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I'm learning about splitting fields but I'm not sure if I am right. Hopefully I can get some insights on whether I have been learning correctly.

The question asks to find the degree of the splitting fields over $\mathbb Q$.

In $x^4+x^3+x+1$, we can factorize it to become $(x+1)^2(x^2-x+1)=(x+1)^2(x-\frac{1+\sqrt 3i}2)(x-\frac {1-\sqrt 3i}{2})$. Am I right to say that we want to find $[\mathbb Q(\sqrt 3i):\mathbb Q]$? Hence the answer is just $2$.

In $x^4+x^2+x+1$, we can factorize it to become $(x-\frac 14\sqrt5+\frac 14-\frac14i\sqrt2\sqrt{5+\sqrt5})(x+\frac 14\sqrt5+\frac 14-\frac14i\sqrt2\sqrt{5-\sqrt5})\\(x+\frac 14\sqrt5+\frac 14+\frac14i\sqrt2\sqrt{5-\sqrt5})(x-\frac 14\sqrt5+\frac 14+\frac14i\sqrt2\sqrt{5+\sqrt5})$.

Hence, we want to find $\mathbb Q(\sqrt5,i\sqrt2\sqrt{5+\sqrt5},i\sqrt2\sqrt{5-\sqrt5})$. But how do we go about doing this?

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  • $\begingroup$ Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $\mathbb Q(\sqrt5,i\sqrt2\sqrt{5+\sqrt5},i\sqrt2\sqrt{5-\sqrt5})$? $\endgroup$ – Dietrich Burde Jan 9 at 12:11
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Your field is $\mathbb{Q}(\sqrt{5})(i\sqrt{10-2\sqrt{5}})$, because the product of $i\sqrt{2}\sqrt{5 \pm \sqrt{5}}$ is $-2(5-\sqrt{5})\in \mathbb{Q}(\sqrt{5})$.

So you have a biquadratic extension and the degree is $4$.

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  • $\begingroup$ How did we get the degree to be 4? Should it be 8? $\endgroup$ – Icycarus Jan 9 at 12:03
  • $\begingroup$ I edited, now there shouldn’t be any reason why the degree is $8$. $\endgroup$ – Mindlack Jan 9 at 12:06
  • $\begingroup$ ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well? $\endgroup$ – Icycarus Jan 9 at 12:10
  • $\begingroup$ I think your own answer to the first point is correct. $\endgroup$ – Mindlack Jan 9 at 12:11

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