Laws of composition

A non-empty subset H of M is called a stable part with respect to the law of composition "*" if $\forall x,y\in H,x*y\in H$.

Properties. The law of composition $M\times M\to M, (x,y)\to x*y$

• Is commutative if: $x*y=y*x, \forall x,y\in M$

• Is associative if: $(x*y)*z=x*(y*z),\forall x,y,z\in M$

• Has an identity element if: $\forall x\in M e*x=x*e=x$. The identity element, if it exists, is unique.

• An associative law with an identity element has invertible elements if there exists $x'\in M$ such that $x'*x=x*x'=e$

If $x,y\in M$ are invertible with respect to a law of composition "*" (associative and with an identity element), then x*y and $x'$ are invertible. Moreover:

1) $(x*y)'=y'*x'$ 2) $(x')'=x$

Let $n\in \mathbb{N},n\ge2$. On we define the operations called addition and multiplication of residue classes modulo n as follows: with .

Addition of residue classes modulo n is associative and commutative, has $0$ as identity element, and every residue class has an opposite. Multiplication of residue classes modulo n is associative, commutative, has $1$ as identity element, and is distributive over addition:

Groups

A pair (G,*), consisting of a non-empty set G and a law of composition "*" on G, is a group if the following conditions hold:

G₁) $\forall x,y,z\in G,(x*y)*z=x*(y*z)$

G2) $\exists e\in G$, such that e*x=x*e=x, $\forall x\in G$

G3) $\forall x\in G,\exists x'\in G$ with $x'*x=x*x'=e$

If in addition axiom G₄ $\forall x,y\in G,x*y=y*x$ holds, then G is called a commutative or abelian group.

A pair (M,*) consisting of a non-empty set M and a law of composition "*" on M is called a monoid if:

M₁: $\forall x,y,z\in M,(x*y)*z=x*(y*z)$

M₂: $\exists e\in M$ such that $e*x=x*e=x,\forall x\in M$

Cancellation rules in a group*.*

Let (G,*) be a group. For any $a,b,c\in G$ we have:

1) $a*b=a*c\Rightarrow b=c$ 2) $b*a=c*a\Rightarrow b=c$

Let (G,*) be a group, $a,b\in G$ and $a'$ the inverse of a. The equation a*x=b has the unique solution $x=a'*b$ in G, and the equation y*a=b has the unique solution $y=b*a'$ in G.

Let $(G,\circ)$ and $(G',*)$ be two groups. A function $f:G\to G'$ is called a group isomorphism if:

1) $f(x\circ y)=f(x)*f(y) \forall x,y\in G$ 2) f is bijective

The group G is isomorphic to the group $G'(G\simeq G')$ if there exists an isomorphism $f:G\to G'$.

Let $(G,\circ)$ and $(G',*)$ be two groups. If $f:G\to G'$ is an isomorphism, then is an isomorphism.

Let the groups $(G,\circ)$ and $(G',*)$. The function $f:G\to G'$ is called a group homomorphism if:

$f(x\circ y)=f(x)*f(y), \forall x,y\in G$

Let the groups $(G,\circ)$ and $(G',*)$ have identity elements e and $e'$. If $f:G\to G'$ is a group homomorphism, then:

1) $f(e)=e'$

2)

Let (G,*) be a group and H a subset of G.

(H,*) is called a subgroup of G if (H,*) is a group.

Let (G,*) be a group with identity element e and H a subgroup of G. Then $e\in H$.

Let (G,*) be a group; $H\subset G,H\neq \emptyset.(H,*)$ is a subgroup of G if and only if $\forall x,y\in H,x*y\in H$, where $y'$ is the inverse of y in G.

Let (G, ) be a group with identity element e and $a\in G$. We say a is an element of finite order of the group G if there exists m>0 such that a^m=e.

If a is of finite order, then the smallest number m>0 with the property a^m=e is called the order of a, written orda=m.

If a is an element of order m of the group G, then is a subgroup of order m of G.

Let (G, ) be a group, $a\in G$ and . The following are equivalent:

1. orda=m

2. and a^m=e.

Rings and fields

A triple (R,+, ), where R is a non-empty set and "+" and "" are two laws of composition on R (called addition and multiplication), is a ring if:

(G) (R,+) is an abelian group

(G) (R, ) is a monoid

(D) multiplication is distributive over addition:

$\forall x,y,z\in \mathbb{R}$,

The ring R has no zero divisors if x≠0, y≠0

A ring R is called commutative if it also satisfies the axiom: (M₃) xy=yx, $\forall x,y\in \mathbb{R}$.

A commutative ring R with at least two elements and no zero divisors is called an integral domain.

