3
$\begingroup$

I need to determine in which of the cases (a)-(b) the field $F$ is a splitting field of some polynomial $f \in \mathbb{Q}[x]$ over $\mathbb{Q}$?

$$(a)\qquad F = \mathbb{Q}(\sqrt{2}+\sqrt{3}) \\ (b) \qquad \qquad \qquad \quad F = \mathbb{C}$$

I said that the field in $(a)$ is a splitting field, and exhibited the polynomial $f = x^{4}-5x^{2}+6 \in \mathbb{Q}[x]$ to make my point.

For part (b), however, of course, every polynomial $\mathbb{Q}[x]$ splits over $\mathbb{C}$, but I said it was not a splitting field because any polynomial $f \in \mathbb{Q}[x]\setminus \{0\}$ splits completely in the set of algebraic numbers $\mathbb{A}$, which is a proper subset of $\mathbb{C}$ (since, for example, $\pi \in \mathbb{C}$, but $\pi \notin \mathbb{A})$, and by definition, a splitting field is the "smallest" (in terms of inclusion) field over which a polynomial splits, is it not?

However, I am not sure if I answered this question correctly, and was wondering if someone out there might not mind serving as an extra set of eyes to let me know whether I did so, and if not, how I can fix this so that it is answered correctly.

Thank you for your time and patience! :)

$\endgroup$
5
$\begingroup$

Everything looks good.

Once one has shown$^\dagger$ that $\mathbb{Q}(\sqrt{3} + \sqrt{2}) = \mathbb{Q}(\sqrt{3}, \sqrt{2})$, it becomes more obvious that $f(x) = x^4 -5x^2 + 6 = (x^2-2)(x^2-3)$ splits in this field. You might want to discuss why this is the smallest field inside which $f$ splits, but this isn't difficult.

Next, as you've argued, $\mathbb{C}$ cannot be a splitting field for any polynomial: we'll always have $\mathbb{Q} \Big( \{ \alpha_k \}_{k=1}^n \Big) \subsetneq \mathbb{C}$ where the $\alpha_k$'s are the roots of the polynomial, so $\mathbb{C}$ cannot be the smallest field inside which that polynomial splits.


$\dagger$ See arguments here or here for further discussion.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ actually, I'm not entirely sure how to show that it is the smallest field inside which $f$ splits. Do you have any hints? I mean, I know it needs a splitting field of dimension $4$ over $\mathbb{Q}$ because of the fact that $\mathbb{Q}(\sqrt{3} + \sqrt{2}) = \mathbb{Q}(\sqrt{3}, \sqrt{2})$ (which was proven in our lecture notes, so I'm allowed to assume it as a given in this case). And I know that its splitting field needs to contain all the polynomial's roots. But I can't say I know how to show it is the smallest. $\endgroup$ – ALannister Aug 17 '17 at 20:07
  • 1
    $\begingroup$ @ALannister, the splitting field of a polynomial in $\mathbb{Q}[x]$ with roots $\{a_k\}_{k=1}^n$ will always be $\mathbb{Q} \Big( \{a_k\}_{k=1}^n \Big)$ as one equivalent way of thinking about $F(S)$ is "the smallest field containing both $F$ and $S$". The trick here is that we want to reduce the "$F(S)$" notation down so that the "S" is as easy to work with as possible. For example, we could say that the splitting field for $x^4 -2$ is $\mathbb{Q}(\sqrt[4]{2}, w \sqrt[4]{2}, w^2 \sqrt[4]{2}, w^3\sqrt[4]{2})$ where $w$ is a fourth root of unity and be technically correct....... $\endgroup$ – Kaj Hansen Aug 17 '17 at 20:16
  • $\begingroup$ However, the "best" answer for the splitting field for $x^4 - 2$ would be $\mathbb{Q}(\sqrt[4]{2}, i)$, even though this field is the same as the one I wrote above. Now, in this problem, the roots are $\pm \sqrt{2}, \pm \sqrt{3}$, so our splitting field is necessarily $\mathbb{Q}( \pm \sqrt{2}, \pm \sqrt{3})$. But we want this in a "simplest form". Of course, this will be equal to the field $\mathbb{Q}(\sqrt{3}, \sqrt{2})$. At this point, we're pretty good to go; $\sqrt{2}$ and $\sqrt{3}$ are algebraically independent, so we won't be able to do much better than that. $\endgroup$ – Kaj Hansen Aug 17 '17 at 20:20

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.