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Let $T$ be a splitting field of polynomial $$f(x)=x^3+18x+3\in\mathbb{Q}[x].$$ What is the degree $[T:\mathbb{Q}]$?

My thoughts: the polynomial $f$ is irreducible over $\mathbb{Q}$, therefore the $[\mathbb{Q}(x_i):\mathbb{Q}]=3$, where $x_i$ is a root of $f$. Now I would need to get $[\mathbb{Q}(x_1,x_2):\mathbb{Q}(x_1)]$ and then $[\mathbb{Q}(x_1,x_2,x_3):\mathbb{Q}(x_1, x_2)]$. If $x_2, x_3$ were the complex roots, than $[\mathbb{Q}(x_1,x_2,x_3):\mathbb{Q}(x_1, x_2)]=1$.

So actually I need just $[\mathbb{Q}(x_1,x_2):\mathbb{Q}(x_1)]$. Any ideas?

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If you denote $\alpha$ the real root of $f(x)$, you obtain the minimal polynomial of the complex roots of $f(x)$ dividing it by $x-\alpha$. It will have degree $2$ (actually it is $x^2+\alpha x+\alpha^2+18$), hence the degree of the splitting field over $\mathbf Q$ is $6$.

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By the Rational Root Theorem, the only possible rational roots are $\pm 1, \pm 3$, but substituting shows that none of these are roots. Hence (since $\deg f < 4$), $f$ is irreducible, so $\operatorname{Gal}(f)$ is transitive and thus $3 \mid \#\operatorname{Gal}(f) = [T : \Bbb Q]$.

On the other hand, the discriminant of $f$ is $$\Delta_f = 4(18)^3 - 27(3^2) < 0 ,$$ so $f$ has two complex roots, and so complex conjugation is a nontrivial automorphism of $T / \Bbb Q$ of order $2$ , and hence $2 \mid \#\operatorname{Gal}(f)$. Since the order of the extension satisfies $[T : \Bbb Q] \mid (\deg f)! = 6$, we must have $[T : \Bbb Q] = 6$.

In fact, this generalizes to an efficient algorithm for computing the degree of the extension for any cubic (over $\Bbb Q$): If a cubic polynomial $g$ over $\Bbb Q$ is irreducible, then its splitting field is $\Bbb Q(\alpha, \sqrt{\Delta_g})$ for any root $\alpha$ of $g$ and either root $\sqrt{\Delta_g}$ of $\Delta_g$. Thus, $[T : \Bbb Q] = 3$ (and $\operatorname{Gal}(g) \cong A_3 \cong \Bbb Z_3$) if $\Delta_g$ is a square in $\Bbb Q$ and $[T : \Bbb Q] = 6$ (and $\operatorname{Gal}(g) \cong S_3$).

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    $\begingroup$ As a general question, is there a surefire method to find the degree of the extension of (any) polynomial over $\mathbb{Q}$? Or is it always a mixture of guesses, some theorems (could you list the most useful ones?), and intuition? Thanks! $\endgroup$ – Siddharth Bhat May 1 '16 at 16:58
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    $\begingroup$ It suffices to find the Galois group, of course, and general methods for computing those can be found in any text that covers Galois Theory (I learned some of this from Dummit & Foot's Abstract Algebra). One can find explicit algorithms for the case $\deg g = 4$ and maybe even $\deg g = 5$. $\endgroup$ – Travis May 1 '16 at 17:03
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    $\begingroup$ These notes treat the cubic and the quartic case: math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf $\endgroup$ – Travis May 1 '16 at 17:04
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    $\begingroup$ Ah, I'm self studying from the same book as well, and I found the exercises on "compute the degree of the field extension" way easier than the question here, which piqued my curiosity. Do you have a reference for harder questions on field extensions? $\endgroup$ – Siddharth Bhat May 1 '16 at 17:04
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    $\begingroup$ You might find this thread interesting: math.stackexchange.com/questions/45893/… $\endgroup$ – Travis May 1 '16 at 17:06

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