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5

Suppose $x=p/q$ is a rational root, with $p$ and $q$ coprime. Then since $x(x+1)^2=1$, $\frac pq(\frac pq+1)^2=1$. Thus, $p(p+q)^2=q^3$. Let $r$ be a prime factor of $q$. Then since $r|p(p+q)^2$, $r|p$ or $r|p+q$. Either way, $r|p$, a contradiction. Thus, we must have $q=1$, but there are no integer roots.

4

An automorphism must fix $\mathbb Q$, since it fixes $1$, and thus all integers, and thus all inverses of integers, and thus all products of integers and inverses of integers, which covers all rational numbers. Now consider the polynomial $f=X^2-d$. If $f(\alpha)=0$, then $f(\tau(\alpha))=\tau(f(\alpha))=\tau(0)=0$, so any automorphism has to send roots of $... 3 Not always. For instance, if$X$is a conic with$X(k)=\varnothing$, the image of$\deg$is$2\mathbb{Z}$. 2 With "free group" replaced by "free profinite group," this is either false or an open problem. If it were known to be true it would imply that every finite group$G$occurs as the Galois group of a finite Galois extension of$\mathbb{Q}$and this is not known. 2 This is way too hard, even if all the$c_i$are algebraic. Every finite extension of$\mathbb{Q}$can be realized as$\mathbb{Q}(c)$for some complex number$c$; this is the Primitive Element Theorem. In particular, every finite Galois extension of$\mathbb{Q}$is of this form. The Inverse Galois Problem asks whether every finite group occurs as the Galois ... 2 If$e^{2\pi i/5} \in \mathbb{Q}(\sqrt[4]{2},i)$, then$\textrm{Gal}(\mathbb{Q}(\sqrt[4]{2},i)/\mathbb{Q}) \cong D_4$has a quotient isomorphic to$\textrm{Gal}(\mathbb{Q}(e^{2\pi i/5})/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}$. But you can check that$D_4$does not have any cyclic quotients of order$4$. 2 I mean if$\lambda$is a root of the given polynomial, then you can view it as$a^{(1/n)}\zeta$, and here we are taking the principal$n^{th}$root for$a$. Any$\sigma \in G$will raise$\zeta$to some power, so$\sigma(\lambda)= a^{(1/n)}\zeta^{r}$, and say$\tau(\lambda)=a^{1/n}\zeta^{s}$. So$\phi(\sigma \circ \tau)(\lambda)=\sigma\circ\tau(\lambda)\...

2

$G$ is cyclic generated by $g(\lambda)=\zeta^m \lambda$. Take $\sigma=g^a,\tau=g^b$. For a primitive $n$-th root of unity to exist $F(\lambda)/F$ is separable (thus Galois) $\lambda$'s $F$ minimal polynomial is $\prod_{d=1}^{|G|} (x-\zeta^{md} \lambda)=x^{|G|} -\lambda^{|G|}$ where $|G|=n/\gcd(n,m)$ must be the least integer such that $\lambda^r\in F$.

2

We have to show that $a^9=a^4+a^3$. $\mathbb{F}_2[x]/(x^6+x+1)$ is the field of polynomials with coefficients in $\mathbb{F}_2$ (in $\mathbb{Z}_2$, the possible coefficients of polynomials are $0$ and $1$) modulo the ideal $(x^6+x+1):=\{p(x)(x^6+x+1)|p(x) \in \mathbb{F_2}[x]\}$. Now take $a^9$ divided by $a^6+a+1$. This gives you : $$a^9=a^4+a^3+a^3(a^6+a+1)\... 2 Clearly we have a^6+a+1=0 by construction. Since in \Bbb F_2 we have 1+1=0, it implies f+f=0 for any polynomial f\in\Bbb F_2[X]. Then adding a+1 to both sides yields$$a^6=a+1$$Then just multiply it by a^3. 1$$x^5+x^3+x^2+1=x^3(x^2+1)+x^2+1=(x^3+1)(x^2+1)==(x+1)(x^2-x+1)(x^2+2x+1)=(x+1)^3(x^2+x+1). $x^2+x+1$ is irreducible because $1$ and $0$ are not roots of this polynomial.

