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I'm working on answering this question but I'm unsure about alot the way I'm going about with the answers.

The question is:

Let L be a subfield of $\mathbb{C}$.

a) Show that $\mathbb{Q}\subseteq L$

I approached this question by assuming that $\mathbb{Q}\nsubseteq L$, which implies that there exists a rational number $\frac{a}{b}$ not in L, where $a,b\in\mathbb{Z}$. But then by closure under multiplication and the multiplicative inverse properties of a field, either $a\notin L$ or $b\notin L$. Since if both are in L, then $a*(\frac{1}{b})$ must be in L. Let's say a∉L, then $\mathbb{Z}\nsubseteq L$. However, all fields must have $\mathbb{Z}$ as a subset, since 1 is contained in every field by the multiplicative identity property, and every other integer must also be in the field by repeatedly using the closure under addition property, and negative numbers are in the field by the additive inverse property. Hence $\mathbb{Z}\subseteq L$ for any field, therefore contradicting that $\mathbb{Q}\nsubseteq L$.

Have I missed anything in my proof, also I'm not completely confident with my answer since I'm not sure whether 1 is an element of every field.

b) Suppose $L/ \mathbb{Q}$ is a finite galois extension. Then for any $\alpha\in L$, let $\bar{\alpha}$ be the complex conjugate of $\alpha$, then $\bar{\alpha}\in L$

I'm not sure on how I would start with this one or how I could use the property that $L/ \mathbb{Q}$ is a Galois extension.

c) Let $\sigma:L\to L$, such that $\sigma (\alpha)=\bar{\alpha}$. Show that $\sigma\in Aut(L/ \mathbb{Q})$, and has order 1 or 2.

Is the question simply asking to show that $\sigma (\alpha)=\sigma(a+ib)=a+\sigma(i)b$. If so, then I can just say that since $(\sigma(i))^2=\sigma(-1)=-1$, thus $\sigma(i)=\pm i$, so by letting $\sigma(i)=- i$, we get $\sigma(\alpha)=\bar{\alpha}$. s this what the question is asking for, and if so, have I explained well enough how $\sigma\in Aut(L/ \mathbb{Q})$. I' unsure about what the question means by the order of the function, if someone could please explain it, I'd be very grateful.

d)Show that $\sigma$ has order 2 $\iff$ L$\nsubseteq\mathbb{R}$

Again, I'm not sure what it means by the order. If someone can explain what it means by order, can you also help me with how I would start this proof.

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  • $\begingroup$ Correction in (1): all the subfields of characteristic zero must contain an isomorphic copy of $\;\Bbb Z\;$ (and thus of $\;\Bbb Q\;$ , of course). $\endgroup$ – Timbuc Nov 4 '14 at 20:23
  • $\begingroup$ @Timbuc 2 years late, but ... What are you talking about? It's a subfield of $\mathbb{C}$ so of course it has characteristic zero. $\endgroup$ – 6005 Sep 19 '16 at 19:42
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    $\begingroup$ In general, stick to one question per post. Rather than one vague title "Galois extensions questions" with 4 specific questions, ask 4 specific questions, each with a specific title. $\endgroup$ – 6005 Sep 19 '16 at 19:43
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b) If $\;\alpha\neq\overline\alpha\;$ , then the minimal polynomial of $\;\alpha\;$ over the rationals has at least degree two (why?), and since this polynomial has real (in fact, rational) coefficients, $\;z\in\Bbb C\;$ is a root of it iff $\;\overline z\;$ is also a root of it i.e., complex roots of real polynomials come as conjugate pairs of roots. Since $\;L/\Bbb Q\;$ is in particular a normal extension, it must contain all the roots of any irreducible polynomial with one root in $\;L\;$ , and thus $\;\overline\alpha\in L\;$

c) You must show the map is a ring (fields) homomorphis, that is: $\;\begin{cases}\sigma(a+b)=\sigma a+\sigma b\\{}\\\sigma(ab)=\sigma a\cdot\sigma b\end{cases}$ , and this follows at once from basic properties of the complex conjugate.

You must also show $\;\sigma(a)=0\iff a =0\;$ (injectivity), and this is trivial, and also that $\;\sigma\;$ is surjective (onto), and this is also trivial because of (b).

d) observe that in fact

$$\deg\sigma =2\iff \exists\,\ell\in L\;\;s.t.\;\;\sigma\ell=\overline\ell\neq\ell\iff\ell\in\Bbb C\setminus\Bbb R\iff L\rlap{\;\,/}\subset\Bbb R$$

the last relation follow from the fact that anyway $\;\sigma^2=Id|_L\;$ .

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  • $\begingroup$ I checked with my tutor, and the order of $\sigma$ actually refers to the minimum number of times to recurringly use the function $\sigma$ on an $\alpha$ to get $\alpha$ as the output. That is, $\sigma (\sigma(...\sigma(\alpha)))=\alpha$. I've managed to answer part c accordingly, and your answer to d was very helpful. $\endgroup$ – Andrew Brick Nov 12 '14 at 1:46
  • $\begingroup$ I've stumbled onto part e of the question, which I must have missed out when writing the question. It asks to prove that if $M=L^{<\sigma>}$, then |$M:L$|$=1$ or $2$, according to whether $L\subset \mathbb{R}$ or $L\not\subset \mathbb{R}$. $\endgroup$ – Andrew Brick Nov 12 '14 at 1:51
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c) You need to show that $\sigma^2(\alpha)=id$. In general, an element $\gamma$ has order n if $\gamma^n=id$ (where n is the smallest integer for which this is true).

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