Let $(\mathbb{Z}[i],+,*)$ be called the ring of Gaussian integers.

Let $(\mathbb{R},+,*)$ be a ring. The operation $y-z=y+(-z), y,z\in \mathbb{R}$ is called subtraction.

In a ring $(R,+,*)$ the following properties hold:

1) $\forall x\in \mathbb{R},x0=0x=0$

2) In a ring with at least two elements we have 1≠0.

3) The rule of signs: $\forall x,y\in \mathbb{R},(-x)y=x(-y)=-xy$ and (-x)(-y)=xy

4) Distributivity of multiplication over subtraction: $\forall x,y,z\in \mathbb{R},x(y-z)=xy-xz$ and $(y-z)x=yx-zx$.

5) In a ring R without zero divisors we may cancel by elements different from 0, that is $\forall x,y,z\in \mathbb{R}, x\neq0, xy=xz$ or $yx=zx\Rightarrow y=z$

The invertible elements of a ring R are called the units of R. We write U(R) for the set of units of the ring R. U(R) is a group under the operation induced by the multiplication of R, called the group of units of R.

If R is an integral domain, then R[X] is an integral domain and $\forall f, g\in R[X], f\neq0, g\neq0$ degfg=degf+degg.

Let R be a commutative ring, and $\alpha \in R$. The element is called the value of the polynomial f at α. The value of the sum and of the product of the polynomials $f,g\in R$ at $\alpha \in R$ equals the sum, respectively the product, of the values of f and g at α: $(f+g)(\alpha)=f(\alpha)+g(\alpha),.$ $(fg)(\alpha)=f(\alpha)g(\alpha)$.

Let $f\in \mathbb{R}[x].$. The function $f*:R\to R$ defined by $f*(x)=f(x)\in \mathbb{R},\forall x\in \mathbb{R}$ is the polynomial function associated with the polynomial f. We shall also write f for the function $f*$.

The zeros of the polynomial function f are called the roots (in R) of the polynomial f. Thus an element $\alpha \in R$ is a root (in R) of the polynomial $f\in \mathbb{R}[x].$ if f(α)=0.

Let R be a commutative ring, and $\alpha \in R$. There exist uniquely determined and r in R with . Moreover, r=f(α).

The remainder theorem. The remainder on dividing the polynomial $f\in \mathbb{R}[X]$ by $X-\alpha \in \mathbb{R}[X]$ is f(α).

Bézout's theorem. The polynomial $f\in \mathbb{R}[X]$ is divisible by $X-\alpha \in \mathbb{R}[X]$ if and only if f(α)*=*0.

A ring K is called a field if 0≠1 and every non-zero element of K is invertible with respect to multiplication. If multiplication is commutative, K is called a commutative field.

Properties

1. A field has no zero divisors

2. The non-zero elements of a field form a group under multiplication

3. Every finite integral domain is a field

Let the rings (R,+,*) and $(R',\oplus,\odot)$. A function $f:R\to R'$ is called a ring homomorphism if, $\forall x,y\in R$;

1) $f(x+y)=f(x)\oplus f(y)$ 2) $f(x*y)=f(x)\odot f(y)$ 3) $f(1)=1'$

where 1 is the unit of the ring R and $1'$ the unit of $R'$.

A bijective ring homomorphism is called an isomorphism. The ring R is isomorphic to the ring $R'$.

Let $f:R\to R'$ be a ring homomorphism. Then:

• $f(0)=0'$, 0 being the zero element of R and $0'$ that of $R'$.

• $f(-x)=-f(x), \forall x\in R$.

• If $x\in \mathbb{R}$ is invertible in the ring R, then f(x) is an invertible element of the ring $R'$ and .

A function $f:K\to K'$ from a field K to $K'$ is called a field homomorphism if it is a homomorphism from K to $K'$ regarded as rings.

Every field homomorphism $f:K\to K'$ is injective.

There exist uniquely determined polynomials $q,r\in K[X]$ satisfying the division equation $f=g\cdot q+r$, where degr<degg if $r\neq0$

• f is divisible by g, written or $g\vdots f$, if there exists $h\in K[X]$ with $f=gh$

• let $a\in K$ and $n\in \mathbb{N}$, n≥2; a is a root of multiplicity n if and $f$ is not divisible by

Let K be a commutative field, $f\in K[X]$, $a\in K$ and $n\in \mathbb{N}$, n≥2. The polynomial f has a root of multiplicity n if and only if f(a)=0;

Let K be a commutative field, $f\in K[X]$ and $a,b\in K[X]$,$a\neq b$. The polynomial $f\vdots(X-a)(X-b)$ if and only if f(a)=f(b)=0.

