Abstract Algebra: Groups, Rings and Fields: Difficult Problems with Solutions

Solution:
A class [tex]k\in\mathbb{Z}_{16}[/tex] is invertible if and only if [tex]\gcd(k,16)=1[/tex], so the invertible elements are the odd classes [tex]1, 3, 5, 7, 9, 11, 13, 15[/tex].
They form a commutative group under multiplication, so in the product we can pair each element with its inverse; every such pair contributes [tex]x\times x^{-1}=1[/tex]. Only the elements with [tex]x=x^{-1}[/tex], i.e. [tex]x^2\equiv 1 \pmod{16}[/tex], remain unpaired. We compute the squares:
[tex]1^2=1,\quad 3^2=9,\quad 5^2=25\equiv 9,\quad 7^2=49\equiv 1[/tex]
[tex]9^2=81\equiv 1,\quad 11^2=121\equiv 9,\quad 13^2=169\equiv 9,\quad 15^2=225\equiv 1[/tex]
So the elements that are their own inverses are [tex]1, 7, 9, 15[/tex], and the others pair up:
[tex]3\times 11=33\equiv 1,\qquad 5\times 13=65\equiv 1 \pmod{16}[/tex]
Hence the product of all invertible elements is
[tex]1\times 3\times 5\times 7\times 9\times 11\times 13\times 15\equiv 1\times 7\times 9\times 15=945=59\times 16+1\equiv 1 \pmod{16}[/tex]

Solution:
Composing both sides of [tex]g_2=h\circ g_1[/tex] on the right with [tex]g_1^{-1}[/tex], we get
[tex]h=g_2\circ g_1^{-1}[/tex]
The inverse of a cycle is the same cycle traversed backwards, so
[tex]g_1^{-1}=(5\,4\,3)(2\,1)(9\,8\,7\,6)[/tex]
i.e.
[tex]g_1^{-1}\colon\ 1\mapsto 2,\ 2\mapsto 1,\ 3\mapsto 5,\ 4\mapsto 3,\ 5\mapsto 4,\ 6\mapsto 9,\ 7\mapsto 6,\ 8\mapsto 7,\ 9\mapsto 8[/tex]
Now we compute [tex]h(x)=g_2\left(g_1^{-1}(x)\right)[/tex], following the cycles of [tex]h[/tex]:
[tex]h(1)=g_2(2)=4,\quad h(4)=g_2(3)=5,\quad h(5)=g_2(4)=8,\quad h(8)=g_2(7)=1[/tex]
[tex]h(2)=g_2(1)=3,\quad h(3)=g_2(5)=7,\quad h(7)=g_2(6)=9,\quad h(9)=g_2(8)=2[/tex]
[tex]h(6)=g_2(9)=6[/tex]
Hence ([tex]6[/tex] is a fixed point)
[tex]h=(1\,4\,5\,8)(2\,3\,7\,9)[/tex]
Check on one element: [tex]h\left(g_1(1)\right)=h(2)=3=g_2(1)[/tex]; the other elements are checked in the same way.

Solution:
[tex]K[/tex] is indeed a field: the polynomial [tex]f(x)=x^3+2x+1[/tex] has no roots in [tex]\mathbb{Z}_3[/tex], because
[tex]f(0)=1,\qquad f(1)=4\equiv 1,\qquad f(2)=13\equiv 1 \pmod 3[/tex]
A polynomial of degree [tex]3[/tex] without roots has no linear factor, so it is irreducible, and the quotient of [tex]\mathbb{Z}_3[x][/tex] by an irreducible polynomial is a field. Let [tex]a(x)=2x^2+x+2[/tex]. We apply the Euclidean algorithm in [tex]\mathbb{Z}_3[x][/tex] (note that [tex]2^{-1}=2[/tex]):
[tex]f-2x\,a=(x^3+2x+1)-(x^3+2x^2+x)=x^2+x+1[/tex]
[tex](x^2+x+1)-2a=(x^2+x+1)-(x^2+2x+1)=2x\ \Rightarrow\ 2x=f-(2x+2)\,a[/tex]
[tex]a=2x\,(x+2)+2[/tex]
Substituting the expression for [tex]2x[/tex]:
[tex]2=a-2x\,(x+2)=a-\bigl(f-(2x+2)\,a\bigr)(x+2)=\bigl(1+(x+2)(2x+2)\bigr)a-(x+2)f[/tex]
[tex](x+2)(2x+2)=2x^2+6x+4\equiv 2x^2+1\ \Rightarrow\ 2\equiv(2x^2+2)\,a \pmod f[/tex]
Multiplying by [tex]2^{-1}=2[/tex]:
[tex]1\equiv(4x^2+4)\,a\equiv(x^2+1)\,a \pmod f[/tex]
Hence the inverse of [tex]2x^2+x+2[/tex] in [tex]K[/tex] is [tex]x^2+1[/tex]. Check: in [tex]K[/tex] we have [tex]x^3=x+2[/tex] and [tex]x^4=x^2+2x[/tex], so
[tex](2x^2+x+2)(x^2+1)=2x^4+x^3+4x^2+x+2\equiv 2x^4+x^3+x^2+x+2[/tex]
[tex]2x^4+x^3+x^2+x+2=(2x^2+x)+(x+2)+x^2+x+2=3x^2+3x+4\equiv 1[/tex]

