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5

Homomorphism part is correct. However, you can't use the logarithm on complex numbers, as it is a multivalued function. Injectivity Part It is enough to check the kernel of the homomorphism is trivial. $e^{in\theta}=1$ implies $n\theta=2k\pi$ for some $k\in\mathbb{Z}$. So, the map is injective iff $\theta$ is not a rational multiple of $\pi$.

4

There exists a free module $F$ and a surjection $p:F\rightarrow P$, you have the exact sequence $0\rightarrow Ker(p)\rightarrow F\rightarrow P\rightarrow 0$, thus the sequence $0\rightarrow Hom(P,Ker(p))\rightarrow Hom(P,F)\rightarrow Hom(P,P)\rightarrow 0$ is exact, thus the morphism $Hom(P,F)\rightarrow Hom(P,P)$ is surjective. We deduce that there exists ...

4

By the Berlekamp algorithm we obtain $$x^4+3x^3+x+4=(x^2 + 4x + 2)^2$$ over $\mathbb{F}_5$. Hence the quotient is not a field, because it has zero divisors.

3

Notice your element satisfies $$(x^2-6)^2-27=0\iff x^4-12x^2+9=0$$ Then if you adjoin the square root of $3$ you get it to split as $$x^2-6=3\sqrt3$$ i.e. the degree of the total extension is $4$, so the Galois group has order $4$. But there are only two groups of order $4$, and since you can exhibit more than one element of order $2$, you know the ...

3

Clearly $o(n) = o(n^{-1})$ so this means there must be exactly one element of order $2$. If you mean this for all $n$, then this means the group must just be $\Bbb Z/2\Bbb Z$ which is isomorphic to $S_2$. So $1$, $3$, $4$ all hold. On the other hand if you just mean this to be true for some $n$, I must think you're crazy since again we see that $n$ is $2$, ...

3

You are correct that $x=2^{1/3}+2^{1/2}\in \Bbb Q( \sqrt[3]{2} , \sqrt{3})$. You are also correct that the degree of the minimal polynomial $f_x$ will equal the extension degree $[\Bbb Q(x):\Bbb Q]$, and hence $\deg f_x\mid [\Bbb Q( \sqrt[3]{2} , \sqrt{3}):\Bbb Q]=6$. However, there might be more intermediate fields $\Bbb Q( \sqrt[3]{2} , \sqrt{3})\supset ... 3 Let$a=\sqrt[3]{2} + \sqrt{3}$. Notice that $$(a-\sqrt{3})^3=2=a^3-3\sqrt 3 a^2+9a-3\sqrt 3 = a^3+9a-\sqrt 3 (3a^2+3) \tag 1$$ therefore $$\sqrt 3 = \frac{a^3+9a-2}{3a^2+3} \tag 2$$ In particular,$\Bbb Q(a)$contains$\Bbb Q(\sqrt 3)$and also contains$\Bbb Q(a-\sqrt 3) = \Bbb Q(\sqrt[3]{2})$. Therefore your intuition is correct: the degree of ... 2 By the Schur-Zassenhaus theorem, every extension $$1\rightarrow M\rightarrow E\rightarrow G\rightarrow 1$$ with$gcd(|M|,|G|)=1$is split, i.e.,$E\cong M\rtimes G$. In other words, we have$H^2(G,M)=1$. We can apply this here with$G=Aff(q)/H\cong \mathbb{F}_q^*$of order$q-1$and$M=H\cong \mathbb{F}_q$of order$q$, because$gcd(q,q-1)=1$. So there ... 2 No:$\alpha$is the set of all polynomials of the form$x+f(x)(x^3+x^2+1), \quad f(x)\in \mathbf Z_5[x]$. Similarly$\alpha^2$is the set of all polynomials of the form$x^2+f(x)(x^3+x^2+1), \quad f(x)\in \mathbf Z_5[x]$. 2 The morphism$x\rightarrow t, y\rightarrow t^2$,$z\rightarrow 1$induces a morphism You have$f:k[x,y,z)/(x^2-yz,z-1)\rightarrow k[t]$defined by$f([x])=t, f([y])=t^2, f([z])=1$where$[x]$is the class of$x$Consider$g:k[t]\rightarrow k[x,y,z]/(x^2-yz,z-1)$defined by$g(t)=[x]$, you have$f(g(t)=f([x])=t$.$g(f([x]))=g(t)=[x]$, ... 1 1) Check that a subgroup$\;H\;$of a group is a normal subgroup iff it is the (disjoint) union of conjugacy classes. 2) With (1), or directly, check that$\;V:=\left\{\,(1),\,(12)(34),\,(13)(24)\,,\,(14)(23)\,\right\}\lhd S_4\;$3) Take now the quotient group$\;S_4/V\;$. By Lagrange's theorem, this group's order is six, so it is either$\;S_3\;$or the ... 1 To Prove:$P(n):=(\rho^n)^{-1}=(\rho^{-1})^n$Base Step:$n=1\implies P(1):=\rho^{-1}=\rho^{-1}$. Hence$P(1)$is true. Induction Hypothesis: For some$k\in\mathbb{N}$,$P(k)$is true, i.e.,$(\rho^k)^{-1}=(\rho^{-1})^k$. Induction Step: Now,$$(\rho^{k+1})^{-1}=(\rho^{k}\circ \rho)^{-1}= \rho^{-1}\circ ... 1 A module in which every submodule is an essential submodule is called a uniform module. A ring$R$for which$R_R$is a uniform module is called a right uniform ring. They were notably used in A. W. Goldie's theory of uniform dimension. You can find in Lam's Lectures on modules and rings a chapter devoted to this. 1 No. Such a root cannot exist modulo$p$when$p$is one less than a multiple of$4$. 1 Suppose that$p$is in the kernel. Then by polynomial division you can write$p=q(X^2-Y^3)+r$where no term of$r$has degree$\ge 2$in$X$, and clearly$r$is then also in the kernel. Now$r$consists of terms of the form$aY^n$and terms of the form$aXY^n$. Under$f$the former terms become$at^{2n}$and the latter terms become$at^{2n+3}$. But$t^{2n}\$ ...

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