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I would like to know how to solve part $ii)$ of the following problem:

Let $K /\mathbb{Q}$ be a splitting field for $f(X) =X^4-3X^2+5$.

i) Prove that $f(X)$ is irreducible in $\mathbb{Q}[X]$

ii) Prove that $K$ has degree $8$ over $\mathbb{Q}$.

iii) Determine the Galois group of the extension $K/\mathbb{Q}$ and show how it acts on the roots of $f$.

I've done part i), and have found the roots of $f$ explicitly as:

$$\pm\bigg(\frac{3\pm\sqrt{-11}}{2}\bigg)^{1/2}$$

but am not sure how to show that the extension has degree $8$. If $x_1$ is the root where both of the $\pm$ signs above are $+$ and $x_2$ is the root where only the outer sign is a $+$, then $K = \mathbb{Q}(x_1,x_2)$. By part $i)$, $x_1$ has degree $4$ over $\mathbb{Q}$ and then $x_2$ has degree $1$ or $2$ over $\mathbb{Q}(x_1)$, but I'm not sure how to show that this degree is $2$, or prove the result by other means.

Due to the ordering of the parts, I would expect there to be an answer for ii) that doesn't require computing the entire Galois group of the extension, so would appreciate something along these lines.

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Tom Oldfield
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  • Let $r$ be a root of $f(X)$. Show $[K:\mathbf Q(r)] = 2$ by finding a root $s$ of $f(X)$ that is not in $\mathbf Q(r)$ and checking that all the roots of $f(X)$ are in $\mathbf Q(r,s)$ and that $s$ is quadratic over $\mathbf Q(r)$. – KCd Dec 29 '15 at 23:18
  • If you call one of those roots $\alpha$ then how does the polynomial split over $\Bbb Q[\alpha]$? It has to split into $(x-\alpha)(x+\alpha)g(x)$ where $g(x)$ is a quadratic. If you can show $g$ is irreducible then you're done. Perhaps factor the whole quartic as a quadratic in $x^2$ and you'll see $\pm\alpha$ are the roots of one of the quadratics. Then show the other quadratic is irreducible over $\Bbb Q[\alpha]$. – Gregory Grant Dec 29 '15 at 23:20
  • @KCd I can see that $x_1^2=x_2^2$ and so $x_2$ is degree $1$ or $2$ over $\mathbb{Q}(x_1)$, but the point is that I don't know how to show that $x_2 \not\in \mathbb{Q}(x_1)$. Do you have any suggestions? – Tom Oldfield Dec 29 '15 at 23:21
  • @GregoryGrant I called one of them $x_1$ as in the question, and then showed that $g(X) = X^2-x_1^2$. I'm exactly stuck on showing that this is irreducible over $\mathbb{Q}(x_1)$, which is equivalent to showing that $x_2 \not\in\mathbb{Q}(x_1)$. Do you have any suggestions? – Tom Oldfield Dec 29 '15 at 23:23
  • It is false that $x_1^2 = x_2^2$: otherwise $x_2 = \pm x_1$, but $x_2$ is neither $x_1$ nor $-x_1$. – KCd Dec 29 '15 at 23:30
  • Where is this question coming from (old qualifying exam?), and what is the assumed background for solving this problem, e.g., are you not allowed to use algebraic number theory? – KCd Dec 29 '15 at 23:34
  • @KCd Ah yes, sorry, I had meant $x_1^2 = \bar{x_2}^2$. The problem comes from an old Galois theory prelim exam and so the background is just (fairly basic) Galois theory. I think even conditions on classifying Galois groups based on numerical properties of the quartic (e.g. discriminant, resolvent cubic) wouldn't be what the examiner had in mind. – Tom Oldfield Dec 29 '15 at 23:40
  • Since this is meant to be a "basic" question I am wondering if there might be a typographical error, e.g., perhaps the polynomial should be $x^4 - 3x^2 - 5$. I am suggesting this because in these types of exam questions about Galois groups of quartics in $\mathbf Q[x]$ you can usually do some trick about real vs. nonreal roots to make progress in place of using more machinery. With constant term $-5$ such a trick is available (the quartic has real and nonreal roots), but with constant term $5$ this trick doesn't work. – KCd Dec 29 '15 at 23:48
  • @KCd It's possible, none of the "elementary" things that I've tried doing have worked. I've thought that since $K$ will be a degree two extension over the fixed field of conjugation, it might be an idea to try and show that this fixed field has degree four over $\mathbb{Q}$. It certainly contains $\sqrt{5}$, so if we could find another real square root then we would be done, but I can't seem to find one. – Tom Oldfield Dec 29 '15 at 23:51
  • Related: https://math.stackexchange.com/questions/204709 – Watson Dec 26 '16 at 14:19

