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! :)


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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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