ch17.md

title	author	header-includes	abstract
Abstract Algebra by Pinter, Chapter 17	Amir Taaki	- \usepackage{mathrsfs} - \usepackage{mathtools} - \usepackage{extpfeil} - \DeclareMathOperator\cis{cis} - \DeclareMathOperator\ker{ker} - \DeclareMathOperator\ord{ord} - \DeclareMathOperator\gcd{gcd} - \DeclareMathOperator\lcm{lcm}	Chapter 17 on Rings

A. Examples of Rings

Prove that the following are commutative rings with unity.

Indicate the zero element, the unity and the negative for an $a$.

Ring axioms:

1. $a \oplus b = b \oplus a$
2. $(a \otimes b) \otimes c = a \otimes (b \otimes c)$
3. $a \otimes (b \oplus c) = (a \otimes b) \oplus (a \otimes c)$

Commutative:

1. $a \otimes b = b \otimes a$

With unity:

1. $\exists 1' \in A : a \otimes 1' = a$

Q1

$$a \oplus b = a + b - 1 \qquad a \otimes b = ab - (a + b) + 2$$

Axiom 1 is self evident.

Using sage, we prove axioms 2 and 3.

sage: a = var('a')
sage: b = var('b')
sage: c = var('c')
sage: ab = a*b - (a + b) + 2
sage: ab_c = ab*c - (ab + c) + 2
sage: bc = b*c - (b + c) + 2
sage: a_bc = a*bc - (a + bc) + 2
sage: ab_c.full_simplify()
-(a - 1)*b + ((a - 1)*b - a + 1)*c + a
sage: a_bc.full_simplify()
-(a - 1)*b + ((a - 1)*b - a + 1)*c + a

sage: def mul(a, b):
....:     return a*b - (a + b) + 2
....: 
sage: def add(a, b):
....:     return a + b - 1
....: 
sage: mul(a, add(b, c)).full_simplify()
(a - 1)*b + (a - 1)*c - 2*a + 3
sage: add(mul(a, b), mul(a, c)).full_simplify()
(a - 1)*b + (a - 1)*c - 2*a + 3

To calculate zero and unity:

$$a \oplus 0' = a$$ $$a + b - 1 = a$$ $$b = 1 = 0'$$

$$a \otimes 1' = a$$ $$ab - (a + b) + 2 = a$$ $$b = 2 = 1'$$

Lastly for the negative:

$$a \oplus b = 0'$$ $$a + b - 1 = 1$$ $$b = -a$$

Q2

$$a \oplus b = a + b + 1 \qquad a \otimes b = ab + a + b$$

sage: def add(a, b):
....:     return a + b + 1
....: 
sage: def mul(a, b):
....:     return a*b + a + b

Axiom 1: $a \oplus b = b \oplus a$

Self-evident

Axiom 2: $(a \otimes b) \otimes c = a \otimes (b \otimes c)$

sage: bool(mul(mul(a, b), c) == mul(a, mul(b, c)))
True

Axiom 3: $a \otimes (b \oplus c) = (a \otimes b) \oplus (a \otimes c)$

sage: bool(mul(a, add(b, c)) == add(mul(a, b), mul(a, c)))
True

Commutative: $a \otimes b = b \otimes a$

Self-evident

Zero:

sage: solve(add(a, b) - a, b)
[b == -1]
sage: add(a, -1)
a

Unity:

sage: solve(mul(a, b) - a, b)
[b == 0]
sage: mul(a, 0)
a

Negative $a$:

sage: solve(add(a, b) + 1, b)
[b == -a - 2]
sage: add(a, -a -2)
-1

Q3

$$(a, b) \oplus (c, d) = (a + c, b + d)$$ $$(a, b) \otimes (c, d) = (ac - bd, ad + bc)$$

sage: c = var('c')
sage: d = var('d')
sage: e = var('e')
sage: f = var('f')
sage: def add(ab, cd):
....:     a, b = ab
....:     c, d = cd
....:     return (a + c, b + d)
....: 
sage: def mul(ab, cd):
....:     a, b = ab
....:     c, d = cd
....:     return (a*c - b*d, a*d + b*c)
....: 

