4 Groups, Rings, Fields[1ZD]

We review these definitions.

Definition 161

[1ZF](Solved on 2022-11-15) A group is a set $G$ equipped with a binary operation $*$ , that associates an element $a * b \in G$ to each pair $a, b \in G$ , respecting these properties.

Associative property: for any given $a, b, c \in G$ we have $(a * b) * c = a * (b * c)$ .
Existence of the neutral element: an element denoted by $e$ such that $a * e = e * a = a$ .
Existence of the inverse: each element $a \in G$ is associated with an inverse element $a^{'}$ , such that $a * a^{'} = a^{'} * a = e$ . The inverse of the element $a$ is often denoted by $a^{- 1}$ (or $- a$ if the group is commutative). ¹

A group is said to be commutative (or abelian) if moreover $a * b = b * a$ for each pair $a, b \in G$ .

Definition 162

[1ZG](Solved on 2022-11-15) A ring is a set $A$ with two binary operations

$+$ (called sum or addition) and
$\cdot$ (called ”multiplication”, also indicated by the symbol $\times$ or $*$ , and often omitted),

such that

$A$ $+$ is a commutative group (usually the neutral element is denoted by $0$ );
the operation $\cdot$ has neutral element (usually the neutral element is indicated by $1$ ) and is associative;
multiplication distributes on addition, both on the left

$a \cdot (b + c) = (a \cdot b) + (a \cdot c) \forall a, b, c \in A$

and on the right

$(b + c) \cdot a = (b \cdot a) + (c \cdot a) \forall a, b, c \in A$

A ring is called commutative if multiplication is commutative. (In which case the right or left distributions are equivalent.)

We assume that

0 \neq 1

(otherwise

{0}

would be a ring).

Examples of commutative rings are: integer numbers $ℤ$ , polynomials $A [x]$ with coefficients in a commutative ring $A$ .

An example of a non-commutative ring is given by matrixes $ℝ^{n \times n}$ , with the usual operation of multiplication and addition.

Definition 163

[1ZH](Solved on 2022-11-15) A field $F$ is a ring in which multiplication is commutative, and every element $x \in F$ with $x \neq 0$ has an inverse $x^{- 1}$ for multiplication.

(So $F ⧵ {0}$ is a commutative group for multiplication, see 6).

Some field examples are: rational numbers

ℚ

, the real numbers

ℝ

and the complex numbers

ℂ

Remark 164

[20R](Solved on 2022-11-15) Typically ² you use the notations on the left instead of the writings on the right (where $x, y, z$ are in the field and $n$ is positive integer)

$x - y$	$x + (- y)$
$\frac{x}{y}$	$x \cdot y^{- 1}$
$x + y + z$	$(x + y) + x$
$x y z$	$(x \cdot y) \cdot z$
$n x$	$\underset{n times}{\underset{⏟}{x + \dots + x}}$
$x^{n}$	$\underset{n times}{\underset{⏟}{x \cdot \dots \cdot x}}$
$x^{- n}$	$(x^{- 1})^{n}$

Precisely, $n x$ means ”add $x$ to itself $n$ times”; the operation $n \mapsto n \cdot x$ can be defined recursively setting $0 \cdot x = 0$ and $(n + 1) \cdot x = n \cdot x + x$ . Similarly $x^{n}$ means ”multiply $x$ by itself $n$ times”: see the exercise 14.

Remark 165

[1ZW]Hurwitz’s theorem [ 39 ] asserts that if $V$ is a field and is also a real vector space with a scalar product, then $V = ℝ$ or $V = ℂ$ .

Definition 166

[1ZJ] An ordered ring $F$ is a ring with a total order relation $\leq$ for which, for every $x, y, z \in F$ ,

$x \leq y \Rightarrow x + z \leq y + z$ ;
$x, y \geq 0 \Rightarrow x \cdot y \geq 0$ .

Due to 6, if $F$ is a field, in the second hypothesis we may equivalently write $x, y > 0 \Rightarrow x \cdot y > 0$ . (Regarding the second hypothesis, see also 7) For further informations see the references in [ 32 ] . We will assume that in an ordered ring the multiplication is commutative.

Examples of ordered field are: rational numbers

ℚ

the real numbers

ℝ

. The complex numbers

ℂ

do not allow an ordering satisfying the above properties (see exercise 12).

Definition 167

[1ZK]An ordered field $F$ is archimedean if $\forall x, y \in F$ with $x > 0, y > 0$ there is a $n \in ℕ$ for which $n x > y$ . (See 164 for the definition of $n x$ ).

E167

[1ZM]The neutral element of a group is unique. Hidden solution: [UNACCESSIBLE UUID ’1ZN’]

E167

[1ZP] In a group, the inverse of an element is unique. Hidden solution: [UNACCESSIBLE UUID ’1ZQ’]

E167

[29C]Having fixed an element $g \in G$ in a group, the left and right multiplications $L_{g} : G \to G$ and $R_{g} : G \to G$

L_{g} (h) = g * h, R_{g} (h) = h * g

are bijections.

E167

[1ZR]Prove ⁴ that in a group:

If $x + y = x + z$ then $y = z$ .
If $x + y = x$ then $y = 0$ .
If $x + y = 0$ then $y = - x$ .
$- (- x) = x$ .

E167

[1ZS]Prove ⁵ that in a ring:

$0 \cdot x = 0$
$(- x) y = - (x y) = x (- y)$ .
$(- x) (- y) = x y$ .
$(- 1) x = - x$ .

