4 Groups, Rings, Fields[1ZD]
We review these definitions.
[1ZF](Solved on 2022-11-15) A group is a set \(G\) equipped with a binary operation \(*\), that associates an element \(a*b∈ G\) to each pair \(a,b∈ G\), respecting these properties.
Associative property: for any given \(a, b, c∈ G\) 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 \(a∈ G\) 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 \(a,b∈ G\).
[1ZG](Solved on 2022-11-15) A ring is a set \(A\) with two binary operations
\(+\) (called sum or addition) and
\(⋅\) (called ”multiplication”, also indicated by the symbol \(×\) or \(*\), and often omitted),
such that
\(A\) \(+\) is a commutative group (usually the neutral element is denoted by \(0\));
the operation \(·\) has neutral element (usually the neutral element is indicated by \(1\)) and is associative;
multiplication distributes on addition, both on the left
\[ a ⋅ (b + c) = (a · b) + (a · c) \quad ∀ a, b, c ∈ A \]and on the right
\[ (b + c) · a = (b · a) + (c · a) \quad ∀ a, b, c ∈ A \]
A ring is called commutative if multiplication is commutative. (In which case the right or left distributions are equivalent.)
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 \(ℝ^{n× n}\), with the usual operation of multiplication and addition.
[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)
|
\(x-y\) |
\(x+ (-y)\) |
|
\(\frac{x} y\) |
\(x ⋅ y^{-1}\) |
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\(x+y+z\) |
\((x+y)+x\) |
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\(x y z\) |
\((x⋅ y)⋅ z\) |
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\(n x\) |
\(\underbrace{x+ \ldots +x}_{n~ \hbox{times}}\) |
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\(x^ n\) |
\(\underbrace{x⋅ \ldots ⋅ x}_{n~ \hbox{times}}\) |
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\(x^{-n}\) |
\((x^{-1}) ^ n\) |
Precisely, \(n x\) means ”add \(x\) to itself \(n\) times”; the operation \(n↦ n⋅ x\) can be defined recursively setting \(0⋅ x=0\) and \((n+1)⋅ x= n⋅ x + x\). Similarly \(x^ n\) means ”multiply \(x\) by itself \(n\) times”: see the exercise 14.
[1ZJ] An ordered ring \(F\) is a ring with a total order relation \(≤\) for which, for every \(x, y, z ∈ F\),
\(x ≤ y ⇒ x + z ≤ y + z\);
\(x, y ≥ 0 ⇒ x · y ≥ 0 \) .
Due to 6, if \(F\) is a field, in the second hypothesis we may equivalently write \(x, y {\gt} 0 ⇒ x · y {\gt} 0 \) . (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.
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[1ZM]The neutral element of a group is unique. Hidden solution: [UNACCESSIBLE UUID ’1ZN’]
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[1ZP] In a group, the inverse of an element is unique. Hidden solution: [UNACCESSIBLE UUID ’1ZQ’]
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[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.
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If \(x + y= x + z\) then \(y= z\).
If \(x + y= x\) then \(y=0\).
If \(x + y=0\) then \(y= -x\).
\(-(-x) = x\).
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\(0 · x = 0\)
\((-x)y = -(xy) = x(-y)\).
\((- x)(- y) = xy\).
\((- 1)x = -x\).
Hidden solution: [UNACCESSIBLE UUID ’299’]
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[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’]
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[1ZT] Suppose that in a ring \(A\) there is a total ordering \(≤\) such that for every \(x, y, z ∈ A\) you have \(x ≤ y ⇒ x + z ≤ y + z\); then show that these are equivalent
\(x ≤ y \, ∧\, 0 ≤ z \quad ⇒\quad x · z ≤ y · z\);
\(x≥ 0∧ y ≥ 0 \quad ⇒\quad x · y ≥ 0 \) .
