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Let $M = \mathbb{Q}(\sqrt{3}, i\sqrt[4]{5})$ be an extension of $\mathbb{Q}$. Then work out the basis of $M$ over $\mathbb{Q}$ and show that the extension $M/\mathbb{Q}$ is not a normal extension. So this is how I showed it and just wanted to check that it is actually a correct way of going about it:

So I calculated the basis correctly I think, but my concern is regarding the normal extension bit. So I deduced that $M/\mathbb{Q}$ is separable since its finite and characteristic of $\mathbb{Q}$ is $0$ so to show that M isn't a normal extension, I just showed that $M$ isnt the splitting field of the following polynomial $f = (x^2 - 3)(x^4 - 5)$. $f$ is separable since it has distinct roots but its splitting field is not equal to $M$, so hence we can deduce that $M/\mathbb{Q}$ is not a normal extension. Is this correct?

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  • $\begingroup$ Close. M just has to be the splitting field of some polynomial, not any one that you choose. But there is a theorem about normal extensions containing a single root of an irreducible polynomial. $\endgroup$ – John Brevik Apr 21 '15 at 16:37
  • $\begingroup$ could you maybe expand on that a bit please? $\endgroup$ – user1314 Apr 21 '15 at 16:39
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You are trying to show the right thing, but be careful. There are of course many polynomlas $M$ is not the splitting field of. And even if a polnomial $f$ happens to split in $M$ we can still multiply it with suitable $g$ such that $fg$ does not split. Therefore, to show that $M$ is normal, it is best to exhibit a polynomial $f$ such that

  • $f$ is irrducible over $\mathbb Q$
  • $f$ has some root in $M$
  • $f$ does not have all roots in $M$ (i.e., doesn't cplit completely)

Your choice of an - obviously - reducible polynomial is therefore "suspicious".

Since $\sqrt 3\in M$ implies $-\sqrt 3\in M$, the polynomial $X^2-3$ is not suitable for this task. However, the other generator $i\sqrt[4]5$ can helkp us: It is certainly a root of $f(X)=X^4-5$. So try to answer the three bullit points for this $f$.

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  • $\begingroup$ So taking $f(x) = X^4 - 5$, we know its roots are $\sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, -i\sqrt[4]{5}$ and obviously $\sqrt[4]{5}, -\sqrt[4]{5}$ don't belong to $M$ and so does this suffice to prove that $M/\mathbb{Q}$ isn't normal? $\endgroup$ – user1314 Apr 21 '15 at 16:50
  • $\begingroup$ @user1314 That shows the second and the third of my bullit points (if you possibly expand a bit on the "obviously") $\endgroup$ – Hagen von Eitzen Apr 21 '15 at 16:54
  • $\begingroup$ Well its easy to show irreducibility because none of the roots of $x^4 - 5$ belong to $\mathbb{Q}$ and no quadratic factors of $x^4 - 5$ belong to $\mathbb{Q}[X]$ either so we can deduce it is irreducible? $\endgroup$ – user1314 Apr 21 '15 at 16:55
  • $\begingroup$ Yes, though Eisenstein is quicker :) $\endgroup$ – John Brevik Apr 21 '15 at 18:49

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