Hidden solution: [UNACCESSIBLE UUID ’299’]

E167

[203] Consider the property

\forall x, y \in A, x \cdot y = 0 \Rightarrow x = 0 \lor y = 0

this property may be false in a ring $A$ ; if it holds in a specific ring, then this ring is said to be an integral domain [ 41 ] .

Show that a field $F$ is always an integral domain. Consequently $F ⧵ {0}$ is a commutative group for multiplication. Hidden solution: [UNACCESSIBLE UUID ’204’]

E167

[1ZT] Suppose that in a ring $A$ there is a total ordering $\leq$ such that for every $x, y, z \in A$ you have $x \leq y \Rightarrow x + z \leq y + z$ ; then show that these are equivalent

$x \leq y \land 0 \leq z \Rightarrow x \cdot z \leq y \cdot z$ ;
$x \geq 0 \land y \geq 0 \Rightarrow x \cdot y \geq 0$ .

E167

[1ZV]Prerequisites:3,5,7. Prove ⁶ than in an ordered ring $F$ :

for each $x \in F, x^{2} \geq 0$ , in particular $1 = 1^{2} > 0$ ;
$x > 0 \Rightarrow - x < 0$
$y > x \Rightarrow - y < - x$ ;
$x \leq y \land a \leq 0 \Rightarrow a \cdot x \geq a \cdot y$ ;
$x \geq a \land y \geq b \Rightarrow x + y \geq a + b$ ;
$x > a \land y \geq b \Rightarrow x + y > a + b$ ;
$x \geq a \geq 0 \land y \geq b \geq 0 \Rightarrow x \cdot y \geq a \cdot b$ ;

Prove than in an ordered field $F$ :

$x > a > 0 \land y > b \geq 0 \Rightarrow x \cdot y > a \cdot b$ ;
$x > 0 \Rightarrow x^{- 1} > 0$ ;
$y > x > 0 \Rightarrow x^{- 1} > y^{- 1} > 0$ ;
$x \cdot y > 0$ if and only if $x$ and $y$ agree on sign (i.e. either both > 0 or both < 0);

Hidden solution: [UNACCESSIBLE UUID ’29B’]

E167

[1ZX] In an ordered field $F$ we call $P = {x \in F : x \geq 0}$ the set of positive (or zero) numbers; it satisfies the following properties: ⁷

$x, y \in P \Rightarrow x + y, x \cdot y \in P$ ,
$P \cap (- P) = {0}$ and
$P \cup (- P) = F$ .

vice versa if in a field $F$ we can find a set $P \subseteq F$ that satisfies them, then $F$ is an ordered field by defining $x \leq y \Leftrightarrow y - x \in P$ .

E167

[1ZY]Not all fields are infinite sets. Consider $X = {0, 1}$ and operations $0 + 0 = 1 + 1 = 0$ , $0 + 1 = 1 + 0 = 1$ , $0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0$ and $1 \cdot 1 = 1$ . Check that it is a field. Show that it cannot be an ordered field.

E167

[1ZZ]Consider the ring of matrixes $ℝ^{2 \times 2}$ let’s define

A = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}), B = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}),

then check that

A B = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}), B A = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix});

you conclude that the ring of matrixes is not commutative.

E167

[08V] Show that there is no ordering $\leq$ on $ℂ$ such that $(ℂ, \leq)$ is an ordered field. Hidden solution: [UNACCESSIBLE UUID ’20S’]

E167

[200] Let’s fix an integer $N \geq 2$ that it is not a perfect square. Consider the subset $F$ of $ℝ$ given by the numbers $x$ that can be written as $x = a + b \sqrt{N}$ , with $a, b \in ℚ$ ; we associate the operations of $ℝ$ : show that $F$ is a field. Hidden solution: [UNACCESSIBLE UUID ’201’]

E167

[202] Let $F$ be a field; given $𝛼 \neq 0$ and $h \in ℕ$ consider the recursive definition of exponentiation $𝛼^{h}$ defined from $𝛼^{0} = 1$ and $𝛼^{(n + 1)} = 𝛼^{n} \cdot 𝛼$ ; then prove that $𝛼^{h + k} = 𝛼^{h} 𝛼^{k}$ and $(𝛼^{h})^{k} = 𝛼^{(h k)}$ for every $k, h \in ℕ$ .

E167

[20T]Prerequisites:14.Given $𝛼 \neq 0$ in a field, define that $𝛼^{0} = 1$ and let $𝛼^{- n}$ be the multiplicative inverse of $𝛼^{n}$ when $n \geq 1$ natural. (Use 14). For $n, m \in ℤ$ show that

𝛼^{n} 𝛼^{m} = 𝛼^{n + m}, (𝛼^{h})^{k} = 𝛼^{(h k)};

if the field is ordered and $𝛼 > 1$ show that $n \mapsto 𝛼^{n}$ is strictly monotonic increasing.

E167

[205] Let $F$ be a commutative ring, $a, b \in F$ , $n \in ℕ$ then

(a + b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k}

where the factor

(\binom{n}{k}) \dot{=} \frac{n!}{k! (n - k)!}

is called the ”binomial coefficient”. (This result is known as the binomial theorem, Newton’s formula, Newton’s binomial). To prove it by induction, check that

(\binom{n + 1}{k + 1}) = (\binom{n}{k + 1}) + (\binom{n}{k})

for $0 \leq k, k + 1 \leq n$ .

[UNACCESSIBLE UUID ’08W’]