Let K be a commutative field and $f\in K[X]$ a polynomial of degree f=n. We say the polynomial f is reducible over K if there exist polynomials $g,h\in K[X]$, of degrees strictly less than n, with $f=gh$. Otherwise we say f is irreducible over K.

Let K be a commutative field and $f,h\in K[X]$. We say f is associated in divisibility with g, written $f\sim g$, if and .

Every polynomial $f\in \mathbb{C}[X]$ of degf>0 can be written as a finite product of degree-1 polynomials from $\mathbb{C}[X]$, uniquely determined up to the order of the factors and association in divisibility.

Vector spaces

Let V and K be non-empty sets. A map $\psi>K\times V\to V$ is called an external law of composition on V with scalars (or operators) in K.

Let (K,+,*) be a commutative field. A vector space over K is an abelian group (V,+) equipped with an external law of composition with scalars in K, $(\alpha,u)\to \alpha u$, satisfying the axioms:

S₁) $\forall \alpha,\beta \in K, \forall u\in V, (\alpha+\beta)u=\alpha u+\beta u$ (distributivity of scalar multiplication over addition of scalars)

S₂) $\forall \alpha \in K, \forall u,v\in V, \alpha(u+v)=\alpha u+\alpha v$ (distributivity of scalar multiplication over addition of vectors)

S₃) $\forall \alpha, \beta \in K, \forall u\in V, \alpha(\beta u)=(\alpha \beta)u$ (commutativity of multiplication of scalars and vectors)

S₄) $\forall u\in V, 1\cdot u=u$ (1, the unit of K, is the identity element for scalar multiplication)

Properties

1. Let $\alpha \in K$ and $v\in V, \alpha v=0\Leftrightarrow \alpha=0$ or v=0

2. For any $\alpha \in K$ and $v\in V$, we have: $(-\alpha)v=\alpha(-v)=-\alpha v,(-\alpha)(-v)=\alpha v$ (the rule of signs)

3. For any $\alpha,\beta \in K$ and $u,v\in V$ we have: $(\alpha-\beta)v=\alpha v-\beta v,\alpha(u-v)=\alpha u-\alpha v$

Let V be a vector space over the field K, $v_{1},v_{2},...,v_{p} din V$ and $\lambda_{1},\lambda_{2},...,\lambda_{p}\in K$. A vector of the form $v=\lambda_{1}v_{1}+\lambda_{2}v_{2}+...+\lambda_{p}v_{p}$ is a linear combination of the vectors $v_{1},v_{2},...,v_{p}$.

The vectors $v_{1},v_{2},...,v_{p}$ are linearly independent if $\forall$ $\lambda_{1},\lambda_{2},...,\lambda_{p}\in K$, $\lambda_{1}v_{1}+\lambda_{2}v_{2}+...+\lambda_{p}v_{p}=0$ implies $\lambda_{1}=...=\lambda_{p}=0$.

Let V be a vector space over K. A system $B=(v_{1}...v_{n})$ of vectors $v_{i}\in V,1\le i\le n$ is a basis of V if:

1) $(\forall)x\in V, (\exists)\lambda_{1},\lambda_{2},...,\lambda_{p}\in K$ such that $x=\lambda_{1}v_{1}+\lambda_{2}v_{2}+...+\lambda_{n}v_{n}$;

2) $v_{1},v_{2},...,v_{n}$ are linearly independent.

In this case, every vector v of V has a unique representation as a linear combination of the basis vectors B, $v=\lambda_{1}v_{1}+\lambda_{2}v_{2}+...+\lambda_{n}v_{n}$.

Let V and $V'$ be two vector spaces over the same field K. A map $f:V\to V'$ is called a linear map from the vector space V to $V'$ if:

1) $f(x+y)=f(x)+f(y) \forall x,y\in V$ 2) $f(\lambda x)=\lambda f(x), \forall \lambda \in K, \forall x\in V$

If V=$V'$, then f is also called a linear operator on V, or an endomorphism of V.

If $f:V\to V'$ is a linear map, then f is in particular a homomorphism from the group (V,+) to the group ($V'$,+). Then:

$f(0)=0'$

Let V and $V'$ be vector spaces over a field K, $\lambda \in K$ and $f:V\to V'$, $g:V\to V'$ linear maps. The map $f+g:V\to V'$, $(f+g)(x)=f(x)+g(x)$ is the sum of f and g, and , $(\lambda f)(x)=\lambda f(x)$ is the product of $\lambda$ with f.

If V, $V',V''$ are vector spaces over K and $g:V\to V'$,$f:V'\to V''$ are linear maps, then the map $f\circ g:V\to V''$ defined by $(f\circ g)(x)=f(g(x))$ is the composition of f with g.