Axiom 1: $a \oplus b = b \oplus a$

sage: bool(add((a, b), (c, d)) == add((c, d), (a, b)))
True

Axiom 2: $(a \otimes b) \otimes c = a \otimes (b \otimes c)$

sage: bool(mul(mul((a, b), (c, d)), (e, f)) == mul((a, b), mul((c, d), (e, f))))
True

Axiom 3: $a \otimes (b \oplus c) = (a \otimes b) \oplus (a \otimes c)$

sage: bool(mul((a, b), add((c, d), (e, f))) == add(mul((a, b), (c, d)), mul((a, b), (e, f))))
True

Commutative: $a \otimes b = b \otimes a$

Self-evident

Zero:

sage: ab_plus_cd = add((a, b), (c, d))
sage: solve(ab_plus_cd[0] - a, c)
[c == 0]
sage: solve(ab_plus_cd[1] - b, d)
[d == 0]
sage: add((a, b), (0, 0))
(a, b)

Unity:

sage: ab_mul_cd = mul((a, b), (c, d))
sage: solve([ab_mul_cd[0] - a, ab_mul_cd[1] - b], c, d)
[[c == 1, d == 0]]
sage: mul((a, b), (1, 0))
(a, b)

Negative $a$:

Since $0' = (0, 0)$ then the negative for $(a, b)$ is simply $(-a, -b)$.

Q4

$$A = { x + y\sqrt{2} : x, y \in \mathbb{Z} }$$

Since normal algebraic operations are defined on A, then 1, 2 and 3 pass. It is also commutative.

Zero: 0

Unity: 1

Negative: $-x -y\sqrt{2}$

Q5

Prove the ring in part 1 is an integral domain.

We show that it has the cancellation property.

Assume $a \otimes b = a \otimes c$.

$$ab - (a + b) + 2 = ac - (a + c) + 2$$ $$ab - b = ac - c$$

Therefore $b = c$, and the ring has the cancellation property.

Q6

Prove the ring in part 2 is a field.

A field is a commutative ring with unity in which every nonzero element is invertible.

$$0' = -1$$ $$1' = 0$$

Thus

$$a \otimes b = 1'$$ $$ab + a + b = 0$$

We solve for b as follows

sage: def mul(a, b):
....:         return a*b + a + b
....: 
sage: solve(mul(a, b), b)
[b == -a/(a + 1)]

(Excluding the $0'$ element which is equal to $-1$)

Q7

Find the inverse for the ring in part 3.

sage: def mul(ab, cd):
....:     a, b = ab
....:     c, d = cd
....:     return (a*c - b*d, a*d + b*c)
....: 
sage: ab_mul_cd = mul((a, b), (c, d))
sage: solve([ab_mul_cd[0] - 1, ab_mul_cd[1]], c, d)
[[c == a/(a^2 + b^2), d == -b/(a^2 + b^2)]]

B. Ring of Real Functions

Q1

Let $a, b \in \mathcal{F}(\mathbb{R})$

Ring axioms:

1. $ab = ba$
2. $(ab)c = a(bc)$
3. $a(b + c) = ab + ac$

Commutative:

1. ab = ba

Zero: $f(x) = 0$

Unity: $f(x) = 1$

Negative: $-f(x)$

Q2

Divisors of zero, are any two functions which when $f(x) \neq 0$ then $g(x) = 0$ but in general $f(x) \neq 0$ and $g(x) \neq 0$.

See more here

Q3

Any functions which are one to one and have an inverse. That is $f(x) = x^3$ but not $f(x) = x^2$.

Q4

A field must have every element invertible. So the ring is not a field.

Ring has divisors of zero, so it does not have the cancellation property $\implies$ ring is not an integral domain.

