Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists b...

Question

Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists $b \in \mathbb{Q}(i)$ which is a root of $a(x)=-1+8x^2+4x^4$.

I have looked at the minimum polynomial for $u$, and I can easily show why $u$ is not actually in $\mathbb{Q}(i)$, just by showing that the degree of $min(i,\mathbb{Q})$ does not equal the degree of $min(u,\mathbb{Q})$. I can't figure out how $a(x)$ is even related.

Keep in mind this is only my third term of abstract, so I only know the basics.

score 3 · Answer 1 · answered May 29 '14 at 21:21

3

Well, if $u\in\Bbb Q(i),$ then $u=x+iy$ for some $x,y\in\Bbb Q,$ yes? Now, squaring both sides, we have $$i+2=x^2-y^2+2xyi.$$ Hence, $x^2-y^2=2$ and $y=\frac1{2x},$ so....

answered May 29 '14 at 21:21

Cameron Buie

102,994

1

so ... $4x^4-8x^2-1=0$, but $a(x)=4x^4+8x^2-1$, so ... – Nicky Hekster May 29 '14 at 21:29
@NickyHekster: consider $ix$ – robjohn May 29 '14 at 21:38
@robjohn, exactly, you should have added that to the OP, considering his last remark. – Nicky Hekster May 29 '14 at 21:41
@NickyHekster: I'm sorry; I wasn't sure if your comment was simply a continuation of, or trying to point out a problem with, Cameron's hint. – robjohn May 29 '14 at 21:49
I am sorry, but I can't tell where this is going. Why would 4x^4-8x^2-1=0 follow right from y=1/(2x)? I had basically done the things that have happened so far, but I can't figure out what to do with the pieces. – user148884 May 30 '14 at 07:19
Oh my lanta never mind I don't know what is wrong with me. Thank you you all have been gems. – user148884 May 30 '14 at 07:23
@user148884: Glad you figured it out! – Cameron Buie May 30 '14 at 15:30

Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists b...

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