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Determine all the subfields of the splitting fields of this polynomial.

I chose this problem because I think to complete it in great detail will be a great study tool for all of the last chapter, as well. (I'm using Dummit & Foote).

Can I have help outlining the method? I need to first find the splitting field, correct? Is there a more elegant observation than noting that I can use difference of squares factoring and rewrite the polynomial as $(x- \sqrt{2})(x+\sqrt{2})(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})$, where none of $\sqrt{2}, \sqrt{3}, \sqrt{5}$ are in the field $\mathbb{Q}$? (The problem doesn't say this is the base field, but I'm assuming.) So the splitting field is $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$. Is this how you recommend finding the splitting field, i.e. explicitly calculating the roots and ensuring your extension contains them? I wish there were a more theoretical way of doing this that would apply to any polynomial.

Now, to find the Galois group, do you exhaustively compute automorphisms and just observe which elements the automorphism fixes in order to find the corresponding fixed field? This seems to get complicated very quickly. Is there some theoretical machinery to aid in this, as well?

Once finding the fixed fields are accomplished, from the Fundamental Theorem of Galois Theory I know there exists a correspondence between the fixed fields and the subfields, so if I find all fixed fields of automorphisms, I will have found all subfields.

Thank you for any help.

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Note that any automorphism $\sigma$ must send $\alpha\in \mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ to another root of the minimal polynomial of $\alpha$. In particular, $\sigma\sqrt2=\pm \sqrt2,\sigma\sqrt3=\pm\sqrt3,$ and $\sigma\sqrt5=\pm\sqrt5$. Since $\{\sqrt2,\sqrt3,\sqrt5\}$ is a generating set for $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ as a field over $\mathbb Q$, we see that $\sigma$ is uniquely determined by $\sigma\sqrt2,\sigma\sqrt3,$ and $\sigma\sqrt5$. Thus the only possible automorphisms are $$\begin{align} a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a+b\sqrt2+c\sqrt3+d\sqrt5\\ a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5\\ &\vdots\\ a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a-b\sqrt2-c\sqrt3-d\sqrt5\\ \end{align}$$ and in order to show that these are all automorphisms, it suffices to show that there are exactly $8$ automorphisms of $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$. Since $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ is the splitting field of a polynomial, we know that it is Galois, so the number of automorphisms it has is equal to its degree. Recall that $$[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]=[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)][\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt2)][\mathbb Q(\sqrt2):\mathbb Q]$$ and so it suffices to show that each degree on the RHS is $2$. Clearly they are at most $2$, as each extension is obtained by adjoining a root of a quadratic polynomial. Thus it suffices to show that each extension on the RHS is nontrivial, i.e. that $\sqrt2\notin \mathbb Q,\sqrt3\notin \mathbb Q(\sqrt2)$ and $\sqrt5\notin \mathbb Q(\sqrt2,\sqrt3)$. The first is a famous theorem. Since $\mathbb Q(\sqrt2)$ has basis $\{1,\sqrt2\}$, if $\sqrt3\in \mathbb Q(\sqrt2)$ we would have $\sqrt3=a+b\sqrt2$ with $a,b\in\mathbb Q$, so $3=a^2+2b^2+2ab\sqrt2$. Since $\sqrt2$ is irrational we must have $2ab=0$, so $a=0$ or $b=0$, thus we need only observe that $3$ and $3/2$ are not squares in $\mathbb Q$. The same technique (with some additional effort) works to show that $\sqrt5\notin \mathbb Q (\sqrt2,\sqrt3)$, observing that $\{1,\sqrt2,\sqrt3,\sqrt6\}$ is a basis for $\mathbb Q(\sqrt2,\sqrt3)$.

Once you see that these are the automorphisms, it should be relatively easy to see what their fixed fields are. For example, the map $$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5$$ has fixed field $\mathbb Q(\sqrt3,\sqrt5)$ while the map $$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2-c\sqrt3+d\sqrt5$$ has fixed field $\mathbb Q(\sqrt6,\sqrt5)$ (why $\sqrt6$?). These fixed fields are all the maximal subfields of $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$, and the remaining intersections are pairwise intersections of these subfields (since the only other nontrivial subgroups of the Galois group are generated by pairs of automorphisms), which are easy to determine.