C. Ring of $2 \times 2$ Matrices

Q1

sage: a = var('a')
sage: b = var('b')
sage: c = var('c')
sage: d = var('d')
sage: r = var('r')
sage: s = var('s')
sage: t = var('t')
sage: u = var('u')
sage: w = var('w')
sage: x = var('x')
sage: y = var('y')
sage: z = var('z')
sage: 
sage: def add(abcd, rstu):
....:     a, b, c, d = abcd
....:     r, s, t, u = rstu
....:     return (a + r, b + s, c + t, d + u)
....: 
sage: def mul(abcd, rstu):
....:     a, b, c, d = abcd
....:     r, s, t, u = rstu
....:     return (a*r + b*t, a*s + b*u, c*r + d*t, c*s + d*u)

Axiom 1:

Self evident.

Axiom 2:

sage: bool(mul((a,b,c,d), mul((r,s,t,u), (w,x,y,z))) == mul(mul((a,b,c,d), (r,s,t,u)), (w,x,y,z
....: )))
True

Axiom 3:

sage: bool(mul((a,b,c,d), add((r,s,t,u), (w,x,y,z))) == add(mul((a,b,c,d), (r,s,t,u)), mul((a,b
....: ,c,d), (w,x,y,z))))
True

Q2

sage: bool(mul((a,b,c,d), (r,s,t,u)) == mul((r,s,t,u), (a,b,c,d)))
False

Unity: $(a,b,c,d)(r,s,t,u) = (a, b, c,d)$

sage: solve([x_mul_y[0] - a, x_mul_y[1] - b, x_mul_y[2] - c, x_mul_y[3] - d], r,s,t,u)
[[r == 1, s == 0, t == 0, u == 1]]

$$ I = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$

Q3

Matrices don't have the cancellation property.

For example $ar_1 + bt_1 = ar_2 + bt_2$ does not imply that $r_1 = r_2$ and $t_1 = t_2$.

Thus is not an integral domain.

Not all matrices are invertible, for example when $det(A) = 0$. See more info here. Hence they $\mathcal{M}_2(\mathbb{R})$ is not a field either.

D. Rings of Subsets of a Set

$$A + B = (A - B) \cup (B - A)$$ $$AB = A \cap B$$

Q1

Ring axioms:

1. \begin{align*} A + B &= (A - B) \cup (B - A) \ &= B + A \end{align*}
2. $$(AB)C = (A \cap B)\cap C = A \cap (B \cap C) = A(BC)$$
3. \begin{align*} A(B+C) &= A\cap [(B - C) \cup (C -B)] \ &= [A \cap (B - C)] \cup [A \cap (C - B) ] \ &= (AB - AC) \cup (AC - AB) \ AB + AC &= (AB - AC) \cup (AC - AB) \end{align*}

Commutativity:

$$AB = A \cap B = BA$$

Unity:

$$AB = A \implies B = D$$

Zero:

$$A + B = A \implies B = \varnothing$$

Q2

All elements of $P_D$ with non-overlapping regions are divisors of zero.

$$X \in P_D, X^2 = \varnothing$$

Q3

$$1' = D$$ $$AB = D \implies A \cap B = D$$ Thus $A = D$ and $B = D$

Q4

There exist non-zero non-invertible elements in $P_D$, hence it is not a field.

$AB = AC$ does not imply $B = C$, hence cancellation property does not hold, and $P_D$ is not an integral domain.

Q5

\begin{align*} e = \varnothing \ a = {a} \ b = {b} \ c = {c} \ ab = {a, b} \ ac = {a, c} \ bc = {b, c} \ abc = {a, b, c} \end{align*}

\begin{tabular}{c | c c c c c c c c c} $\oplus$ & e & a & b & c & ab & ac & bc & abc \ \cline{1-9} e & e & a & b & c & ab & ac & bc & abc \ a & a & e & ab & ac & b & c & abc & bc \ b & b & ab & e & bc & a & abc & c & ac \ c & c & ac & bc & e & abc & a & b & ab \ ab & ab & b & a & abc & e & bc & ac & c \ ac & ac & c & abc & a & bc & e & ab & b \ bc & bc & abc & c & b & ac & ab & e & a \ abc & abc & bc & ac & ab & c & b & a & e \ \end{tabular}