4 Answers4

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After some thought, I've found a very short answer that uses a minimal amount of computation:

With notation as in the question, we have: $$x_1x_2 = \sqrt{5}, x_1^2 = \frac{3+\sqrt{-11}}{2}.$$ Thus $K$ contains the subfield $F = \mathbb{Q}(\sqrt{5},\sqrt{-11})$ which is Galois and degree $4$ over $\mathbb{Q}$, with Galois group $G'\cong V_4$, generated by $\sigma$ and $\tau$, where $\sigma$ fixes $\sqrt{5}$ and permutes $\pm\sqrt{-11}$ and $\tau$ fixes $\sqrt{-11}$ and permutes $\pm\sqrt{5}$.

If $F=K$, then $x_i \in F$ and then the relations above immediately give that $\sigma\tau(x_1) = \pm x_2$ and $\sigma\tau(x_2)=\mp x_1$. But then $\sigma\tau \in G'$ has order $4$, a contradiction, so we must have that $K$ is strictly larger than $F$, so must be of degree $8$ over $\mathbb{Q}$ as desired.

Tom Oldfield
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Over $\mathbb{Q}(\sqrt{-11})$ you have two polynomials $$x^2-3-\sqrt{-11}=0$$ and $$y^2-3+\sqrt{-11}=0$$

Now verify that $$(xy)^2=20$$ so that if $y\in\mathbb{Q}(\sqrt{-11},x)$ then $\sqrt{5} \in \mathbb{Q}(\sqrt{-11},x)$. So it only remains to verify that this cannot happen.

For this one approach is to assume $$\sqrt{5} =a+b\sqrt{-11}+(c+d\sqrt{-11})x$$ and solving this for $x$ we are reduced to $x\in \mathbb{Q}(\sqrt{-11},\sqrt{5})$.

Which would mean $$\sqrt{3+\sqrt{-11}}=a+b\sqrt{-11}+c\sqrt{5}+d\sqrt{-11}\sqrt{5}$$

lets write this as $$\sqrt{3+\sqrt{-11}}=p+q\sqrt{5}$$ where $p=a+b\sqrt{-11}$ and $q=c+d\sqrt{-11}$ elements of $\mathbb{Q}(\sqrt{-11})$.

Squaring we have $$3+\sqrt{-11}=p^2+5q^2+2qp\sqrt{5}$$ if $pq\neq 0$ then $\sqrt{5}\in \mathbb{Q}(\sqrt{-11})$. So $pq=0$. If $q=0$ then $\sqrt{3+\sqrt{-11}}\in \mathbb{Q}(\sqrt{-11})$. And if $p=0$ then $\sqrt{3+\sqrt{-11}}=(c+d\sqrt{-11})\sqrt{5}$ which is also impossible.