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  • $\begingroup$ In this particular problem it seems obvious, but how do I know that $\sqrt{2}$ and $-\sqrt{2}$ are the only roots of the polynomial $(x^2-2)$, such that the automorphism can only permute the root two ways? Put another way, how do I know there are only two roots to that equation, when I haven't proved the Fundament Theorem of Algebra yet. $\endgroup$ – Jared Jun 5 '13 at 17:22
  • $\begingroup$ @Questioneer That any polynomial of degree $n$ has at most $n$ roots is a much easier fact to prove; it follows from the factor theorem, which states that if $c$ is a root of $f(x)$ then $(x-c)\mid f(x)$. Induction shows that $f(x)$ has at most $\mathrm{deg}(f)$ roots. $\endgroup$ – Alex Becker Jun 5 '13 at 17:25
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    $\begingroup$ @AlexBecker How do you know a generic element of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ looks like $a+ b\sqrt{2} + c\sqrt{3} + d\sqrt{5}$? For example, why is $\sqrt{6}$ expressible in such a form? $\endgroup$ – Sam Y. Apr 4 '17 at 0:24
  • $\begingroup$ @SamY.: I do not think that Alex Becker is claiming that generic elements are expressible as a $\Bbb Q$-linear combination of $1,\sqrt 2,\sqrt 3, \sqrt 5$, but rather using a shorthand to write down the $8$ distinct $\Bbb Q$-automorphisms since it suffices to say what such an automorphism does to $\sqrt 2,\sqrt 3, \sqrt 5$. Indeed, since $K=\Bbb Q(\sqrt 2,\sqrt 3,\sqrt 5)$ has degree $8$ over $\Bbb Q$, a $\Bbb Q$-basis of $K$ is $1,\sqrt 2,\sqrt 3, \sqrt 5, \sqrt 6, \sqrt{10}, \sqrt{15}, \sqrt{30}$, so a generic element is a combination of these $8$ elements. $\endgroup$ – Alex Ortiz Mar 18 '19 at 1:05
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So for your first question, the splitting field will be generated by the roots of the polynomial you are trying to split. Sometimes, some of the roots might have relations and make some other roots redundant, and also, there will be times that it is "nicer" to find some other generators for your polynomial (for instance the splitting field for $x^4-10$ would be $\mathbb{Q}(\omega, \sqrt[4]{10})$, where $\omega$ is the primitive fourth root of unity).

Back to your problem, take your polynomial and factor it into irreducible monic polynomials. In your case you would get that $(x^2-2)(x^2-3)(x^2-5)$ is already factored into irreducible parts. Note that $x^2-2$ is the minimal polynomial of $\sqrt{2}$, so $\sqrt{2}$ can only be mapped to $\pm\sqrt{2}$, similarly for $\sqrt{3}$ and $\sqrt{5}$. This reduces the possibilities for your mappings a lot. You have that you can send $\sqrt{2}$ to $\pm\sqrt{3}$, $\sqrt{3}$ to $\pm\sqrt{3}$, and $\sqrt{5}$ to $\pm\sqrt{5}$. Each combination of possibilities gives you an automorphism, so your Galois Group will be of order $2^3=8$, also note that every map is of order $2$, so the only group with order $8$ and every element of order $2$ is $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

From here is it straight forward to compute the fixed fields. Say you have the automorphism that negates $\sqrt{2}$ and $\sqrt{3}$ but fixes $\sqrt{5}$, then the fixed field would be just $\mathbb{Q}(\sqrt{5})$ (do you see why?), and it's the same for every automorphism.

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  • $\begingroup$ I don't quite get what you mean by "Sometimes, some of the roots might have relations and make some other roots redundant, and also, there will be times that it is "nicer" to find some other generators for your polynomial". Actually any finite extension of $\mathbb{Q}$ can be generated by one element. I think what is behind your example is that sometimes there are larger groups of conjugate roots (due to an irreducible factor of higher degree, in your case 4). $\endgroup$ – Ash GX Jun 5 '13 at 17:43
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To add to Mr. Becker's answer, an easier way of seeing that $\sqrt 5 \notin \mathbb Q(\sqrt2,\sqrt3)$ (once it has been determined that that $\sqrt 2, \sqrt 3$, and $\sqrt 6$ are linearly independent over $\mathbb Q$) is to notice that the Galois group of $\mathbb Q(\sqrt2,\sqrt3)$ is the Klein-4 group, which has 3 subgroups of order 2. By the fundamental theorem of Galois theory, these subgroups correspond to the quadratic extensions $\mathbb Q(\sqrt 2), \mathbb Q(\sqrt3)$ and $\mathbb Q(\sqrt 6)$. By following the same steps used to show that $\sqrt 2 \notin \mathbb Q(\sqrt 3)$, one can see that $\sqrt 5$ does not belong to any of the aforementioned quadratic extensions. So if $\sqrt 5 \in \mathbb Q(\sqrt2,\sqrt3)$, then $\mathbb Q(\sqrt 5)$ would constitute a fourth quadratic extension, corresponding to a fourth subgroup of degree 2 in the Klein-4 group, a contradiction.

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