\begin{tabular}{c | c c c c c c c c c} $\otimes$ & e & a & b & c & ab & ac & bc & abc \ \cline{1-9} e & e & a & b & c & ab & ac & bc & abc \ a & a & a & ab & ac & ab & ac & abc & abc \ b & b & ab & b & bc & ab & abc & bc & abc \ c & c & ac & bc & e & abc & a & b & abc \ ab & ab & ab & ab & abc & ab & abc & abc & abc \ ac & ac & ac & abc & ac & abc & ac & abc & abc \ bc & bc & abc & bc & bc & abc & abc & bc & abc \ abc & abc & abc & abc & abc & abc & abc & abc & abc \ \end{tabular}

E. Ring of Quaternions

Q1

Unity:

sage: a = var('a')
sage: b = var('b')
sage: c = var('c')
sage: d = var('d')
sage: matrix([[a + b*I, c + d*I], [-c + d*I, a - b*I]])
[ a + I*b  c + I*d]
[-c + I*d  a - I*b]
sage: alpha = matrix([[a + b*I, c + d*I], [-c + d*I, a - b*I]])
sage: matrix([[1, 0], [0, 1]]) * alpha
[ a + I*b  c + I*d]
[-c + I*d  a - I*b]

Distributive law:

sage: bb = var('e f g h')
sage: cc = var('i j k l')
sage: def make_matrix(xx):
....:     return matrix([[xx[0] + I*xx[1], xx[2] + xx[3]*I], [-xx[2] + xx[3]*I, xx[0] - xx[1]*I]])
....: 
sage: bool(alpha*(make_matrix(bb) + make_matrix(cc)) == (alpha*make_matrix(bb) + alpha*make_matrix(cc)))
True

Non-commutative:

sage: bool(alpha*make_matrix(bb) == make_matrix(bb)*alpha)
False

Q2

$$ \mathbf{1} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$

$$ \mathbf{i} = \begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix} $$

$$ \mathbf{j} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} $$

$$ \mathbf{k} = \begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix} $$

\begin{align*} \alpha &= a \mathbf{1} + b \mathbf{i} + c \mathbf{j} + d \mathbf{k} \ &=
\begin{pmatrix} a + bi & c + di\ -c + di & a - bi \end{pmatrix} \end{align*}

Q3

For the formula $\mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = -\mathbf{1}$

sage: ii = matrix([[I, 0], [0, -I]])
sage: ii*ii
[-1  0]
[ 0 -1]
sage: -ii*ii
[1 0]
[0 1]
sage: jj = matrix([[0, 1], [-1, 0]])
sage: jj*jj
[-1  0]
[ 0 -1]
sage: kk = matrix([[0, I], [I, 0]])
sage: kk*kk
[-1  0]
[ 0 -1]
sage: bool(ii**2 == jj**2)
True
sage: bool(ii**2 == kk**2)
True

$\mathbf{ij} = -\mathbf{ji} = \mathbf{k}$

sage: bool(ii*jj == -jj*ii)
True
sage: bool(ii*jj == kk)
True

$\mathbf{jk} = -\mathbf{kj} = \mathbf{i}$

sage: bool(jj*kk == -kk*jj)
True
sage: bool(jj*kk == ii)
True

$\mathbf{ki} = -\mathbf{ik} = \mathbf{j}$

sage: bool(kk*ii == -ii*kk)
True
sage: bool(kk*ii == jj)
True

Q4

$$\bar{\alpha} = \begin{pmatrix} a - bi & -c - di\\ c - di & a + bi \end{pmatrix} $$

$$||\alpha|| = a^2 + b^2 + c^2 + d^2 = t$$

Show that

$$ \bar{\alpha}\alpha = \alpha\bar{\alpha} = \begin{pmatrix} t & 0\\ 0 & t \end{pmatrix} $$

sage: alpha
[ a + I*b  c + I*d]
[-c + I*d  a - I*b]
sage: alpha_bar = matrix([[a - b*I, -c - d*I], [c - d*I, a + b*I]])
sage: bool(alpha_bar*alpha == alpha*alpha_bar)
True
sage: alpha_bar*alpha
[(a + I*b)*(a - I*b) + (c + I*d)*(c - I*d)                                         0]
[                                        0 (a + I*b)*(a - I*b) + (c + I*d)*(c - I*d)]

Note that $(a + ib)(a - ib) = a^2 + b^2$ and the same for $c$ and $d$.