Solution:
We factor the divisor: [tex]g(1)=0[/tex] and
[tex]g(x)=(x-1)(x^2+2x-3)=(x-1)^2(x+3)[/tex]
so [tex]1[/tex] is a double root of [tex]g[/tex] and [tex]-3[/tex] is a simple root. The remainder has degree less than [tex]3[/tex], so
[tex]f(x)=q(x)\,g(x)+r(x),\qquad r(x)=ax^2+bx+c[/tex]
Substituting the roots of [tex]g[/tex] gives [tex]f(1)=r(1)[/tex] and [tex]f(-3)=r(-3)[/tex]. Differentiating, [tex]f'=q'g+qg'+r'[/tex], and since [tex]g(1)=g'(1)=0[/tex] (double root), also [tex]f'(1)=r'(1)[/tex]. We compute:
[tex]f(1)=2+6+7=15[/tex]
[tex]f(-3)=2(-3)^n+6(-3)^{n-1}-21=(-3)^{n-1}(-6+6)-21=-21[/tex]
[tex]f'(x)=2nx^{n-1}+6(n-1)x^{n-2}+7\ \Rightarrow\ f'(1)=2n+6(n-1)+7=8n+1[/tex]
This gives the system
[tex]\begin{cases}a+b+c=15\\9a-3b+c=-21\\2a+b=8n+1\end{cases}[/tex]
Subtracting the first equation from the second gives [tex]8a-4b=-36[/tex], i.e. [tex]2a-b=-9[/tex]. Adding this to the third equation:
[tex]4a=8n-8\ \Rightarrow\ a=2n-2,\qquad b=8n+1-2a=4n+5,\qquad c=15-a-b=12-6n[/tex]
Hence the remainder is
[tex]r(x)=(2n-2)x^2+(4n+5)x+12-6n[/tex]
Check for [tex]n=2[/tex]: [tex]f(x)=2x^2+13x[/tex] has degree [tex]2[/tex], so it is its own remainder, and the formula gives [tex]2x^2+13x+0[/tex].

Solution:
Consider the map [tex]\varphi:R\to R[/tex], [tex]\varphi(x)=1-x[/tex]. Since [tex]\varphi(\varphi(x))=1-(1-x)=x[/tex], the map [tex]\varphi[/tex] is a bijection and is its own inverse. It transforms the new operations into the original ones:
[tex]\varphi(a\oplus b)=1-(a+b-1)=(1-a)+(1-b)=\varphi(a)+\varphi(b)[/tex]
[tex]\varphi(a\odot b)=1-a-b+ab=(1-a)(1-b)=\varphi(a)\varphi(b)[/tex]
The order of the factors is preserved, so commutativity of [tex]R[/tex] is not needed. Because [tex]\varphi[/tex] is injective, every ring axiom carries over from the original operations to the new ones. For example, for the associativity of [tex]\odot[/tex]:
[tex]\varphi((a\odot b)\odot c)=(\varphi(a)\varphi(b))\varphi(c)=\varphi(a)(\varphi(b)\varphi(c))=\varphi(a\odot(b\odot c))[/tex]
hence [tex](a\odot b)\odot c=a\odot(b\odot c)[/tex]. Similarly, for the left distributive law:
[tex]\varphi(a\odot(b\oplus c))=\varphi(a)(\varphi(b)+\varphi(c))=\varphi(a)\varphi(b)+\varphi(a)\varphi(c)=\varphi((a\odot b)\oplus(a\odot c))[/tex]
so [tex]a\odot(b\oplus c)=(a\odot b)\oplus(a\odot c)[/tex]. The right distributive law and the associativity and commutativity of [tex]\oplus[/tex] are checked in the same way. The zero of the new ring is [tex]\varphi^{-1}(0)=1[/tex], and the opposite of [tex]a[/tex] is [tex]\varphi^{-1}(-\varphi(a))=2-a[/tex]:
[tex]a\oplus 1=a+1-1=a,\qquad a\oplus(2-a)=a+2-a-1=1[/tex]
Hence [tex]R[/tex] is a ring with respect to [tex]\oplus[/tex] and [tex]\odot[/tex], and [tex]\varphi[/tex] is an isomorphism from it onto the original ring. Moreover, the new ring has an identity element [tex]\varphi^{-1}(1)=0[/tex], because [tex]a\odot 0=0\odot a=a[/tex].

