Understanding $\mathbb{Q}[x]/(x^{2}+1)^{2}$ contains…..

Question

This is a question about an answer to this question:

Is $\mathbb{Z}[x] / \langle (x^2 + 1)^2 \rangle$ isomorphic to a familiar ring?

Here the answer says that the ring $\mathbb{Q}[x]/(x^{2}+1)^{2}$ contains a square root of unity

$i:=\frac{1}{2}x(x^{2}+3)$

I'm not sure what that means. Also why was Newton's method mentioned?

score 3 · Accepted Answer · answered Nov 05 '14 at 14:01

3

$(\frac{1}{2}x(x^2+3))^2=\frac{1}{4}(x^2(x^4+6x^2+9)=\frac{1}{4}(-1(1-6+9))=-1$

answered Nov 05 '14 at 14:01

you are welcome :) @user135520 – Nov 05 '14 at 14:08
why mister profull o took $x^2=-1$? – Myshkin Dec 07 '14 at 18:25
@UneFemmeDouce Who is mister profull o? – awllower Dec 08 '14 at 00:41
@awllower I am mister profullo... – Dec 08 '14 at 18:45
Then why did you take $x^2=-1?$ – awllower Dec 08 '14 at 23:37
@awllower : Original question is different from this edited question... please look at the original question.. – Dec 09 '14 at 17:51
I know that (Cf. my answer). But, after the question was changed, per chance you should edit the answer as well? – awllower Dec 09 '14 at 23:47
1

@awllower : I am a bit busy these days and could not edit that... – Dec 10 '14 at 05:59

awllower · Answer 2 · 2014-11-05T14:15:04.753

3

$$x^2+3\equiv2\pmod{x^2+1},$$ so $$(x(x^2+3)/2)^2\equiv x^2\equiv-1\pmod{x^2+1},$$
and we can think of $x(x^2+3)/2$ as a square root of $-1.$
P.S. the element $x$ suffices for our purpose.
Hope this helps.
Edit:
I found that you wrote the question wrongly: the ring in question should be $\mathbb Q(x)/((x^2+1)^2).$
Then we shall see that $$(x^2+3)^2=x^4+6x^2+9\equiv 4x^2+8\pmod{(x^2+1)^2},$$
thus $$\frac{x^2(x^2+3)^2}{4}\equiv\frac{x^2(4x^2+8)}{4}=x^4+2x^2\equiv-1\pmod{(x^2+1)^2},$$ and this is why the element $x$ does not suffice in that linked answer.
Hope this clarifies some unclear points.

edited Nov 05 '14 at 14:15

answered Nov 05 '14 at 14:04

awllower

16,926

Thanks, I just saw this. The answer below showed why it was a square root of negative one. That's a good way to do it. – user135520 Nov 05 '14 at 14:06
Your question body contains a typo, and I added an answer for the supposedly corrected question. And thanks for the appreciation. – awllower Nov 05 '14 at 14:16
Corrected the typo. Thanks – user135520 Nov 05 '14 at 14:56

Understanding $\mathbb{Q}[x]/(x^{2}+1)^{2}$ contains…..

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