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I am working on a problem from A Book Of Abstract Algebra by Pinter Chapter 31.A.4 which is

Explain: $\mathbb{Q}(i,\sqrt{2})$ is the root field of $x^4 - 2x^2 + 9$ over $\mathbb{Q}$ and is the root field of $x^2 - 2\sqrt{2}x + 3$ over $\mathbb{Q}(\sqrt{2})$.

The root field of a polynomial over a field is the same thing as I have seen elsewhere referred to as a splitting field of a polynomial over a field. I believe they are equivalent definitions but maybe there is a subtle difference?

I can find the roots of $x^4 - 2x^2 + 9$ by factoring into $(x^2 -1)^2 + 8$ so that the roots are $\pm \sqrt{1 \pm 2i\sqrt{2}}$ so that the root field is $\mathbb{Q}(\sqrt{1 \pm 2i\sqrt{2}})$

I am not sure how to argue that this root field is equal to $\mathbb{Q}(i,\sqrt{2})$ I assume it is using the closure properties of the field but can't work the details. I think the square roots are throwing me off?

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  • $\begingroup$ Did you mean $\mathbb{Q}(i,\sqrt{2})$ where you typed $\mathbb{Q}(1,\sqrt{2})$? $\endgroup$ – J. W. Tanner Apr 20 '20 at 14:16
  • $\begingroup$ Yes, thanks. I'll fix it. $\endgroup$ – topoquestion Apr 20 '20 at 14:17
  • $\begingroup$ Try. thinking of succesive adjunctions $\endgroup$ – Chris Leary Apr 20 '20 at 14:31
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We have $$X^4-2X^2+9=(X^2-2\sqrt2 X+3)(X^2+2\sqrt 2X+3).$$ The roots of these quadratic polynomials are $\sqrt{2}\pm i$ and $-\sqrt{2}\pm i$. It is then clear that the splitting field of $X^4-2X^2+9$ over $\mathbf{Q}$ (or "root field" as Pinter calls it) as well as the splitting field of $X^2-2\sqrt{2}X+3$ over $\mathbf{Q}(\sqrt{2})$ is indeed $\mathbf{Q}(\sqrt{2},i)$.

The thing you didn't see was that in fact $$(\sqrt{2}\pm i)^2=2\pm 2\sqrt{2}i-1=1\pm 2\sqrt{2}i,$$ so one can simplify the nested radical $\sqrt{1+2\sqrt{2}i}=\sqrt{2}+i$.

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  • $\begingroup$ That's. I didn't see that simplification. I'm not clear fully on how to compare root fields but I think theorem 2 of this chapter shows $\mathbb{Q}(i + \sqrt{2}) = \mathbb{Q}(i,\sqrt{2})$ $\endgroup$ – topoquestion Apr 20 '20 at 17:08
  • $\begingroup$ As you say. Thanks. $\endgroup$ – topoquestion Apr 20 '20 at 17:09
  • $\begingroup$ That is true, but the reason here is the following. The splitting field is the field extension of $\mathbf{Q}$ obtained by adjoining all roots: $\mathbf{Q}(i+\sqrt{2},-i+\sqrt{2},i-\sqrt{2},-i-\sqrt{2})=\mathbf{Q}(i+\sqrt{2},i-\sqrt{2})=\mathbf{Q}(i,\sqrt{2})$. The last equality is true because $\boxed{\subset}$ is obvious and $\boxed{\supset}$ is true because $\frac12((i+\sqrt{2})+(i-\sqrt{2}))=i$ and $\frac12((i+\sqrt{2})-(i-\sqrt{2}))=\sqrt{2}$. Is is clearer now? :) $\endgroup$ – rae306 Apr 20 '20 at 20:30
  • $\begingroup$ That makes sense, thanks. I missed the $i-\sqrt(2)$ part and couldn't get $i$ and $\sqrt{2}$ separated without it. $\endgroup$ – topoquestion Apr 20 '20 at 20:41

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