Rene Schipperus
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  • Thank you for your answer, I had thought along these lines myself and concluded that it suffices to prove that $\sqrt{5}\not\in\mathbb{Q}(x_1)$, where $\mathbb{Q}(x_1)$ is what you've decided to call $\mathbb{Q}(\sqrt{-11},x)$, but I can't see why this should be true... Any suggestions? – Tom Oldfield Dec 29 '15 at 23:44
  • I added to my answer. – Rene Schipperus Dec 29 '15 at 23:53
  • For the record this requires showing first that $\mathbf Q(\sqrt{5},\sqrt{-11})$ has degree 4 over $\mathbf Q$, which is equivalent to showing $\sqrt{-11} \not\in \mathbf Q(\sqrt{5})$, and that follows from $\mathbf Q(\sqrt{5})$ having a real embedding while $\sqrt{-11}$ of course can't be embedded into $\mathbf R$. – KCd Dec 30 '15 at 00:00
  • @KCd But $\sqrt{11} \not \in \mathbb{Q}(\sqrt{5})$ and there are no non real imbeddings. – Rene Schipperus Dec 30 '15 at 00:02
  • It seems that squaring the relation gives a bunch of cross terms, so whilst there is no linear relationship between the three square roots you describe, this gives a system of four non-linear equations which would then have to be solved to deduce $a,b,c,d$ are all $0$, which looks somewhat lengthy. Might there be an easier way to reach the conclusion? – Tom Oldfield Dec 30 '15 at 00:07
  • Concerning $\sqrt{11}$ not being in $\mathbf Q(\sqrt{5})$ also, my point was that what I said follows from the lack of a real embedding of $\sqrt{-11}$, not that it is equivalent to this. So your comment is not inconsistent with mine. – KCd Dec 30 '15 at 00:07
  • I agree with Tom that the squaring of a sum of 4 terms starts to lead into some nasty algebra, and since we shouldn't bring in ideas that are beyond the scope of a basic Galois theory course or are too computationally intensive for an exam (trace mappings, quartic discriminants, ramification in number fields, etc.) I suspect this is not the kind of solution that would be used on a prelim exam. – KCd Dec 30 '15 at 00:12
  • @TomOldfield If you assume $a+b\sqrt{5}+c\sqrt{-11}+d\sqrt{5}\sqrt{-11}=0$ then you could solve for say $\sqrt{-11}$, remembering that we can divide in these fields we get $\sqrt{-11}=a+b\sqrt{5}$ and squaring this we see that $\sqrt{5}$ is rational. – Rene Schipperus Dec 30 '15 at 00:15
  • @ReneSchipperus Yes, but my point wasn't that these things aren't linearly dependent, but that the given equation doesn't immediately imply that $a,b,c,d = 0$ without more work. You get four equations from squaring out, one of which is $a^2 -11b^2+5c^2-55d^2-3 = 0$, and solving these simultaneously is hard. – Tom Oldfield Dec 30 '15 at 00:21
  • @TomOldfield Sorry, I am not sure what you are refering to at the moment, (my fautly I used $a,b,c,d$ for everything). I think my point is that you dont need to solve such complicated equations, but rather solve for certain terms and then reformulate. – Rene Schipperus Dec 30 '15 at 00:31
  • @Rene My point is that where you say "and if we square this we get a linear relationship between..." there is still quite a lot of computation to do before you can conclude $a=b=c=d=0$, which is the result that you want to end up with. – Tom Oldfield Dec 30 '15 at 00:33
  • @TomOldfield So after you square you just say this gives a linear relation between the three square roots, and now that is impossible. – Rene Schipperus Dec 30 '15 at 00:37
  • @ReneSchipperus It's not impossible, it just implies that all of your coefficients are $0$. This gives you the four equations I mention. Perhaps you should square out the expression and then collect like terms to see what I mean. – Tom Oldfield Dec 30 '15 at 00:38
  • @TomOldfield All coefficients cant be zero since the left side is non zero. Do you mean that I have to prove that the linear relation is non trivial ? – Rene Schipperus Dec 30 '15 at 00:43
  • @ReneSchipperus Obviously there's some miscommunication going on here, but I don't know any way to explain myself better, so I think it would be best to let the issue lie. Thanks for your time. – Tom Oldfield Dec 30 '15 at 00:47
  • @TomOldfield Yeah ok no worries I hope the method makes sense to you in time. Best, Rene – Rene Schipperus Dec 30 '15 at 00:49
  • @ReneSchipperus To clarify, I understand the method, my point is that there seems to be more work to it than you seem to realise, which is why I'm not such a big fan. But anyway, thanks again. – Tom Oldfield Dec 30 '15 at 00:57
  • Thanks for the updates, that does seem to be less hassle than I had thought (although there's still too many computations for my liking!). – Tom Oldfield Jan 11 '16 at 18:52
  • @TomOldfield Rough ideas often need to be worked over and refined. I think that is the case here. Anyway, the following remark should be used systematically: $\sqrt{p}\in k(\sqrt{q})$ if and only if $pq$ is a square in $k$, assuming of course that $q$ is not a square in $k$. – Rene Schipperus Jan 11 '16 at 21:07
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Rightly $$ x_1=\sqrt{\frac{3+\sqrt{-11}}{2}}= \frac{1}{2}(\sqrt{\sqrt{20}+3}+i\sqrt{\sqrt{20}-3}) $$ and $$ x_2=\sqrt{\frac{3-\sqrt{-11}}{2}}= \frac{1}{2}(\sqrt{\sqrt{20}+3}-i\sqrt{\sqrt{20}-3}) $$ (any determination thereof) suffice to generate the splitting field.