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[1ZV]Prerequisites:3,5,7. Prove 6 than in an ordered ring \(F\):
for each \(x ∈ F, x^ 2 ≥ 0\) , in particular \(1 = 1^ 2{\gt}0\);
\(x {\gt} 0 ⇒ -x {\lt} 0\)
\(y {\gt} x ⇒ -y {\lt} -x \) ;
\(x ≤ y \land a ≤ 0 ⇒ a · x ≥ a · y\) ;
\(x \ge a \land y\ge b ⇒ x + y ≥ a + b\) ;
\(x {\gt} a \land y\ge b ⇒ x + y {\gt} a + b\) ;
\(x \ge a \ge 0 \land y\ge b\ge 0 ⇒ x · y ≥ a · b\) ;
Prove than in an ordered field \(F\):
\(x {\gt} a {\gt} 0 \land y{\gt} b\ge 0 ⇒ x · y {\gt} a · b\) ;
\(x {\gt} 0 ⇒ x ^{−1} {\gt} 0\) ;
\(y {\gt} x {\gt} 0 ⇒ x ^{−1} {\gt} y ^{−1} {\gt} 0\) ;
\(x · y {\gt} 0\) if and only if \(x\) and \(y\) agree on sign (i.e. either both > 0 or both < 0);
Hidden solution: [UNACCESSIBLE UUID ’29B’]
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[1ZX] In an ordered field \(F\) we call \(P = \{ x ∈ F : x ≥ 0\} \) the set of positive (or zero) numbers; it satisfies the following properties: 7
\(x, y ∈ P ⇒ x + y , x · y ∈ P \),
\(P ∩ (−P ) = \{ 0 \} \) and
\(P ∪ (−P ) = F \).
vice versa if in a field \(F\) we can find a set \(P⊆ F\) that satisfies them, then \(F\) is an ordered field by defining \(x ≤ y ⇔ y−x ∈ P\).
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[1ZY]Not all fields are infinite sets. Consider \(X=\{ 0,1\} \) and operations \(0+0=1+1=0\), \(0+1=1+0=1\), \(0⋅0=0⋅1=1⋅0=0\) and \(1⋅1=1\). Check that it is a field. Show that it cannot be an ordered field.
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[1ZZ]Consider the ring of matrixes \(ℝ^{2× 2}\) let’s define
\[ A={\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} }\quad ,\quad B={\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} }\quad , \]then check that
\[ AB={\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} } \quad ,\quad BA={\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} }\quad ; \]you conclude that the ring of matrixes is not commutative.
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[08V] Show that there is no ordering \(≤\) on \(ℂ\) such that \((ℂ,≤)\) is an ordered field. Hidden solution: [UNACCESSIBLE UUID ’20S’]
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[200] Let’s fix an integer \(N≥ 2\) 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 \(a,b∈ℚ\); we associate the operations of \(ℝ\): show that \(F\) is a field. Hidden solution: [UNACCESSIBLE UUID ’201’]
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[202] Let \(F\) be a field; given \(𝛼≠ 0\) and \(h∈ℕ\) consider the recursive definition of exponentiation \(𝛼^ h\) defined from \(𝛼^ 0=1\) and \(𝛼^{(n+1)}= 𝛼^ n ⋅ 𝛼\); then prove that \(𝛼^{h+k}=𝛼^ h𝛼^ k\) and \((𝛼^ h)^ k=𝛼^{(hk)}\) for every \(k,h∈ℕ\).
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[20T]Prerequisites:14.Given \(𝛼≠ 0\) in a field, define that \(𝛼^ 0=1\) and let \(𝛼^{-n}\) be the multiplicative inverse of \(𝛼^ n\) when \(n≥ 1\) natural. (Use 14). For \(n,m∈ℤ\) show that
\[ 𝛼^{n}𝛼^ m=𝛼^{n+m}\quad ,\quad (𝛼^ h)^ k=𝛼^{(hk)}\quad ; \]if the field is ordered and \(𝛼{\gt}1\) show that \(n↦ 𝛼^ n\) is strictly monotonic increasing.
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[205] Let \(F\) be a commutative ring, \(a,b∈ F\), \(n∈ℕ\) then
\[ (a+b)^{n}=∑ _{k=0}^{n}\binom {n}{k}a^{{n-k}}b^{k} \]where the factor
\[ \binom n k {\stackrel{.}{=}}\frac{n!}{k!(n-k)!} \]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 \(0≤ k,k+1≤ n\).