Earlier we found the identity is

$$ \mathbf{1} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$

Thus the multiplicative inverse (both on the left and right) such that $\alpha\beta = \beta\alpha = \mathbf{1}$ is given by $(1/t)\bar{\alpha}$.

Q5

From part 4 we show there is a multiplicative inverse. Thus by the definition, $\mathcal{L}$ is a skew field.

F. Ring of Endomorphisms

Q1

Let $f, g, h \in End(G)$

1. $f + g = g + f$
2. $(f \cdot g) \cdot h = f \cdot (g \cdot h)$
3. $f \cdot (g + h) = f \cdot g + f \cdot h$

Q2

For a homomorphism $f(0) = 0$

Applying the rule $f(a + b) = f(a) + f(b)$

$$ e = \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 & 1 & 2 & 3 \end{pmatrix} $$

$$ a = \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 & 2 & 0 & 2 \end{pmatrix} $$

$$ b = \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 & 3 & 2 & 1 \end{pmatrix} $$

$$ c = \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 0 \end{pmatrix} $$

\begin{tabular}{c | c c c c c c c c c} $+$ & e & a & b & c \ \cline{1-5} e & a & b & c & e \ a & b & c & e & a \ b & c & e & a & b \ c & e & a & b & c \ \end{tabular}

\begin{tabular}{c | c c c c c c c c c} $\times$ & e & a & b & c \ \cline{1-5} e & e & a & b & c \ a & a & c & a & c \ b & b & a & e & c \ c & c & c & c & c \ \end{tabular}

G. Direct Product of Rings

$$(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)$$ $$(x_1, y_1) \cdot (x_2, y_2) = (x_1 x_2, y_1 y_2)$$

Q1

1. \begin{align*} (x_1, y_1) + (x_2, y_2) &= (x_1 + x_2, y_1 + y_2) \ &= (x_2 + x_1, y_2 + y_1) \ &= (x_2, y_2) + (x_1, y_1) \end{align*}
2. \begin{align*} (x_1, y_1) \cdot [(x_2, y_2) \cdot (x_3, y_3)] &= (x_1 x_2 x_3, y_1 y_2 y_3) \ &= [(x_1, y_1) \cdot (x_2, y_2)] \cdot (x_3, y_3) \end{align*}
3. \begin{align*} (x_1, y_1) \cdot [(x_2, y_2) + (x_3, y_3)] &= (x_1, y_1) \cdot (x_2 + x_3, y_2 + y_3) \ &= (x_1 \cdot (x_2 + x_3), y_1 \cdot (y_2 + y_3)) \ &= (x_1 x_2 + x_1 x_3, y_1 y_2 + y_1 y_3) \ &= (x_1, y_1) \cdot (x_2, y_2) + (x_1, y_1) \cdot (x_3, y_3) \end{align*}

Q2

\begin{align*} (x_1, y_1) \cdot (x_2, y_2) &= (x_1 x_2, y_1 y_2) \ &= (x_2 x_1, y_2 y_1) \ &= (x_2, y_2) \cdot (x_1, y_1) \end{align*}

\begin{align*} (1, 1) \cdot (x_1, y_1) &= (1x, 1y) \ &= (x, y) \end{align*}

Q3

Divisors of 0 are $x_1, x_2$ and $y_1, y_2$ such that $x_1 x_2 = 0_x'$ and $y_1 y_2 = 0_y'$ where for any $x \in A$ and $y \in B$ $x + 0_x' = x$ and $y + 0_y' = y$.

$(x, 0)$ and $(0, y)$ are zero divisors of $A \times B$.

Q4

$(a, b)$ is an invertible elemtn of $A \times B$ iff there is an ordered pair $(c, d)$ in $A \times B$ satisfying $(a, b) \cdot (c, d) = (1, 1)$.

