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I am trying to do problem 3.28 from the Algebra questions on this site. It says the following:

How would you find the Galois group of $x^3+2x+1$? Adjoin a root to $\mathbb Q$. Can you say something about the roots of $x^3+3x+1$ in this extension?

I can easily find that the Galois group of $x^3+2x+1$ is $S_3$. Let $\theta$ be a root of $x^3+2x+1$. The second question is a bit vague, but I assume the intended answer is that $x^3+3x+1$ has no roots in $\mathbb Q(\theta)$. Indeed, if $x^3+3x+1$ had a root in $\mathbb Q(\theta)$, then it would have a root in the splitting field of $x^3+2x+1$. However, since this splitting field is Galois, this would imply that all of the roots of $x^3+3x+1$ lie in the splitting field of $x^3+2x+1$. We can also see easily that the Galois group of $x^3+3x+1$ is $S_3$, so this would mean that the splitting fields of $x^3+2x+1$ and $x^3+3x+1$ are actually the same.

So my question is:

How can we show that the splitting fields of $x^3+2x+1$ and $x^3+3x+1$ are different fields?

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    $\begingroup$ Look at their discriminants. The discriminant of the first is $-59$ and of the second is $-135$. Therefore, $\sqrt{59}i$ and $\sqrt{15}i$ belong to their splitting fields, respectively. If the fields were the same then its degree over $\mathbb{Q}$ would be multiple of $4$, since it would contain the extension by the roots of $x^2+15$ and then the further extension by the roots of $x^2+59$, which doesn't split in the first. $\endgroup$ – YAlexandrov Mar 9 '18 at 14:33
  • $\begingroup$ Ahh yes that works. Thank you! $\endgroup$ – Colin Defant Mar 9 '18 at 14:44
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    $\begingroup$ @YAlexandrov It may be something you would rather not do, but IMHO you fleshing out that comment to an answer would be welcomed. Your call, of course. $\endgroup$ – Jyrki Lahtonen Mar 9 '18 at 21:49

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