If $x_2\in\mathbb{Q}(x_1)$, also $x_1+x_2=\sqrt{\sqrt{20}+3}$ belongs to $\mathbb{Q}(x_1)$; however, $\alpha=\sqrt{\sqrt{20}+3}$ has degree $4$ over $\mathbb{Q}$ and therefore $\mathbb{Q}(x_1)=\mathbb{Q}(\alpha)$ would be a subset of the reals, which it is not.

Why does $\alpha$ have degree $4$? We clearly have $\alpha^2\in\mathbb{Q}(\sqrt{5})$, so we just need to show that $\alpha$ cannot be written as $$ \alpha=a+b\sqrt{5} $$ for rational $a$ and $b$. This means $$ 2\sqrt{5}+3=a^2+5b^2+2ab\sqrt{5} $$ and so $b=a^{-1}$, hence $$ a^4-3a^2+5=0 $$ which is exactly the equation we started with and that has no rational root.

egreg
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  • OK, that's a type of "real number trick" that you invariably wind up using with these exam-type problems where the full classification of Galois groups of quartics is unreasonable to bring to bear. – KCd Dec 30 '15 at 00:04
  • @KCd The roots are pairwise conjugate, and this path should be explored. – egreg Dec 30 '15 at 00:07
  • Nice answer, one could simply remark that one defines $x_2=\overline{x_1}$ and then find the poly for $x_1+x_2$ over $\mathbb{Q}$ and show its irreducible. Not sure its less work though. – Rene Schipperus Dec 30 '15 at 00:10
  • Thanks for the answer, I very much like the approach and the fact that it makes use of the roots being conjugate. That said, I'm not sure how you determined the real and imaginary parts. Did you just set equal to $a+bi$ and then set $(a+bi)^2 = \frac{1}{2}(3+\sqrt{-11})$? – Tom Oldfield Dec 30 '15 at 00:12
  • @TomOldfield Yes, that's the trick. Actually, I used $(a+bi)^2=6+2i\sqrt{11}$. – egreg Dec 30 '15 at 00:13
  • Yeah if you have $x_1^2=3+\sqrt{-11}$ and $x_2^2=3-\sqrt{-11}$ then $(x_1+x_2)^2=6+2xy$ and we know that $xy=\pm \sqrt{20}$ thus we have $\mathbb{Q}(x_1+x_2)$ is a real field of order $4$. Well, its just a restatement of your answer. – Rene Schipperus Dec 30 '15 at 00:26
  • What would be the best way to show that $\alpha$ is degree $4$ over $\mathbb{Q}$? It seems "obvious ", but then things like $\sqrt{\sqrt{20}+6}$ end up being only degree $2$... The minimal polynomial seems not obviously irreducible, although you can explicitly describe the (irrational) roots, so you can just show that it doesn't split as a product of quadratics, which isn't so bad, but then the whole part becomes a lot of computation when put together... – Tom Oldfield Dec 30 '15 at 00:45
  • @TomOldfield The proof is not difficult, I added it. – egreg Dec 30 '15 at 09:14
  • Ah, that's great, the direct proof works out, thanks a lot. It's possibly worth noting that you don't actually need to compute real and imaginary parts, once you know that $x_1$ and $x_2$ are conjugates, the argument shows that $x_1+x_2$ has degree $4$ over $\mathbb{Q}$ without writing it out explicitly. – Tom Oldfield Dec 30 '15 at 17:39
  • @egreg Sorry to unaccept, I did like this proof, but in the end the answer I came up with seemed to use fewer computations and can be applied to other situations (including when your base field doesn't necessarily lie in $\mathbb{C}$). – Tom Oldfield Jan 11 '16 at 18:48
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Consider the polynomial $Z^2 - 3Z + 5$. It has two roots, and they both belong to the quadratic extension $L = \mathbb{Q}(\sqrt{-11})$. The roots of your original polynomial are the square roots of these guys, so they each belong to two quadratic extensionf of $L$, and all four of them belong to one extension $K$ of $L$ of degree at most $4$ (over $L$), so degree at most $8$ over $\mathbb{Q}$.

