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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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