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I am trying to figure out some things regarding field extensions and some questions have arisen on the way.

Let $a$ be a positive integer which doesn't have a rational $nth$ root:

  1. Is the splitting field of $x^n−a$ always equal to the splitting field of $x^n+a$? If not, when is this the case?

  2. When is the $nth$ cyclotomic polynomial irreducible over $Q(\sqrt[n]{a})$? If $gcd(a,n)=1$ is the cyclotomic polynomial then always irrecucible?

  3. Letting $a=1$. If $n\geq 3$ is the splitting field of $x^n-1$ equal to the splitting field of $x^n+1$?

Comment on 3: For example, the minimal polynomial of the roots of $x^3-1$ (the 3rd roots of unity) is $x^2+x+1$ and the minimal polynomial of the roots of $x^3+1$ is $x^2-x+1$. Since $x^2-x+1$ is the minimal polynomial of the 6th roots of unity we should have that the splitting field of $x^3-1$ is contained in the splittingfield of $x^3+1$. Since the two fields have the same degree over $\mathbb{Q}$ this would imply that they are the same. Is this true in general?

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  • $\begingroup$ The case a=1 is treated at point 3. I should have added that $a>0$ $\endgroup$
    – harajm
    Apr 30, 2013 at 3:16

2 Answers 2

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Powers of $2$ are sometimes easy to ignore. The splitting field of $x^2-2$ is very different from the splitting field of $x^2+2$, and the splitting fields of $x^4-1$ and $x^4+1$ are different as well.

Your question #2 is not so easily answered as the others, I think.

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  • $\begingroup$ Is there any general rule for questions 1 and 3? When can we know that the splitting fields are the same? $\endgroup$
    – harajm
    Apr 30, 2013 at 3:19
  • $\begingroup$ When $n$ is odd, $x^n-a$ and $x^n+a$ can be transformed into each other by $x\mapsto-x$, after all. $\endgroup$
    – Lubin
    Apr 30, 2013 at 3:23
  • $\begingroup$ Can you make the argument explicit, I'm slow. Why does this imply that the splitting fields are the same? $\endgroup$
    – harajm
    Apr 30, 2013 at 3:26
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Let's assume we are working over $\mathbb{Q}$ for simplicity with $\overline{\mathbb{Q}}$ a fixed algebraic closure.

Then the splitting field of $x^n-a$ is the smallest extension of $\mathbb{Q}$ containing all the $n$th roots of $a$, which are $a^{1/n}, \zeta_n a^{1/n}, \ldots , \zeta_n^{n-1} a^{1/n}$, where $a^{1/n}$ is some $n$th root of $a$ and $zeta_n$ is a primitive $n$th root of unity. On the other hand, the splitting field of $x^n + a$ is the smallest extension of $\mathbb{Q}$ containing all the $n$th roots of $-a$. Notice that $(-a)^{1/n} = (-1)^{1/n} a^{1/n}$. The difference is an $n$th root of $-1$. When does $-1$ have an $n$th root? When $n$ is odd, $(-1)^n = -1$. Hence, when $n$ is odd, $\mathbb{Q}$ already has an $n$th root of $-1$ and thus we need add nothing more to the splitting field of $x^n-a$ to get a splitting field of $x^n + a$.

This fact is nicely summarized in the question "Which roots of unity lie in $\mathbb{Q}(\zeta_n)$?" When $n$ is odd, then $\mathbb{Q}(\zeta_n)$ contains all the $2n$th roots of unity, and when $n$ is even it contains just the $n$th roots of unity. This should answer your third question.

So, to completely answer your first question: the splitting fields of $x^n - a$ and $x^n + a$ are the same if $n$ is odd. When $n$ is even, the latter polynomial splits in a quadratic extension of the former polynomials splitting field. In particular, $K = \mathbb{Q}(a^{1/n},\zeta_n)$ is a splitting field of $x^n-a$ and $K(\sqrt{\zeta_n})$ is a splitting field of $x^n + a$.

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  • $\begingroup$ Great answer! But, the sentence: So, to completely answer your first question: the splitting fields of $x^n−a$ and $x^n+a$ are the same iff n is odd. Shouldn't you remove one f here, I.e $x^n−a$ and $x^n+a$ are the same if $n$ is odd. I am thinking about the splitting field of $x^4-2$ and $x^4+2$ over $\mathbb{Q}$ for example, which are both equal to $\mathbb{Q}(\sqrt[4]{2},i)$ $\endgroup$
    – harajm
    May 2, 2013 at 17:34
  • $\begingroup$ Yes! Good catch. I made the change to the post. $\endgroup$
    – tghyde
    May 2, 2013 at 20:13

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