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?

  • $\begingroup$ 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

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 $f$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.