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STATEMENT: Let $\alpha$ be the real positive fourth root of 2. Find all intermediate fields in the extension $\mathbb{Q}(\alpha)$ of $\mathbb{Q}$.

QUESTION: I basically used the tower law to show that $\mathbb{Q}(\alpha),\mathbb{Q}(\alpha^2),\mathbb{Q}$ are all intermediate fields by tower law, and $\mathbb{Q}(\alpha^3)=\mathbb{Q}(\alpha)$. I am just not sure how to show that these are all of the intermediate fields. I would really appreciate a hint or suggestion.

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  • $\begingroup$ Do you know that extensions of $\mathbb Q$ with the same degree are isomorphic by means of an isomorphism that fixes $\mathbb Q$? $\endgroup$ – Git Gud Nov 7 '14 at 21:05
  • $\begingroup$ No, I did not know. But with that fact I think I can show that there are only three intermediate fields. Thanks Git. $\endgroup$ – Enigma Nov 8 '14 at 2:43
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Are you familiar with the Galois Correspondence? Because you could compute the Galois group of the polynomial $x^4-2$ and write all its subgroups, and then by the correspondence you would obtain all the subfields between $\mathbb{Q}(\sqrt[4]{2},i)$ (the splitting field of the previous polynomial) and $\mathbb{Q}$: in particular one of the ramifications would be the subfields of $\mathbb{Q}\sqrt[4]{2}) \subset \mathbb{Q}(\sqrt[4]{2},i)$,

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  • $\begingroup$ I'm not familiar with galois theory, but I can read up on it. $\endgroup$ – Enigma Nov 7 '14 at 20:27
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Here is an alternative argument that does not use Galois theory. Instead one can make use of the linear algebraic structure:

Observation 1: $\operatorname{irr}(\alpha,\Bbb{Q},X)=X^4-2$ is the irreducible polynomial of $\alpha$ over $\Bbb{Q}$. Thus if $F$ is an intermediate field of $\Bbb{Q}(\alpha)$ over $\Bbb{Q}$, i.e. if $F$ a field such that

$$\Bbb{Q}< F<\Bbb{Q}(\alpha),$$

then $4=\operatorname{deg}(\operatorname{irr}(\alpha,\Bbb{Q},X))=[\Bbb{Q}(\alpha):\Bbb{Q}]=[\Bbb{Q}(\alpha):F][F:\Bbb{Q}]$, so that

$$[\Bbb{Q}(\alpha):F]=2=[F:\Bbb{Q}].$$

Observation 2: Since $\Bbb{Q}<\Bbb{Q}(\alpha^2)<\Bbb{Q}(\alpha)$ and $\operatorname{irr}(\alpha^2,\Bbb{Q},X)=X^2-2$ is the irreducible polynomial of $\alpha^2$ over $\Bbb{Q}$, we have that $\Bbb{Q}(\alpha^2)$ is an intermediate field.

Observation 3: $\Bbb{Q}(\alpha)$ is a $4$-dimensional $\Bbb{Q}$-vector space with basis $\{1,\alpha,\alpha^2,\alpha^3\}$ and $\Bbb{Q}(\alpha^2)$ is a $2$-dimensional $\Bbb{Q}$-vector space with basis $\{1,\alpha^2\}$. Rearranging the (standard?) basis of $\Bbb{Q}(\alpha)$ so that it is subordinate to the filtration $\Bbb{Q}<\Bbb{Q}(\alpha^2)<\Bbb{Q}(\alpha)$, we have

\begin{align} \Bbb{Q}(\alpha)=\Bbb{Q}\oplus \alpha\Bbb{Q} \oplus \alpha^2\Bbb{Q} \oplus \alpha^3\Bbb{Q}=\Bbb{Q}\oplus \alpha^2\Bbb{Q} \oplus \alpha\Bbb{Q} \oplus \alpha^3\Bbb{Q} = \Bbb{Q}(\alpha^2) \oplus \alpha\Bbb{Q} \oplus \alpha^3\Bbb{Q}. \end{align}


Claim: $\Bbb{Q}(\alpha^2)$ is the only intermediate field.

Proof: Suppose there is another intermediate field $F$. Then $F\cap \Bbb{Q}(\alpha^2)=\Bbb{Q}$, and since $F$ is a $2$-dimensional $\Bbb{Q}$-subspace of $\Bbb{Q}(\alpha)$, $\exists a,b,c\in\Bbb{Q}$ not all $0$ such that

$$F=\Bbb{Q}\oplus (a\alpha+b\alpha^2+c\alpha^3)\Bbb{Q}.$$

Regarding the values of $a,b$ and $c$ there are exactly three possibilities:

  1. $b=0$.
  2. $b\neq0,$ WLOG $a=1$.
  3. $b\neq0,$ WLOG $c=1$.

In the first case,

\begin{align} F\ni &(a\alpha+c\alpha^3)^2=(4ac)+(a^2+2c^2)\alpha^2\in \Bbb{Q}\oplus \alpha^2\Bbb{Q}=\Bbb{Q}(\alpha^2)\\ &\implies (4ac)+(a^2+2c^2)\alpha^2\in F\cap\Bbb{Q}(\alpha^2)=\Bbb{Q}\\ &\implies (a^2+2c^2)=0 \stackrel{a,c\in\Bbb{Q}}{\implies} a=0=c,{\bf\large\unicode{x21af}}. \end{align}

The other two cases follow similarly: In the second case we have

\begin{align} F\ni &(\alpha+b\alpha^2+c\alpha^3)^2=(4c+2b^2)+(4bc)\alpha+(1+2c^2)\alpha^2 +(2b)\alpha^3\\ &\implies (4bc)\alpha+(1+2c^2)\alpha^2+(2b)\alpha^3 \in (\alpha+b\alpha^2+c\alpha^3)\Bbb{Q}\\ &\implies 2b=4bc^2 \stackrel{b\neq0}{\implies} c^2=\dfrac{1}{2}\stackrel{c\in\Bbb{Q}}{\implies} \sqrt{2}=\alpha^2\in\Bbb{Q},{\bf\large\unicode{x21af}}; \end{align}

and in the third case we have

\begin{align} F\ni &(a\alpha+b\alpha^2+\alpha^3)^2=(4a+2b^2)+(2b)\alpha+(a^2+2)\alpha^2 +(2ab)\alpha^3\\ &\implies (2b)\alpha+(a^2+2)\alpha^2+(2ab)\alpha^3 \in (a\alpha+b\alpha^2+\alpha^3)\Bbb{Q}\\ &\implies 2b=2a^2b, a^2+2=2ab^2 \stackrel{b\neq0}{\implies} a^2=1,a=\dfrac{3}{2b^2}>0\\ &\implies a=1, b^2=\dfrac{3}{2}\implies \sqrt{\dfrac{3}{2}}=\vert b\vert\in\Bbb{Q},{\bf\large\unicode{x21af}}. \end{align}

Hence the overlap of $F$ and $\Bbb{Q}(\alpha^2)$ can not be $1$-dimensional, $\checkmark$.


Note: For further reference this is Exercise V.6.4 from Lang's Algebra (p. 253 of 3e). Since this exercise is before the (official) Galois chapter in the book I believe an argument that does not (explicitly) use Galois theory is what was asked.

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