ring
Ring defined: Let R be a non-empty set with two operations, if the following condition is satisfied:
I) a R-exchange group constituting the adder for
multiplying the II) R are, for any A, B, C \ (\ in \) R & lt, has (ab &) C = a (BC)
III) for any a, B, C \ (\ in \) R & lt, there are (a + b) c = ac + bc, a (b + c) = ab + ac
called a ring R
In other words, if the summation satisfies the definition of R for the exchange group, for multiplying meet the definition of a wide base, and distributes over, then R is a ring.
Notation:
I) If the multiplication is commutative ring, called R is a commutative ring
ii) If there is an element R = E \ (1_R \) for multiplying the R to \ (\ FORALL A \ R in, A = A A1_R 1_R = \) , R is called unital ring , or said as containing unitary ring
iii) If there are two non-zero elements in a R, b R for the multiplication satisfies ab = 0, R is called zero factor ring
iv) If R is a commutative ring and at the same time a unitary ring-containing, but not zero factor, called R is a complete ring
Properties of the ring
1. For any A \ (\ in \) R & lt, has 0a = a0 = 0
Proof.
\ (\ Because \) 0A = (0 + 0) = A + 0A 0A
\ (\ THEREFORE \) 0A = 0
Similarly available a0 = 0
2. For any A, B \ (\ in \) R & lt, have (-a) b = a (-b ) = - (ab)
证明.
\(\because\) (-a)b+ab=(-a+a)b=0b=0
a(-b)+ab=a(-b+b)=a0=0
\(\therefore\)(-a)b=a(-b)=-(ab)
3. For any A, B \ (\ in \) R & lt, have (-a) (- b) = ab, which is actually a second example of the nature of the
4. For any n \ (\ in \) the Z, a, B \ (\ in \) R & lt, (Na) B = a (Nb) = n (ab &), i.e. a n-multiplied and b, or a n-multiplied by b is equal to the product of n and ab.
5. arbitrary \ (a_i, b_j \ in R \) have
\ ((\ Sigma_ {i = 1} ^ n a_i) (\ Sigma_ {j = 1} ^ m b_j) = \ Sigma_ {i = 1} ^ n \ Sigma_ {j = 1 } ^ m a_i b_j \)
6. Let R be a complete ring, then R has established cancellation law, i.e. when c $ \ ne $ 0, c · a = c · b when there is a = b
ideal
Definition: Let R be a ring, I is the sub-ring R, if for any r \ (\ in \) R and A \ (\ in \) the I, both RA \ (\ in \) the I, called R I is left over, r is arbitrary if \ (\ in \) R and a \ (\ in \) I have Ar \ (\ in \) I, called I R is the right ideal.
If I left at the same time is ideal and right ideal of R, R claimed that I was the ideal.
Notation: {0} over R and R are, called ordinary ideal of R.
Ring R is non-empty set I is necessary and sufficient ideal conditions:
1. For any A, B \ (\ in \) I, there ab & \ (\ in \) I
2. For any R & lt \ (\ in \) R & lt and A \ (\ in \) the I have RA \ (\ in \) the I, Ar \ (\ in \) the I
. Prove
the necessity of clearly established
sufficiency:
seen from the first condition is a subgroup of R I
and then the second condition is apparent from the multiplication I is closed, and I as a subset of R, distributive law of multiplication is satisfied , the sub-ring R I is
at the same time, I R satisfies ideal conditions.
Theorem: set \ (\ lbrace A_j \ rbrace_ { i \ in J} \) are a family of ideal R, the \ (\ bigcap_ {j \ in J} A_j \) is an ideal
Proof.
