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a) Show that $f(x)=x^6-2x^3-10$ is irreducible over $\mathbb{Q}$.

Proof: Using Eisenstein's Criterion, with $p=2$, we show $f(x)$ is irreducible over $\mathbb{Q}$.

b) Let $\alpha=\sqrt[3]{1+\sqrt{11}}$. Show that $|\mathbb{Q}(\alpha):\mathbb{Q}|=6$.

Proof: Note that $\alpha$ is a root of $f(x)$. By part (a), we can conclude that $f(x)$ is the minimal polynomial of $\mathbb{Q}(\alpha)$. Furthermore $|\mathbb{Q}(\alpha):\mathbb{Q}|=6$.

c) Show that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})$ is a splitting field for $f(x)$.

Proof: Note that $\beta=\sqrt[3]{1-\sqrt{11}}$ is also a root of $f(x)$. [Check by plugging it in $f(x)$.] $$ \alpha \cdot \beta = \sqrt[3]{(1-\sqrt{11})(1-\sqrt{11})} = \sqrt[3]{-10} $$

d) Show that $\sqrt[3]{10}\not\in \mathbb{Q}(\sqrt{-3},\sqrt{11})$ and conclude that $|L:\mathbb{Q}|=12$, where $L=\mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})$. Moreover $K=L(\alpha)$ and thus $|K:\mathbb{Q}|=12$ or $36$.

  • How do I show $\sqrt[3]{10}\not\in \mathbb{Q}(\sqrt{-3},\sqrt{11})$?
  • Since the corresponding minimal polynomials of $\sqrt{-3},\sqrt{11},\sqrt[3]{10}$ have degree $2,2,3$, respectively, and share no roots, $|L:\mathbb{Q}|=12$.
  • $L(\alpha)= \mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})(\alpha)$. Does $\alpha$ get consumed by $\sqrt[3]{10}$ from what we shown in part (c) and the first part in (d)? Is this how we get $K=L(\alpha)$?
  • How can $|K:\mathbb{Q}|=36$?
Username Unknown
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  • Hints: A) $(-x)^3=-x^3$. What does that tell you about cube roots? B) Show that if $\gamma$ is any root of $f(x)$ then either $\gamma^3=\alpha^3$ or $\gamma^3=\beta^3$. What do you know about complex third roots of unity? – Jyrki Lahtonen Jul 25 '21 at 06:04
  • Okay. So because $\sqrt[3]{-1}=-1$, I can 'get rid' of the negative. I haven't used third roots of unity in problems like this. – Username Unknown Jul 25 '21 at 06:17
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    I would be surprised if the splitting field of $x^3-2$ has not been given as an example earlier in the course/book. What was the role the third roots of unity played in there? Or, can you find all the six complex zeros of $f(x)$? – Jyrki Lahtonen Jul 25 '21 at 06:20
  • Oh I see. This post helps explain it: https://math.stackexchange.com/questions/400660/determine-splitting-field-k-over-mathbbq-of-the-polynomial-x3-2 – Username Unknown Jul 25 '21 at 06:25
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    Great. Feel free too compile an answer out of those pieces. That way you get more feedback on the details :-) Alternatively, at this point I don't see any harm in you editing the question and adding the next part. If it revolves around this same splitting field, it's fine I think. – Jyrki Lahtonen Jul 25 '21 at 06:26
  • Yes, the key point is that $\sqrt{-3}$ always leads up to $\frac{-1+\sqrt{-3}}{2}$ which is a cube root of unity : so you have a very powerful element in your candidate splitting field. Use it! – Sarvesh Ravichandran Iyer Jul 25 '21 at 06:28
  • I updated the question with the last part. Thank you so far for your help, @JyrkiLahtonen – Username Unknown Jul 25 '21 at 06:36
  • For part d: If you know the degrees $[\Bbb{Q}(\sqrt{11},\sqrt{-3}):\Bbb{Q}]$ and $[\Bbb{Q}(\root3\of{10}):\Bbb{Q}]$ then you can do the first two bullets with the usual trick involving the tower law. In the third bullet I think the point is that by part c) it follows that $\beta\in L(\alpha)$. The earlier parts also show that $\alpha^3\in K$, which gives you a lot of information about the possible values of $[K:L]$. – Jyrki Lahtonen Jul 25 '21 at 06:51
  • It is overkill here, but you may be interested in this classic exercise when pondering about the irreducibility of $x^3-\alpha$ over $K$. – Jyrki Lahtonen Jul 25 '21 at 06:55
  • Adding the following link because I strongly feel that this more general result should also be linked to whenever I link to the prime degree case. The interested reader should check out Bill Dubuque's answer. Caveat: not needed here, but definitely nice to know, and leads to a more comprehensive view of this recurring theme. – Jyrki Lahtonen Jul 25 '21 at 07:01
  • You can also observe that if $F=\mathbb {Q} (\alpha, \sqrt{-3})$ then $|F:\mathbb{Q} |=12$ (as $\alpha$ is real and of degree $6$ and $\sqrt {-3}\notin\mathbb{Q}(\alpha) $). And then $K=F(\sqrt[3]{10})$ and $|K:F|=1 3$ or $|K:F|=3$. – Paramanand Singh Jul 25 '21 at 10:56
  • The last comment has a typo. Please read the last statement as "$|K:F|=1$ or $|K:F|=3$". – Paramanand Singh Jul 25 '21 at 11:34
  • @JyrkiLahtonen Sorry I haven't written it up. I had family visitng this week. I wrote something up. Hopefully it sounds good. Thank you for your help! – Username Unknown Jul 31 '21 at 19:50