Q5

Because $A \times B$ has zero divisors, it is not an integral domain, thus also not a field since every field is an integral domain.

H. Elementary Properties of Rings

Q1

In any ring, $a(b - c) = ab - ac$ and $(b - c)a = ba - ca$.

\begin{align*} a(b - c) &= a(b + (-c)) \ &= ab + a(-c) \ &= ab - ac \end{align*}

$$(b - c)a = ba - ca$$

Q2

In any ring, if $ab = -ba$, then $(a + b)^2 = (a - b)^2 = a^2 + b^2.

\begin{align*} (a + b)^2 &= (a + b)a + (a + b)b \ &= a^2 + ba + ab + b^2 \ &= a^2 + ba + (-ba) + b^2 \ &= a^2 + b^2 \end{align*}

\begin{align*} (a - b)^2 &= (a - b)a - (a - b)b \ &= a^2 - ba - ab - (-b^2) \end{align*}

Now to solve this we prove that $(-x)(-y) = xy$. We make use of 3 facts of rings:

1. $a0 = 0 = 0a$
2. $x + (-x) = 0$
3. $a(x + y) = ax + ay$

\begin{align*} (-x)(-y) &= (-x)(-y) + x(-y + y) \ &= (-x)(-y) + x(-y) + xy \ &= (-x + x)(-y) + xy \ &= 0 + xy \ &= xy \end{align*}

\begin{align*} (a - b)^2 &= a^2 - ba - ab - (-b^2) \ &= a^2 - ba - ab + b^2 \ &= a^2 + ab - ab + b^2 \ &= a^2 + b^2 \end{align*}

Q3

In any integral domain, if $a^2 = b^2$, then $a = \pm b$.

An integral domain is a commutative ring with unity having the cancellation property.

The cancellation property says:

If $ab = ac$ or $ba = ca$, then $b = c$ if $a \neq 0$.

\begin{align*} a^2 - b^2 &= 0 \ &= (a + b)a - (a + b)b \qquad \text{[Note: integral domain is commutative]} \ &= (a + b)(a - b) \end{align*}

Integral domains have no divisors of zero, so $(a + b)(a - b) = 0$ implies that either $a + b = 0$ or $a - b = 0$. In either case, adding or subtracting $b$ from both sides yields $a = \pm b$.

Q4

*In any integral domain, only $1$ and $-1$ are their own multiplicative inverses.$

Note that $x = x^{-1}$ iff $x^2 = 1$

Taking the converse, only $(-1)^2$ and $1^2$ are equal to $1$.

$$a \cdot 1 = 1 \implies a = 1$$ $$a \cdot (-1) = 1 \implies a = -1$$

Q5

Show that the commutative law for addition need not be assumed in defining a ring with unity: it may be proved from the other axioms.

\begin{align*} (a + b)(1 + 1) = (a + b)1 + (a + b)1 &= a(1 + 1) + b(1 + 1) \ a + b + a + b &= a + a + b + b \ (-a) + a + b + a + b &= (-a) + a + a + b + b \ b + a + b &= a + b + b \ b + a + b + (-b) &= a + b + b + (-b) \ b + a &= a + b \end{align*}

Q6

Let $A$ be any ring. Prove that if the additive group of $A$ is cyclic, then $A$ is a commutative ring.

Let $c$ be the additive generator of $A$. Then any element of $A$ can be expressed as repeated addition of $c$ for $n$ times. Then adding two elements of $A$ where $a = nc$ and $b = mc$, then $ab = (m + n)c = ba$.

Q7

Prove if any integral domain if $a^n = 0$ for some integer $n$, then $a = 0$.

$$a^n = a^{n - 1}a = a \cdots a = 0$$

But integral domains have no zero divisors. Thus $a = 0$.

I. Properties of Invertible Elements

Prove parts 1-5 are true in a nontrivial ring with unity.

Q1

If $a$ is invertible and $ab = ac$ then $b = c$.

Pre-multiply by $a^{-1}$ on both sides and by $a^{-1} a = 1$, then $b = c$.