\ (\ Because A_j \) is over
\ (\ therefore \ forall a, b \ in A_j, ab \ in A_j, \) for all J \ (\ in \) J
\ (\ THEREFORE ab & \ in \ bigcap_ {j \ in J} A_j
\) for any of R & lt \ (\ in \) R & lt and any \ (a \ in \ bigcap_ {j \ in J} A_j \) , there is a \ (\ in A_j \ ) , J \ (\ in \) J
as \ (A_j \) is the ideal R, so RA \ (\ in A_j \) , Ar \ (\ in A_j \) , J \ (\ in \) J
therefore \ (RA \ in \ bigcap_ {J \ in J} A_j \) , \ (Ar \ in \ bigcap_ {J \ in J} A_j \)
so \ (\ bigcap_ {j \ in J} A_j \) is ideal of R
Generate ideal
Definition: Let X be a subset of the ring R, provided \ (\ lbrace A_j \ rbrace_ { j \ in J} \) all over the ring R containing X, the new over \ (\ bigcap_ {j \ in J} A_j \) said generated by the ideal X, referred to as (X).
Notation:
1.X is called over the elements (X) of the generator, if = {X- \ (A_1, \ cdots, A_N \) }, it is desirable (X) referred to as ( \ (A_1, \ cdots, A_N \ ) ), called the finitely generated.
2. If X = {a}, which is said to generate over (a) called the primary over
3 over the ring R is called the primary ring, R is ideal if all over the main
Theorem: Let R be a ring, the master over (A) = { \ (RA + Ar '+ Na + \ Sigma_ {I =. 1} ^ mr_ias_i | R & lt, R & lt', r_i, S_I \ in R, m \ in N, n-\ in the Z \) }
wherein
Examples: integral over the ring Z is a primary ring and I = (a) expression is \ (I = (a) = \ lbrace sa | s \ in Z \ rbrace \)
Corollary: Let I = (a) is an integral over the ring Z is an integer of B \ (\ in \) the I necessary and sufficient condition that a | b
Quotient ring
Theorem: Let R be a ring, I is an ideal of R, the R / I for adding: (a + I) + ( b + I) = (a + b) + I, and multiplication: (a + I) (b + I) = (ab) + I form a ring.
And when R is a commutative ring or a ring containing a unitary, commercially ring R / I is a unitary ring-containing or ring switching.
Proof (informal).
Found that commercially group defined by the quotient ring R / I form a pair of adders exchange group.
For multiplication: \ ((A + the I) (B + the I) = (A + i_1) (B + i_2 are used) = ab & + Ai_2 + i_1b + i_1i_2 \)
\ (\ Because \) the I is an ideal of R
\ ( \ THEREFORE Ai_2 \ in the I, i_1b \ in the I \)
\ (\ THEREFORE Ai_2 + i_1b + i_1i_2 \ in the I \) , i.e., (a + I) (b + I) = (ab) + I
can be simply understood as commercially group accompany operations on sets may correspond directly to the operation-membered ring representatives
so quotients also form a ring, called ring supplier
Theorem ring with decomposition
Defined ring homomorphic: homomorphic ring on the basis of the same state group on adds a condition: for G and G 'multiplication, satisfies F (A \ (\ CDOT \) B) = F (A) \ (\ CDOT \) F (B)
corresponding, homomorphism theorem in the ring can be mapped to the same state
Natural mapping: Let f be a ring R to the ring R 'of a same state, the core Ker (f) R is ideal, in turn, if R I is desired, then the map, \ (S: R \ to R / I (a \ mapsto a + I ) \) is the same core as I state.
The proof state with natural and similar groups.
With decomposition (of a homomorphism configured a isomorphism): Let f be a ring R to the ring R 'homomorphism, isomorphism only R / ker (f) the group f (R) then there is \ (F ': A + Ker (F) \ mapsto F (A) \) .
Similarly, it is possible to obtain a mapping conversion relationship: \ (F = I \ CDOT F '\ CDOT s \) , where s is the ring R to the quotient ring R / ker (f) natural homomorphism, \ (I: C \ mapsto c \) is f (R) to R 'identity map. That is:
\ [R & lt \ stackrel {S} {\ to} R & lt / Ker (F) \ stackrel {F '} {\ to} F (G) \ stackrel {I} {\ to} G' \ stackrel {F} {\ leftarrow} G \]