2 Answers2

2

Answers for (a) and (b) are in the post.

Proof of (c): Note that $\beta=\sqrt[3]{1-\sqrt{11}}$ is also a root of $f(x)$. [Check by plugging it in $f(x)$.]

$$\alpha\cdot\beta= \sqrt[3]{(1+\sqrt{11})(1-\sqrt{11})}=\sqrt[3]{-10}=-\sqrt[3]{10}$$ Since $\beta\in\mathbb{Q}(\alpha)$, $-\alpha\beta\in\mathbb{Q}(\alpha)$. Hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\sqrt[3]{10})$.

Recall the idea of roots of unity. The roots of $f(x)$ are $x=\alpha, \omega\alpha, \omega^2\alpha$ where $\omega=\dfrac{-1+i\sqrt{3}}{2}$. It follows that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})=\mathbb{Q}(\alpha,\sqrt[3]{10})$. Furthermore $K$ is a splitting field for $f(x)$.

Proof of (d): Since $|\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}|=3$, and $|\mathbb{Q}(\sqrt{-3},\sqrt{11}):\mathbb{Q}|=4$, $\sqrt[3]{10}\not\in\mathbb{Q}(\sqrt{-3},\sqrt{11})$. (why??) I know we said Tower Law, but I think I need a review of it.

Let $L=\mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})$. By Tower Law, $$ |L:\mathbb{Q}|=|L:\mathbb{Q}(\sqrt{-3},\sqrt{11}):\mathbb{Q}(\sqrt[3]{10})|\cdot |\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}| = 4\cdot 3 =12$$

Let $K=L(\alpha)$, if $\alpha$ is real, then just like the last part, we have $$|K:\mathbb{Q}|=12$$ Recall the $\alpha$ is a root of unity that, then we have 2 more roots to account more, so $$|K:\mathbb{Q}|=12\cdot 3 = 36.$$

Comments on the proofs will be greatly appreciated.