Q2

An element $a$ can have no more than one multiplicative inverse.

This would imply $ab = ac$ where $b \neq c \neq 0$, which is a contradiction.

Q3

If $a^2 = 0$ then $a + 1$ and $a - 1$ are invertible.

\begin{align*} a^2 &= 0 \ a^2 - 1 &= -1 \ (a + 1)(a - 1) &= -1 \ -1(a + 1)(a - 1) &= 1 \end{align*}

Thus the inverse $(a + 1)^{-1} = -(a - 1)$ and $(a - 1)^{-1} = -(a + 1)$.

Q4

If $a$ and $b$ are invertible, their product $ab$ is invertible.

\begin{align*} ab(ab)^{-1} &= abb^{-1}a^{-1} \ &= aa^{-1} \ &= 1 \end{align*}

Q5

The set $S$ of all the invertible elements in a ring is a multiplicative group.

By above, any $a, b \in S$ where $a$ and $b$ are invertible, then their product $ab$ is also invertible and hence $ab \in S$.

Q6

By part 5, the set of all the nonzero elements in a field is a multiplicative group. Now use Lagrange's theorem to prove that in a finite field with $m$ elements, $x^{m -1} = 1$ for every $x \neq 0$.

By Lagrange's theorem, the order of any element in the group must divide the group's order. Therefore let $\ord(x) = n$, then $m - 1 = qn$ where $|S| = m$. Note we are not counting the zero element as part of the multiplicative group.

$$x^(m - 1) = x^{qn} = (x^n)^q = 1$$

Q7

*If $ax = 1$, $x$ is a right inverse of $a$; if $ya = 1$, $y$ is a left inverse of $a$. Prove if $a$ has a right inverse $x$ and a left inverse $y$, then $a$ is invertible, and its inverse is equal to $x$ and to $y$.

\begin{align*} yaxa &= y(ax)a = 1 \ &= (ya)(xa) &= xa \end{align*}

Thus $ax = xa = 1$, and by similar argument $ay = ya = 1$.

Q8

Prove that in a commutative ring, if $ab$ is invertible, then $a$ and $b$ are both invertible.

$$(ab)(ab)^{-1} = 1 = a \cdot (b(ab)^{-1})$$

Thus $a$ and $b$ are both invertible.

J. Properties of Divisors of Zero

Q1

If $a \neq \pm 1$ and $a^2 = 1$, then $a + 1$ and $a - 1$ are divisors of zero.

$$a^2 - 1 = 0 = (a + 1)(a - 1)$$

Q2

If $ab$ is a divisor of zero, then $a$ or $b$ is a divisor of zero.

$$a \neq 0, abx = 0 = a(bx) = 0$$

Likewise for $b$.

Q3

In a commutative ring with unity, a divisor or zero cannot be invertible.

$$x \neq 0, a^{-1} ax = a^{-1} (ax) = a^{-1} 0 = 0 = (a^{-1} a) x = 1x = x$$

Proof by contradiction.

Q4

Suppose $ab \neq 0$ in a commutative ring. If either $a$ or $b$ is a divisor or zero, so is $ab$.

$$(ax)b = 0b = 0 = abx$$

Same for $b$.

Q5

Suppose $a$ is neither $0$ nor a divisor or zero. If $ab = ac$ then $b = c$.

$$ab - ac = a(b -c) = 0$$

Since $a \neq 0$ and is not a divisor of zero, then $b - c = 0$.

Hence $b - c = 0$ or $b = c$.

Q6

$A \times B$ always has divisors of zero.

$(x, 0)$ and $(0, y)$ are zero divisors of $A \times B$.

K. Boolean Rings

A ring $A$ is a boolean ring if $a^2 = a$ for every $a \in A$. Prove that parts 1 and 2 are true in any boolean ring $A$.

