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:


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.

  • $\begingroup$ 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 $\mathbf Q(r)$. $\endgroup$ – KCd Dec 29 '15 at 23:18
  • $\begingroup$ 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 $\Bbb Q[\alpha]$. $\endgroup$ – Gregory Grant Dec 29 '15 at 23:20
  • $\begingroup$ @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? $\endgroup$ – Tom Oldfield Dec 29 '15 at 23:21
  • $\begingroup$ @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? $\endgroup$ – Tom Oldfield Dec 29 '15 at 23:23
  • $\begingroup$ It is false that $x_1^2 = x_2^2$: otherwise $x_2 = \pm x_1$, but $x_2$ is neither $x_1$ nor $-x_1$. $\endgroup$ – KCd Dec 29 '15 at 23:30

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.

  • $\begingroup$ What if polynomial is replaced by $x^4+x^2-6$ ? $\endgroup$ – Pranita Gupta Nov 6 '17 at 1:27

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.

  • $\begingroup$ 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? $\endgroup$ – Tom Oldfield Dec 29 '15 at 23:44
  • $\begingroup$ I added to my answer. $\endgroup$ – Rene Schipperus Dec 29 '15 at 23:53
  • $\begingroup$ 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 $\mathbf R$. $\endgroup$ – KCd Dec 30 '15 at 0:00
  • $\begingroup$ @KCd But $\sqrt{11} \not \in \mathbb{Q}(\sqrt{5})$ and there are no non real imbeddings. $\endgroup$ – Rene Schipperus Dec 30 '15 at 0:02
  • $\begingroup$ 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? $\endgroup$ – Tom Oldfield Dec 30 '15 at 0:07

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.

  • $\begingroup$ 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. $\endgroup$ – KCd Dec 30 '15 at 0:04
  • $\begingroup$ @KCd The roots are pairwise conjugate, and this path should be explored. $\endgroup$ – egreg Dec 30 '15 at 0:07
  • $\begingroup$ 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. $\endgroup$ – Rene Schipperus Dec 30 '15 at 0:10
  • $\begingroup$ 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 $(a+bi)^2 = \frac{1}{2}(3+\sqrt{-11})$? $\endgroup$ – Tom Oldfield Dec 30 '15 at 0:12
  • $\begingroup$ @TomOldfield Yes, that's the trick. Actually, I used $(a+bi)^2=6+2i\sqrt{11}$. $\endgroup$ – egreg Dec 30 '15 at 0:13

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.

  • $\begingroup$ Kevin, thanks for your answer but the details are exactly what I'm having trouble with. Perhaps you could help fill them in? $\endgroup$ – Tom Oldfield Dec 30 '15 at 1:00
  • $\begingroup$ @TomOldfield I added in the details. $\endgroup$ – user149792 Dec 31 '15 at 23:21
  • $\begingroup$ Thanks for adding the details, this is more or less what I expected (i.e. some explicit computations along the lines of Rene's answer) but it's nice to know I wasn't missing something slicker. $\endgroup$ – Tom Oldfield Jan 11 '16 at 18:50
  • $\begingroup$ What if polynomial is replaced by $x^4+x^2-6$ ? $\endgroup$ – Pranita Gupta Nov 6 '17 at 1:29

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.