ab}\in\Bbb Q$

Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{ab}\ $ all are not in $\rm\,K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\,K$

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K\:,\:$ therefore it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm\color{#C00}{(1)}:\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ which contradicts hypotheses as follows:

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $\color{#C00}{(1)}$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$

$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$

$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{ab}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED

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