Q1

For every $a \in A$, $a = -a$.

\begin{align*} (a + a)^2 &= (a + a) \ &= a(a + a) + a(a + a) = a^2 + a^2 + a^2 + a^2 = a + a + a + a \ a + a + a + a &= a + a \ a + a + a + a + (-a) + (-a) &= a + a + (-a) + (-a) \ a + a &= 0 \ a + a + (-a) &= -a \ a &= -a \end{align*}

Q2

\begin{align*} (a + b) &= (a + b)^2 = a^2 + ab + ba + b^2 = a + ab + ba + b \ ab + ba &= 0 \ ab &= ba \end{align*}

Q3

$$x(x - 1) = x^2 - x = x - x = 0$$

Thus for every $x \notin {0, 1}$, $x$ is a divisor of zero.

Q4

$$aa^{-1} = 1 = a(aa^{-1})a^{-1} = a^2 a^{-1} a^{-1} = (aa^{-1})a^{-1} = a^{-1}$$

Q5

$$a \vee b = a + b + ab$$

$$a \vee bc = a + bc + abc$$ $$(a \vee b)(a \vee c) = (a + b + ab)(a + c + ac) = a^2 + ac + a^2 c + ba + bc + bac + a^2 b + abc + a^2 bc$$

Using the fact $a^2 = a$, $a = -a$ and that $A$ is commutative, we get $$(a \vee b)(a \vee c) = a + bc + abc = a \vee bc$$

$$a \vee (1 + a) = a + 1 + a + a + a^2 = 1$$

$$a \vee a = a + a + a^2 = a$$

$$a (a \vee b) = a^2 + ab + a^2 b = a$$

I. The Binomial Formula

Prove $$ \begin{pmatrix} n\ k \end{pmatrix} + \begin{pmatrix} n\ k - 1 \end{pmatrix}

\begin{pmatrix} n + 1\ k \end{pmatrix} $$

Expansion for $a^{n - k} b^k$ is $$ \begin{pmatrix} n\ k \end{pmatrix} $$

Thus $$ \begin{pmatrix} n - 1\ k \end{pmatrix} + \begin{pmatrix} n - 1\ k - 1 \end{pmatrix}

\begin{pmatrix} n\ k \end{pmatrix} $$

Hence formula is true by induction.

M. Nilpotent and Unipotent Elements

An element $a$ of a ring is nilpotent if $a^n = 0$ for some positive integer $n$.

Q1

In a ring with unity, prove that if $a$ is nilpotent, then $a + 1$ and $a - 1$ are both invertible.

\begin{align*} 1 - a^n &= (1 - a)(1 + a + a^2 + \cdots + a^{n - 1}) \ &= (1 - a)(-1 \dot -1)(1 + a + a^2 + \cdots + a^{n - 1}) \ &= (a - 1)(a^{n - 1} + \cdots + a^2 + a + 1) \ &= (1 + a)(1 - a + a^2 - a^3 + \cdots \pm a^{n - 1}) \ &= (a + 1)(1 - a + a^2 - a^3 + \cdots \pm a^{n - 1}) \ &= 1 \end{align*}

Because $a^n = 0$

Q2

In a commutative ring, prove that any product $xa$ of a nilpotent element $a$ by any element $x$ is nilpotent.

$$(xa)^n = x^n a^n = x^n 0 = 0$$

Q3

In a commutative ring, prove the sum of two nilpotent elements is nilpotent.

$(a + b)^{m + n}$ is nilpotent, because every element of the expansion is zero. When the power of $a$ is less than $m$, then the power of $b$ is greater than $n$ and vice versa.

Q4

In a commutative ring, prove that the product of two unipotent elements $a$ and $b$ is unipotent.

$$(1 - a)^n = 0 \quad \text{and} \quad (1 - b)^m = 0$$ $$(1 - ab)^{m + n} = [(1 - a) + a(1 - b)]^{m + n}$$

From part 3 above.

Q5

In a ring with unity, prove that every unipotent element is invertible.

From part 1 we see $$1 - a^n = (1 - a)(1 + a + \cdots + a^{n - 1}) = 1$$

But $a$ is unipotent hence $(1 - a)^n = 0$, $$1 - (1 - a)^n = (1 - (1 - a))(\cdots) = a(\cdots) = 1$$ Hence $a$ is invertible.