This explicitly constructs a field that includes all of the roots of your polynomial, so it contains the splitting field $K$, but it is not immediately obvious that it is that splitting field. Perhaps a smaller field suffices?

To complete the proof, start with the splitting field $K$ which contains the four roots of the polynomial. Being a field, it must also contain the squares of these roots, and those are none other than the roots of the quadratic. So $K$ does indeed contain $L$. Now, we need to show that it is indeed a degree $4$ extension of it, and not $L$ itself nor a quadratic extension of it. I leave the details to you.

Update. Over $K = \mathbb{Q}(\sqrt{-11})$, i.e. in the field extension of degree $2$, the polynomial $f$ factors as$$\left(X^2 - {{3 + \sqrt{-11}}\over2}\right)\left(X^2 - {{3 - \sqrt{-11}}\over2}\right).$$Obviously, it does not split further over $K$ (check that ${{3 \pm \sqrt{-11}}\over2}$ are not squares in $K$), so we need to adjoin some roots of $f$. Let$$\alpha\text{ be a root of }X^2 - {{3 + \sqrt{-11}}\over2},\text{ }\beta \text{ be a root of }X^2 - {{3 - \sqrt{-11}}\over2}.$$We need to have both of them in the splitting field. Then we will obviously get $-\alpha$ and $-\beta$ there and thus $f$ will split in $K(\alpha, \beta)$. So the only remaining question is as follows: if we adjoin $ \alpha$, could we not automatically drag $\beta$ along, or in other words, is $K(\alpha, \beta) = K(\alpha)$, or equivalently in this case, is $K(\alpha) = K(\beta)$? When $\alpha$ and $\beta$ are roots of polynomials, say $X^2 - 2$ and $X^3 - 3$, we would not even ask this sort of question, since it is pretty clear that adjoining $\sqrt{2}$ would not allow us to get $\sqrt{3}$ as a $\mathbb{Q}$-linear combination of $\sqrt{2}$ and $1$, but this case is slightly less obvious. So is $\beta \in K(\alpha)$, or equivalently, are $\alpha$ and $\beta$ linearly dependent over $K$? Well, assume$$\beta = c\alpha + d,\quad c, \,d \in K.$$Squaring it will give us$$\beta^2 = c^2\alpha^2 + 2cd\alpha + d^2.$$In other words,$$\beta^2 - c^2\alpha^2 - d^2 = 2cd\alpha.$$The left-hand side of the above is an element of $K$ but $\alpha$ is not there, hence the only possibilities are $c = 0$ or $d = 0$. Well, $c = 0$ would imply that $\beta \in K$, which is not true. And $d = 0$ would imply that$$c = {\beta\over\alpha} \in K$$or$$c^2 = {{\beta^2}\over{\alpha^2}} = {{3 - \sqrt{-11}}\over{3 + \sqrt{-11}}} = {{(3 - \sqrt{-11})^2}\over{20}}$$is a square in $K$. That would imply that $20$ is a square in $K$, which is not the case (check this). Thus, we conclude that $K(\alpha, \beta) \supsetneq K(\alpha)$. So we have the following tower of quadratic field extensions$$\mathbb{Q} \subsetneq K \subsetneq K(\alpha) \subsetneq K(\alpha, \beta).$$Since the degree of a field extension is multiplicative, we get$$[K(\alpha, \beta) : \mathbb{Q}] = 2^3 = 8,$$as desired.

user149792
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