Username Unknown
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  • Your answer has an issue in the final stage. One needs more analysis to figure out if $K$ is of degree 12 or 36 over rationals. – Paramanand Singh Aug 01 '21 at 09:29
  • If you can prove that $\alpha\notin L$ then you can conclude that $K$ is of degree 36 over rationals. If $\alpha\in L$ then $K=L$ and is of degree 12 – Paramanand Singh Aug 01 '21 at 09:32
  • Since $\alpha$ is real it is sufficient to prove that $\alpha\notin \mathbb {Q} (\sqrt {11},\sqrt[3]{10})$. – Paramanand Singh Aug 01 '21 at 09:33
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    If $\alpha\in\mathbb {Q} (\sqrt {11},\sqrt[3]{10})=M$ then $\beta=\sqrt[3]{1-\sqrt{11}}=-\sqrt[3]{10}/\alpha\in M$ and hence $\gamma=\alpha+\beta\in M$. But minimal polynomial of $\gamma$ is of degree $9$ over $\mathbb {Q} $. You can use this contradiction to complete the proof that $K$ is of degree 36 over $\mathbb {Q} $. – Paramanand Singh Aug 01 '21 at 14:49
  • @ParamanandSingh Can you elaborate on the 'why' I mention in my proof? – Username Unknown Aug 02 '21 at 01:20
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    Regarding "why", $A=\mathbb {Q} (\sqrt{-3},\sqrt {11})$ is of degree $4$ over $\mathbb {Q} $. If $\sqrt[3]{10}$ lies in this field then $B=\mathbb {Q} (\sqrt[3]{10})$ would be a subfield. Hence by tower law $4=[A:\mathbb {Q}] =[A:B] [B:\mathbb {Q}] =3[A:B]$. But clearly $4$ is not a multiple of $3$ and we get contradiction. – Paramanand Singh Aug 02 '21 at 01:55
  • This is ok for the purposes of answering the questions a-d. As Paramanand Singh pointed out, it is not yet clear that the degree of the splitting field is exactly 36. You can conclude that it is either 12 or 36, depending on whether $1+\sqrt{11}$ has a cube root in $L$ or not. – Jyrki Lahtonen Aug 07 '21 at 10:13
  • Anyway, with the aid of Dedekind's theorem we see that the degree of the splitting field is at least $36$. Hence it is exactly $36$. I don't know whether there is a more elementary proof for this fact. – Jyrki Lahtonen Aug 07 '21 at 10:24
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Supplementing the other answer with a possibly unnecessarily high-browed argument proving that the splitting field has degree $36$. Or, equivalently, that $1+\sqrt{11}$ has no cube root in the field $L$.

The tool I use is Dedekind's theorem, so the argument relies on Galois theory and group actions. Consider the polynomial $f(x)=x^6-2x^3-10$. Let $G$ be the Galois group of this polynomial, viewed as a subgroup of the group of permutations of the six roots, identified with $S_6$.

  • Eisenstein's criterion ($p=2$) tells us that $f(x)$ is irreducible over $\Bbb{Q}$. Therefore $G$ acts transitively on the set of roots, and hence $6\mid |G|$.

Let us fix a root $\alpha$ of $f$ in the splitting field $K$ (unique as a subfield of $\Bbb{C}$). Factorization of $f$ modulo different primes reveals other things about the action of $G$. I will pay particular attention to the subgroup $H=Stab_G(\alpha)$ that keeps the root $\alpha$ fixed.

  • Modulo $7$ we have $f(x)=(x^3+1)(x^3+4)$. The first cubic factors fully, $x^3+1=(x+1)(x+2)(x+4)$, as the field $\Bbb{F}_7$ has three third roots of unity: $1,2,4$. On the other hand, the cubic $x^3+4$ remains irreducible modulo $7$ because the congruence $x^3+4\equiv0\pmod7$ has no integer solutions. So Dedekind tells us that $G$ contains a $3$-cycle $\sigma$. This $3$-cycle fixes three of the zeros. Because $G$ acts transitively on the set of roots, one of the conjugates of $\sigma$ in $G$ has $\alpha$ as a fixed point. It follows that $H$ contains a $3$-cycle, and hence $3\mid |H|$.
  • By the orbit-stabilizer theorem $|G|=6|H|$ is thus a multiple of $18$. The other answer shows that $|G|=[K:\Bbb{Q}]\in\{12,36\}$, so we can conclude that $$|G|=[K:\Bbb{Q}]=36.$$

This was somewhat unsatisfactory in the sense that initially I needed help from a computer program. At some point this becomes necessary as eliminating alternative Galois groups becomes more and more demanding, when the degree grows (barring some special cases).

In the thread inspired by this Paramanand Singh reaches the same conclusion by a different computer aided calculation proving that an element of $K$ has a minimal polynomial of degree nine, implying $9\mid |G|$ and again eliminating $[K:\Bbb{Q}]=12$ as an alternative.

Jyrki Lahtonen
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