1

Hint. First prove that $\mathbb{Q}(\zeta)^H=\mathbb{Q}(\zeta+\zeta^{-1})$ using definitions. Then notice that $\zeta+\zeta^{-1}=2\cos(2\pi/5)$, then deduce that you can take $\lambda=\sqrt{5}$.

1

The natural way to write $E_1E_2$ is by setting each element equal to a sum of quantities in the form $ab$ with $a\in E_1$ and $b\in E_2$, and then do the algebra with all the realtions you can get. In fact, this is what typically children do at school. From a formal standpoint, we are saying that $E_1E_2$, as a $F$-algebra, is a quotient of the tensor ...

1

I'll show that if the element $x\in A$ is not a divisor of zero, then it is invertible. Indeed the hypothesis means that the the $k$-linear multiplication map $m_x:A\to A: y\mapsto x\cdot y$ is injective. Since $A$ is finite-dimensional over $k$, the injective endomorphism $m_x$ is automatically surjective, which implies in particular that $1_A$ is in the ...

1

If $x$ is a zero divisor, we are done, so let us assume $x$ is not a zero divisor in $A$; then $x$ is not nilpotent, for if $m \in \Bbb N$ were the smallest exponent such that $x^m = 0, \tag 1$ then $m \ge 2$ lest $x = 0, \tag 2$ contradicting our assumption that $x$ is not a zero divisor; with $m \ge 2$, we may write $xx^{m - 1} = x^m = 0, \tag 3$ and since ...

1

$x^m-t$ is irreducible over $\Bbb{Q}$, when it stays irreducible over $\Bbb{Q}(\zeta_m)$ it works easily, for example when $\gcd(m,\phi(m))=1$. If $\mathbb{Q}(t'^{1/m})\cong \mathbb{Q}(t^{1/m})$ then $\mathbb{Q}(t'^{1/m})= \mathbb{Q}(\beta)$ for some root $\beta$ of $t^{1/m}$'s minimal polynomial, ie. $\mathbb{Q}(t'^{1/m})= \mathbb{Q}(\zeta_m^r t^{1/m})$ ...

1

This definition means basically that $\alpha$ can be expressed by applying the four field operations $+,-,\cdot,/$ and the operation of 'taking $k$th root', starting out from elements of $F$. For example, $\frac{\sqrt[3]{\sqrt 5-\sqrt[4]7}+\sqrt2}{\sqrt[19]{6-\sqrt3}}$ is a radical element over $\Bbb Q$. Note that, however, the $k$th root operation is not ...

1

Yes, since $K$ is a field, $3$ has an inverse, which you can write as $\frac{1}{3}$. If $K$ is of chacteristic $0$, then $K$ contains a copy of $\mathbb Q$ and this is just the rational number $\frac{1}{3}$. If $K$ has positive characteristic, then it contains a copy of ${\mathbb F}_p$ for some prime $p$, and the inverse of $3$ is in this field; it is then ...

1

From David's example, the factors of $p(x)$ are not going to be helpful, because while they cover some of the roots of $p(x)$, they may not cover enough to get to the same splitting field. Use instead the primitive element theorem. The extension $K / F$ is finite and separable (why?) so by the theorem , $K = F[\alpha]$ for some $\alpha \in K$. Think about ...

1

What is the full, complete, unabridged statement of the exercise? The question as posed is false. Consider $K=\mathbb{C}, a=-1, b=0, \lambda=0, \mu=-\sqrt{2/(3 \sqrt{3})}$. Then $w(x,y) = y^2 - (x^3-x)$ and $t(x,y) = y-\sqrt{2/(3\sqrt{3})}$ intersect in exactly two points: $(2/\sqrt{3}, \sqrt{2/(3 \sqrt{3}))}$ and $(-1/\sqrt{3}, \sqrt{2/(3 \sqrt{3}))}$. ...

1

An integer in a number field of degree $n$ is defined as a root of some polynomial $X^n+a_1X^{n-1}+\ldots+a_n$ where $a_i \in \mathbb{Z}$. But as the degree of a transcendental extension is infinite, it seems to me that this definition does not seem viable for these field. This is imprecise. An algebraic integer can be defined without any reference to a ...

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