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3

Hint: We have $\Bbb Q \leq \Bbb Q(\omega_7 + \omega_7^5) \leq \Bbb Q(\omega_7)$.


2

Let $K=\mathbb{Q}(\frac{\pi^3}{1+\pi})$. Clearly $\pi$ satisfies the relation $$\frac{x^3}{x+1}-\frac{\pi^3}{\pi+1}=0$$ Therefore $\pi$ is a root of the following irreducible polynomial of degree $3$: $$x^3+\left(\frac{\pi^3}{1+\pi}\right)x+\left(\frac{\pi^3}{1+\pi}\right)\in K[x]$$ And to answer the question in the title of your post, yes, $\mathbb{Q}(\pi)$ ...


2

Note that $Gal(E|\mathbb{Q}(a))$ is a subgroup of $Gal(E|\mathbb{Q})$. As $Gal(E|\mathbb{Q})$ is Abelian, $Gal(E|\mathbb{Q}(a))$ is a normal subgroup. So, by the fundamental theorem of Galois theory, $\mathbb{Q}(a)/\mathbb{Q}$ is a normal extension. Thus, all roots of the minimal polynomial of $a$, which happens to be $f(x)$ lie in $\mathbb{Q}(a)$. Thus, the ...


1

not sure how much help you want, but you are probably just over thinking this a little, or have slightly misunderstood the question (unless I have!). I hope the following will help. In particular, try and stick with working directly with the fields. You have by definition that $L(\mathcal{M})$ is the smallest subfield of $M$ containing the elements of $L$ ...


1

Given any $\delta \in Gal(L/K)$, notice that $h(\delta)$ is exactly the restriction of $\delta$ to the set $M$. The reason that $\delta \rvert_M$ maps $M$ to itself is that if we apply $\delta$ to both sides of the equation $f(l_i) = 0$, we see that $\delta$ will fix all the coefficients of $f$, as well as fixing the right hand side $0$, and the resulting ...


1

If you've studied the cyclotomic polynomial the answer becomes quite simple! Since $\omega_7=e^{2\pi i/7}$ is a root of the seventh cyclotomic polynomial $\Phi_7$, and $\Phi_7$ can be shown to be irreducible, the degree of the extension is $\deg\Phi_7 = \varphi(7) = 6$, because $7$ is prime ($\varphi$ denotes the Euler totient function). In this proof we ...


1

Extended hint: The field $\Bbb{Q}(\omega_7+\omega_7^5)$ is contained in the 7th cyclotomic field $\Bbb{Q}(\omega_7)$. That field is an abelian extension of $\Bbb{Q}$, so all intermediate fields are themselves Galois extensions of $\Bbb{Q}$. This follows from Galois correspondence as all subgroups of an abelian group are normal. To make much progress here ...


1

As was already said, solution is the limit of infinite sequence of field extensions of rational numbers with square roots of all primes. But I’d point out there is more explicit description. The smallest ring (and ${\mathbb Q}$-algebra) which contains all square roots of positive rational numbers is, obviously, an infinite(countable)-dimensional vector ...



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