Solution:
The ideal consists of the products [tex](2+2i)(x+yi)[/tex] with [tex]x,y\in\mathbb{Z}[/tex]:
[tex](2+2i)(x+yi)=(2x-2y)+(2x+2y)i[/tex]
Solving [tex]2x-2y=a[/tex], [tex]2x+2y=b[/tex] gives [tex]x=\frac{a+b}{4}[/tex], [tex]y=\frac{b-a}{4}[/tex]. Hence [tex]a+bi\in A[/tex] exactly when [tex]a+b\equiv 0\pmod 4[/tex] and [tex]b-a\equiv 0\pmod 4[/tex], i.e. when [tex]b[/tex] is even and [tex]a\equiv b\pmod 4[/tex].
Every [tex]a+bi[/tex] is congruent modulo [tex]A[/tex] to exactly one of the [tex]8[/tex] elements [tex]a_0+b_0i[/tex] with [tex]a_0\in\{0, 1, 2, 3\}[/tex], [tex]b_0\in\{0, 1\}[/tex]. Indeed, take [tex]b_0\equiv b\pmod 2[/tex] and [tex]a_0\equiv a-(b-b_0)\pmod 4[/tex]; then the difference [tex](a-a_0)+(b-b_0)i[/tex] has an even imaginary part and [tex]a-a_0\equiv b-b_0\pmod 4[/tex], so it lies in [tex]A[/tex].
Two different representatives are never congruent: if [tex](a_0-a_0')+(b_0-b_0')i\in A[/tex], then [tex]b_0-b_0'[/tex] is even, so [tex]b_0=b_0'[/tex], and then [tex]a_0\equiv a_0'\pmod 4[/tex], so [tex]a_0=a_0'[/tex]. Therefore
[tex]\left|\mathbb{Z}[i]/A\right|=8[/tex]
This agrees with the general formula [tex]\left|\mathbb{Z}[i]/(\alpha)\right|=N(\alpha)[/tex]: [tex]N(2+2i)=2^2+2^2=8[/tex].

Solution:
If [tex]z^n=1[/tex], then [tex]|z|^n=1[/tex], so [tex]|z|=1[/tex]; hence [tex]C_n\subseteq U[/tex]. Since [tex]U[/tex] is abelian, [tex]C_n[/tex] is a normal subgroup. Consider the map
[tex]\varphi:U\to U,\qquad \varphi(z)=z^n[/tex]
It is well defined, since [tex]|z^n|=|z|^n=1[/tex], and it is a homomorphism, since [tex]U[/tex] is commutative:
[tex]\varphi(zw)=(zw)^n=z^nw^n=\varphi(z)\varphi(w)[/tex]
It is surjective: every [tex]w\in U[/tex] can be written as [tex]w=\cos\theta+i\sin \theta[/tex], and by de Moivre's formula the number [tex]z=\cos\frac{\theta}{n}+i\sin \frac{\theta}{n}\in U[/tex] satisfies [tex]z^n=w[/tex]. Its kernel is
[tex]\ker\varphi=\{z\in U : z^n=1\}=C_n[/tex]
By the first isomorphism theorem, [tex]U/\ker\varphi\cong\operatorname{Im}\varphi[/tex], i.e.
[tex]U/C_n\cong U[/tex]
and the isomorphism is [tex]zC_n\mapsto z^n[/tex].